Statistics Solution Manual

Solutions Manual

to accompany

STATISTICS FOR ENGINEERS ANDSCIENTISTS, 2nd ed.

Prepared by

William Navidi

PROPRIETARY AND CONFIDENTIAL

This Manual is the proprietary property of The McGraw-Hill Companies, Inc.(“McGraw-Hill”) and protected by copyright and other stateand federal laws. Byopening and using this Manual the user agrees to the following restrictions, and ifthe recipient does not agree to these restrictions, the Manual should be promptlyreturned unopened to McGraw-Hill:This Manual is being provided only to au-thorized professors and instructors for use in preparing for the classes usingthe affiliated textbook. No other use or distribution of this Manual is permit-ted. This Manual may not be sold and may not be distributed to or used byany student or other third party. No part of this Manual may be reproduced,displayed or distributed in any form or by any means, electronic or otherwise,without the prior written permission of McGraw-Hill.

PROPRIETARY MATERIAL. c© The McGraw-Hill Companies, Inc. All rights reserved.No part of this Manualmay be displayed,reproducedor distributedin any form or by any means,without the prior written permissionofthepublisher,or usedbeyondthe limited distribution to teachersandeducatorspermittedby McGraw-Hill for theirindividualcoursepreparation.If youareastudentusingthisManual,youareusingit withoutpermission.

Table of Contents

Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1

Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . 26

Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . 106

Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . 143

Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . 229

Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . 264

Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . 336

Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . 382

Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . 419

Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . 461

PROPRIETARY MATERIAL. c© The McGraw-Hill Companies, Inc. All rights reserved.No part of this Manualmay be displayed,reproducedor distributedin any form or by any means,without the prior written permissionofthepublisher,or usedbeyondthe limited distribution to teachersandeducatorspermittedby McGraw-Hill for theirindividualcoursepreparation.If youareastudentusingthisManual,youareusingit withoutpermission.

SECTION 1.1 1

Chapter 1

Section 1.1

1. (a) The population consists of all the bolts in the shipment. It is tangible.

(b) The population consists of all measurements that could be made on that resistor with that ohmmeter.

It is conceptual.

(c) The population consists of all residents of the town. It is tangible.

(d) The population consists of all welds that could be made bythat process. It is conceptual.

(e) The population consists of all parts manufactured that day. It is tangible.

2. (iii). It is very unlikely that students whose names happen to fall at the top of a page in the phone book willdiffer systematically in height from the population of students as a whole. It is somewhat more likely thatengineering majors will differ, and very likely that students involved with basketball intramurals will differ.

3. (a) False

(b) True

4. (ii) is correct

5. (a) No. What is important is the population proportion of defectives; the sample proportion is only an approx-imation. The population proportion for the new process may in fact be greater or less than that of the oldprocess.

(b) No. The population proportion for the new process may be 0.10 or more, even though the sample proportionwas only 0.09.

(c) Finding 2 defective bottles in the sample.

Page 1PROPRIETARY MATERIAL. c© The McGraw-Hill Companies, Inc. All rights reserved.No part of this Manualmay be displayed,reproducedor distributedin any form or by any means,without the prior written permissionofthepublisher,or usedbeyondthe limited distribution to teachersandeducatorspermittedby McGraw-Hill for theirindividualcoursepreparation.If youareastudentusingthisManual,youareusingit withoutpermission.

2 CHAPTER 1

6. (a) False

(b) True

(c) True

(d) False

7. A good knowledge of the process that generated the data.

Section 1.2

1. False

2. No. In the sample 1, 2, 4 the mean is 7/3, which does not appear at all.

3. No. In the sample 1, 2, 4 the mean is 7/3, which does not appear at all.

4. No. The median of the sample 1, 2, 4, 5 is 3.

5. The sample size can be any odd number.

6. Yes. For example, the list 1, 2, 12 has an average of 5 and a standard deviation of 6.08.

7. Yes. If all the numbers in the list are the same, the standard deviation will equal 0.

8. The mean increases by $50; the standard deviation is unchanged.


SECTION 1.2 3

9. The mean and standard deviation both increase by 5%.

10. (a) LetX1, ...,X1000 denote the 1000 scores.

1000

∑i=1

Xi = 1(0)+3(1)+2(2)+4(3)+25(4)+35(5)+198(6)+367(7)+216(8)+131(9)+18(10)= 7138

X =∑1000

i=1 Xi

1000=

71381000

= 7.138

(b) The sample variance is

s2 =1

999

(1000

∑i=1

X2i −1000X

2

)

=1

999[(1)02 +(3)12+(2)22+(4)32+(25)42+(35)52+(198)62+(367)72+(216)82

+ (131)92+(18)102−1000(7.1382)]

= 1.71867

The standard deviation iss=√

s2 = 1.311.

Alternatively, the sample variance can be computed as

s2 =1

999

1000

∑i=1

(Xi −X)2

=1

999[1(0−7.138)2+3(1−7.138)2+2(2−7.138)2+4(3−7.138)2+25(4−7.138)2+35(5−7.138)2

+ 198(6−7.138)2+367(7−7.138)2+216(8−7.138)2+131(9−7.138)2+18(10−7.138)2]

= 1.71867

(c) The sample median is the average of the 500th and 501st value when arranged in order. Both these values areequal to 7, so the median is 7.

(d) The first quartile is the average of the 250th and 251st value when arranged in order. Both these values areequal to 6, so the first quartile is 6.

(e) Of the 1000 scores, 216 + 131 + 18 = 365 are greater than the mean of 7.138, so the proportion is 365/1000 =0.365.


4 CHAPTER 1

(f) The quantity that is one standard deviation greater thanthe mean is 7.138 + 1.311 = 8.449. Of the 1000 scores,131 + 18 = 149 are greater than 8.449, so the proportion is 149/1000 = 0.149.

(g) The region within one standard deviation of the mean is 7.138±1.311= (5.827,8.449). Of the 1000 scores,198 + 367 + 216 = 781 are in this range, so the proportion is 781/1000 = 0.781.

11. The total number of points scored in the class of 30 students is 30×72= 2160. The total number of pointsscored in the class of 40 students is 40×79= 3160. The total number of points scored in both classes combinedis 2160+3160= 5320. There are 30+40= 70 students in both classes combined. Therefore the mean scorefor the two classes combined is 5320/70= 76.

12. (a) The mean for A is

(18.0+18.0+18.0+20.0+22.0+22.0+22.5+23.0+24.0+24.0+25.0+25.0+25.0+25.0+26.0+26.4)/16= 22.744

The mean for B is

(18.8+18.9+18.9+19.6+20.1+20.4+20.4+20.4+20.4+20.5+21.2+22.0+22.0+22.0+22.0+23.6)/16= 20.700

The mean for C is

(20.2+20.5+20.5+20.7+20.8+20.9+21.0+21.0+21.0+21.0+21.0+21.5+21.5+21.5+21.5+21.6)/16= 20.013

The mean for D is

(20.0+20.0+20.0+20.0+20.2+20.5+20.5+20.7+20.7+20.7+21.0+21.1+21.5+21.6+22.1+22.3)/16= 20.806

(b) The median for A is (23.0 + 24.0)/2 = 23.5. The median for B is (20.4 + 20.4)/2 = 20.4. The median for C is(21.0 + 21.0)/2 = 21.0. The median for D is (20.7 + 20.7)/2 = 20.7.

(c) 0.20(16) = 3.2≈ 3. Trim the 3 highest and 3 lowest observations.

The 20% trimmed mean for A is

(20.0+22.0+22.0+22.5+23.0+24.0+24.0+25.0+25.0+25.0)/10= 23.25

The 20% trimmed mean for B is

(19.6+20.1+20.4+20.4+20.4+20.4+20.5+21.2+22.0+22.0)/10= 20.70


SECTION 1.2 5

The 20% trimmed mean for C is

(20.7+20.8+20.9+21.0+21.0+21.0+21.0+21.0+21.5+21.5)/10= 21.04

The 20% trimmed mean for D is

(20.0+20.2+20.5+20.5+20.7+20.7+20.7+21.0+21.1+21.5)/10= 20.69

(d) 0.25(17) = 4.25. Therefore the first quartile is the average of the numbersin positions 4 and 5. 0.75(17) =12.75. Therefore the third quartile is the average of the numbers in positions 12 and 13.A: Q1 = 21.0, Q3 = 25.0; B: Q1 = 19.85,Q3 = 22.0; C: Q1 = 20.75,Q3 = 21.5; D: Q1 = 20.1, Q3 = 21.3

(e) The variance for A is

s2 =115

[18.02+18.02+18.02+20.02+22.02+22.02+22.52+23.02+24.02

+ 24.02+25.02+25.02+25.02+25.02+26.02+26.42−16(22.7442)] = 8.2506

The standard deviation for A iss=√

8.2506= 2.8724.

The variance for B is

s2 =115

[18.82+18.92+18.92+19.62+20.12+20.42+20.42+20.42+20.42

+ 20.52+21.22+22.02+22.02+22.02+22.02+23.62−16(20.7002)] = 1.8320

The standard deviation for B iss=√

1.8320= 1.3535.

The variance for C is

s2 =115

[20.22+20.52+20.52+20.72+20.82+20.92+21.02+21.02+21.02

+ 21.02+21.02+21.52+21.52+21.52+21.52+21.62−16(20.0132)] = 0.17583

The standard deviation for C iss=√

0.17583= 0.4193.

The variance for D is

s2 =115

[20.02+20.02+20.02+20.02+20.22+20.52+20.52+20.72+20.72

+ 20.72+21.02+21.12+21.52+21.62+22.12+22.32−16(20.8062)] = 0.55529

The standard deviation for D iss=√

0.55529= 0.7542.

(f) Method A has the largest standard deviation. This could be expected, because of the four methods, this isthe crudest. Therefore we could expect to see more variationin the way in which this method is carried out,resulting in more spread in the results.


6 CHAPTER 1

(g) Other things being equal, a smaller standard deviation is better. With any measurement method, the result issomewhat different each time a measurement is made. When thestandard deviation is small, a single measure-ment is more valuable, since we know that subsequent measurements would probably not be much different.

13. (a) All would be divided by 2.54.

(b) Not exactly the same, because the measurements would be alittle different the second time.

14. (a) LetS0 be the sum of the original 10 numbers and letS1 be the sum after the change. ThenS0/10= 20, soS0 = 200. NowS1 = S0−39.27+392.7= 553.43, so the new mean isS1/10= 55.343.

(b) The median is unchanged at 18.

(c) Let X1, ...,X10 be the original 10 numbers. LetT0 = ∑10i=1X2

i . Then the variance is(1/9)[T0 − 10(202)] =52 = 25, soT0 = 4225. LetT1 be the sum of the squares after the change. ThenT1 = T0−39.272 +392.72 =156,896.1571. The new standard deviation is

√(1/9)[T1−10(55.3432)] = 118.45.

15. (a) The sample size isn= 16. The tertiles have cutpoints(1/3)(17)= 5.67 and(2/3)(17)= 11.33. The first tertileis therefore the average of the sample values in positions 5 and 6, which is(44+46)/2= 45. The second tertileis the average of the sample values in positions 11 and 12, which is(76+79)/2= 77.5.

(b) The sample size isn = 16. The quintiles have cutpoints(i/5)(17) for i = 1,2,3,4. The quintiles are thereforethe averages of the sample values in positions 3 and 4, in positions 6 and 7, in positions 10 and 11, and inpositions 13 and 14. The quintiles are therefore(23+41)/2= 32,(46+49)/2= 47.5, (74+76)/2= 75, and(82+89)/2= 85.5.

16. (a) Seems certain to be an error.

(b) Could be correct.


SECTION 1.3 7

Section 1.3

1. (a) Stem Leaf11 612 67813 1367814 1336815 12667889916 12234555617 01334446718 133355819 220 3

(b) Here is one histogram. Other choices for the endpoints are possible.

11 12 13 14 15 16 17 18 19 20 210

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

Re

lative

Fre

qu

en

cy

Weight (oz)

(c) 10 12 14 16 18 20 22 24Weight (ounces)

(d)

10

12

14

16

18

20

22

We

igh

t (o

un

ce

s)

The boxplot shows no outliers.


8 CHAPTER 1

2. (a) Stem Leaf0 12344568891 0001112222334567788992 0234593 037894 05 5867 78 19 6

10 27

(b) Here is one histogram. Other choices for the endpoints are possible.

0 10 20 30 40 50 60 70 80 90 1001100

5

10

15

20

Fre

qu

en

cy

Number of Sites

(c) 0 20 40 60 80 100 120Number of Sites

(d)

0

20

40

60

80

100

120

Nu

mb

er

of

Site

s

The boxplot shows 5 outliers.


SECTION 1.3 9

3. Stem Leaf1 15882 000034683 02345884 03465 22356666896 002334597 1135588 5689 1225

10 11112 213 0614151617 118 619 920212223 3

There are 23 stems in this plot. An advantage of this plot overthe one in Figure 1.6 is that the values are givento the tenths digit instead of to the ones digit. A disadvantage is that there are too many stems, and many ofthem are empty.

4. (a) Here are histograms for each group. Other choices for the endpoints are possible.

0.3 0.4 0.5 0.6 0.7 0.80

0.05

0.1

0.15

0.2

0.25

0.3

Re

lative

Fre

qu

en

cy

Proportion Recovered0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.05

0.1

0.15

0.2

0.25

Re

lative

Fre

qu

en

cy


10 CHAPTER 1

(b)

0.2

0.4

0.6

0.8

1

Fra

ctio

n R

eco

ve

red

Method 1 Method 2

(c) The results of Method 1 are on the whole lower than the results of Method 2. Also Method 1 produces resultsthat are distributed somewhat symmetrically, while the results of Method 2 are somewhat skewed to the right,which can be seen from the median being closer to the first quartile than the third. The results of Method 2 aremore spread out than those of Method 1 as well.

5. (a) Here are histograms for each group. Other choices for the endpoints are possible.

18 19 20 21 22 23 24 25 260

0.05

0.1

0.15

0.2

0.25

Re

lative

Fre

qu

en

cy

Group 1 measurements (cm)15 16 17 18 19 20 21 22 23 24 25

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Re

lative

Fre

qu

en

cy

Group 2 measurements (cm)


SECTION 1.3 11

(b)

10

15

20

25

30

Me

asu

rem

en

t (c

m)

Group 1 Group 2

(c) The measurements in Group 1 are generally larger than those in Group 2. The measurements in Group 1 are notfar from symmetric, although the boxplot suggests a slight skew to the left since the median is closer to the thirdquartile than the first. There are no outliers. Most of the measurements for Group 2 are highly concentrated ina narrow range, and skewed to the left within that range. The remaining four measurements are outliers.

6. (a) The histogram should be skewed to the right. Here is an example.


12 CHAPTER 1

(b) The histogram should be skewed to the left. Here is an example.

(c) The histogram should be approximately symmetric. Here is an example.

7. (a) The proportion is the sum of the relative frequencies (heights) of the rectangles above 130. This sum is approx-imately 0.12+0.045+0.045+0.02+0.005+0.005= 0.24. This is closest to 25%.

(b) The height of the rectangle over the interval 130–135 is greater than the sum of the heights of the rectanglesover the interval 140–150. Therefore there are more women inthe interval 130–135 mm.

8. The relative frequencies of the rectangles shown are 0.2,0.3, 0.15, 0.1, and 0.1. The sum of these relativefrequencies is 0.85. Since the sum of all the relative frequencies must be 1, the missing rectangle has a heightof 0.15.


SECTION 1.3 13

9. (a)

1 3 5 7 9 11 13 15 17 19 21 23 250

3

6

9

12

15

18

Fre

quen

cy

Emissions (g/gal)

(b)

1 3 5 7 9 11 13 15 17 19 21 23 250

0.05

0.1

0.15

Den

sity

Emissions (g/gal)

(c) Yes, the shapes of the histograms are the same.

10. (a)

1 3 5 7 9 11 15 250

0.05

0.1

0.15

0.2

0.25

0.3

Re

lativ

e F

req

ue

ncy

Emissions (g/gal)

(b) No

(c) The class interval widths are unequal.

(d) The classes 11–<15 and 15–<25


14 CHAPTER 1

11. Any point more than 1.5 IQR (interquartile range) below the first quartile or above the third quartile is labeledan outlier. To find the IQR, arrange the values in order: 4, 10,20, 25, 31, 36, 37, 41, 44, 68, 82. There aren = 11 values. The first quartile is the value in position 0.25(n+1) = 3, which is 20. The third quartile is thevalue in position 0.75(n+1) = 9, which is 44. The interquartile range is 44−20= 24. So 1.5 IQR is equal to(1.5)(24) = 36. There are no points less than 20−36= −16, so there are no outliers on the low side. There isone point, 82, that is greater than 44+36= 80. Therefore 82 is the only outlier.

12. The mean. The median, and the first and third quartiles areindicated directly on a boxplot, and the interquartilerange can be computed as the difference between the first and third quartiles.

13. The figure on the left is a sketch of separate histograms for each group. The histogram on the right is a sketchof a histogram for the two groups combined. There is more spread in the combined histogram than in either ofthe separate ones. Therefore the standard deviation of all 200 heights is greater than 2.5 in. The answer is (ii).

14. (a) False

(b) True

(c) False

(d) False

(e) True

(f) False


SECTION 1.3 15

15. (a) IQR = 3rd quartile− 1st quartile. A: IQR = 6.02−1.42= 4.60, B: IQR = 9.13−5.27= 3.86

(b) Yes, since the minimum is within 1.5 IQR of the first quartile and themaximum is within 1.5 IQR of the third quartile, there are no outliers,and the given numbers specify the boundaries of the box and the ends ofthe whiskers.

0

2

4

6

8

10

12

(c) No. The minimum value of−2.235 is an “outlier,” since it is more than 1.5 times the interquartile range belowthe first quartile. The lower whisker should extend to the smallest point that is not an outlier, but the value ofthis point is not given.

16. (a) (4)

(b) (2)

(c) (1)

(d) (3)

17. (a)

0

100

200

300

400

500

Fra

ctu

re S

tre

ss (

MP

a)


16 CHAPTER 1

(b) The boxplot indicates that the value 470 is an outlier.

(c) 0 100 200 300 400 500Fracture Strength (MPa)

(d) The dotplot indicates that the value 384 is detached fromthe bulk of the data, and thus could be considered tobe an outlier.

18. (a) iii

(b) i

(c) iv

(d) ii

19. (a)

0 5 10 150

10

20

30

40

50

60

x

y The relationship is non-linear.

(b)x 1.4 2.4 4.0 4.9 5.7 6.3 7.8 9.0 9.3 11.0

lny 0.83 1.31 1.74 2.29 1.93 2.76 2.73 3.61 3.54 3.97


SUPPLEMENTARY EXERCISES FOR CHAPTER 1 17

0 5 10 150.5

1

1.5

2

2.5

3

3.5

4

x

ln y The relationship is approximately linear.

(c) It would be easier to work withx and lny, because the relationship is approximately linear.

Supplementary Exercises for Chapter 1

1. (a) The mean will be divided by 2.2.

(b) The standard deviation will be divided by 2.2.

2. (a) The mean will increase by 50 g.

(b) The standard deviation will be unchanged.

3. (a) False. The true percentage could be greater than 5%, with the observation of 4 out of 100 due to samplingvariation.

(b) True

(c) False. If the result differs greatly from 5%, it is unlikely to be due to sampling variation.


18 CHAPTER 1

(d) True. If the result differs greatly from 5%, it is unlikely to be due to sampling variation.

4. (a) No. This could well be sampling variation.

(b) Yes. It is virtually impossible for sampling variation to be this large.

5. (a) It is not possible to tell by how much the mean changes, because the sample size is not known.

(b) If there are more than two numbers on the list, the median is unchanged. If there are only two numbers on thelist, the median is changed, but we cannot tell by how much.

(c) It is not possible to tell by how much the standard deviation changes, both because the sample size is unknownand because the original standard deviation is unknown.

6. (a) The sum of the numbers decreases by 12.9−1.29= 11.61, so the mean decreases by 11.61/15 = 0.774.

(b) No, it is not possible to determine the value of the mean after the change, since the original mean is unknown.

(c) The median is the eighth number when the list is arranged in order, and this is unchanged.

(d) It is not possible to tell by how much the standard deviation changes, because the original standard deviation isunknown.

7. (a) The mean decreases by 0.774.

(b) The value of the mean after the change is 25−0.774= 24.226.



(c) The median is unchanged.

(d) It is not possible to tell by how much the standard deviation changes, because the original standard deviation isunknown.

8. (a) The sum of the numbers 284.34, 173.01, 229.55, 312.95,215.34, 188.72, 144.39, 172.79, 139.38, 197.81,303.28, 256.02, 658.38, 105.14, 295.24, 170.41 is 3846.75.The mean is therefore 3846.75/16 = 240.4219.

(b) The 16 values arranged in increasing order are:

105.14,139.38,144.39,170.41,172.79,173.01,188.72,197.81,

215.34,229.55,256.02,284.34,295.24,303.28,312.95,658.38

The median is the average of the 8th and 9th numbers, which is(197.81+215.34)/2= 206.575.

(c) 0.25(17) = 4.25, so the first quartile is the average of the4th and 5th numbers, which is(170.41+172.79)/2= 171.60.

(d) 0.75(17) = 12.75, so the third quartile is the average of the 12th and 13th numbers, which is(284.34+295.24)/2= 289.79.

(e)

100

200

300

400

500

600

700

mg/lThe median is closer to the first quartile than to the third quar-tile, which indicates that the sample is skewed a bit to the right.In addition, the sample contains an outlier.

9. Statement (i) is true. The sample is skewed to the right.


20 CHAPTER 1

10. (a) False. The length of the whiskers is at most 1.5 IQR.

(b) False. The length of the whiskers is at most 1.5 IQR.

(c) True. A whisker extends to the most extreme data point that is within 1.5 IQR of the first or third quartile.

(d) True. A whisker extends to the most extreme data point that is within 1.5 IQR of the first or third quartile.

11. (a) Incorrect, the total area is greater than 1.

(b) Correct. The total area is equal to 1.

(c) Incorrect. The total area is less than 1.

(d) Correct. The total area is equal to 1.

12. (i) It would be skewed to the right. The mean is greater than the median. Also note that half the values arebetween 0 and 0.10, so the left-hand tail is very short.

13. (a) Skewed to the left. The 85th percentile is much closerto the median (50th percentile) than the 15th percentileis. Therefore the histogram is likely to have a longer left-hand tail than right-hand tail.

(b) Skewed to the right. The 15th percentile is much closer tothe median (50th percentile) than the 85th percentileis. Therefore the histogram is likely to have a longer right-hand tail than left-hand tail.



14. CumulativeClass Relative Cumulative Relative

Interval Frequency Frequency Frequency Frequency11–< 12 1 0.02 1 0.0212–< 13 3 0.06 4 0.0813–< 14 5 0.10 9 0.1814–< 15 5 0.10 14 0.2815–< 16 9 0.18 23 0.4616–< 17 9 0.18 32 0.6417–< 18 9 0.18 41 0.8218–< 19 7 0.14 48 0.9619–< 20 1 0.02 49 0.9820–< 21 1 0.02 50 1.00

15. (a)

6 9 12131415161718 20 230

0.05

0.1

0.15

0.2

0.25

Den

sity

Log2 population

(b) 0.14

(c) Approximately symmetric


22 CHAPTER 1

(d)

0 2 4 6 8 0

0.05

0.1

0.15

0.2

0.25

Den

sity

Population (millions)

The data on the raw scale are skewed so much to the right that itis impossible to see the features of thehistogram.

16. (a) The mean is

123

(2099+528+2030+1350+1018+384+1499+1265+375+424+789+810

+ 522+513+488+200+215+486+257+557+260+461+500)= 740.43

(b) The variance is

s2 =122

[20992+5282+20302+13502+10182+3842+14992+12652+3752+4242+7892+8102

+ 5222+5132+4882+2002+2152+4862+2572+5572+2602+4612+5002−23(740.432)]

= 302320.26

The standard deviation iss=√

302320.26= 549.84.

(c) The 23 values, arranged in increasing order, are:

200,215,257,260,375,384,424,461,486,488,500,513,522,528,557,789,810,1018,1265,1350,1499,2030,2099

The median is the 12th value, which is 513.

(d)0 500 1000 1500 2000 2500

Electrical Conductivity (µS/cm)



(e) Since(0.1)(23) ≈ 2, the 10% trimmed mean is computed by deleting the two highest and two lowest values,and averaging the rest.

The 10% trimmed mean is119

(257+260+375+384+424+461+486+488+500+513

+522+528+557+789+810+1018+1265+1350+1499)= 657.16

(f) 0.25(24) = 6. Therefore, when the numbers are arranged in increasing order, the first quartile is the number inposition 6, which is 384.

(g) 0.75(24) = 18. Therefore, when the numbers are arranged in increasing order, the third quartile is the numberin position 18, which is 1018.

(h) IQR = 3rd quartile− 1st quartile = 1018−384= 634.

(i)

0

500

1000

1500

2000

2500

Ele

ctr

ica

l C

on

du

ctivity (

µS

/cm

)

(j) The points 2030 and 2099 are outliers.

(k) skewed to the right


24 CHAPTER 1

17. (a)

0 2 4 10 15 20 25 30 500

0.05

0.1

0.15

0.2

0.25

Den

sity

Number of shares owned

(b) The sample size is 651, so the median is approximated by the point at which the area to the left is 0.5 =325.5/651. The area to the left of 3 is 295/651, and the area tothe left of 4 is 382/651. The point at which thearea to the left is 325.5/651 is 3+(325.5−295)/(382−295)= 3.35.

(c) The sample size is 651, so the first quartile is approximated by the point at which the area to the left is 0.25 =162.75/651. The area to the left of 1 is 18/651, and the area tothe left of 2 is 183/651. The point at which thearea to the left is 162.75/651 is 1+(162.75−18)/(183−18)= 1.88.

(d) The sample size is 651, so the third quartile is approximated by the point at which the area to the left is 0.75 =488.25/651. The area to the left of 5 is 425/651, and the area to the left of 10 is 542/651. The point at whichthe area to the left is 488.25/651 is 5+(10−5)(488.25−425)/(542−425)= 7.70.

18. (a)

0 1 2 3 4 5 6 7 8 90

10

20

30

40

Freq

uenc

y

Time (months)

(b) The sample size is 171, so the median is the value in position (171+1)/2 = 86 when the values are arrangedin order. There are 45+17+18= 80 values less than or equal to 3, and 80+19= 99 values less than or equalto 4. Therefore the class interval 3 –< 4 contains the median.



(c) The sample size is 171, so the first quartile is the value inposition 0.25(171+ 1) = 43 when the values arearranged in order. There are 45 values in the first class interval 0 –< 1. Therefore the class interval 0 –< 1contains the first quartile.

(d) The sample size is 171, so the third quartile is the value in position 0.75(171+ 1) = 129 when the values arearranged in order. There are 45+17+18+19+12+14= 125 values less than or equal to 6, and 125+13= 138values less than or equal to 7. Therefore the class interval 6– < 7 contains the third quartile.

19. (a)

0

10

20

30

40

50

60

70

Lo

ad

(kg

)

Sacaton Gila Plain Casa Grande

(b) Each sample contains one outlier.

(c) In the Sacaton boxplot, the median is about midway between the first and third quartiles, suggesting that thedata between these quartiles are fairly symmetric. The upper whisker of the box is much longer than the lowerwhisker, and there is an outlier on the upper side. This indicates that the data as a whole are skewed to the right.In the Gila Plain boxplot data, the median is about midway between the first and third quartiles, suggestingthat the data between these quartiles are fairly symmetric.The upper whisker is slightly longer than the lowerwhisker, and there is an outlier on the upper side. This suggest that the data as a whole are somewhat skewedto the right. In the Casa Grande boxplot, the median is very close to the first quartile. This suggests that thereare several values very close to each other about one-fourthof the way through the data. The two whiskers areof about equal length, which suggests that the tails are about equal, except for the outlier on the upper side.


26 CHAPTER 2

Chapter 2

Section 2.1

1. P(not defective) = 1−P(defective) = 1−0.08= 0.92

2. (a)1, 2, 3, 4

(b) P(even number) = P(2)+P(4) = 3/8+1/8= 1/2

(c) No, the set of possible outcomes is still1, 2, 3, 4.

(d) Yes, a list of equally likely outcomes is then1, 1, 2, 2, 2, 3, 3, 4, 4, soP(even) = P(2)+P(4) = 3/9+2/9= 5/9.

3. (a) The outcomes are the 16 different strings of 4 true-false answers. These areTTTT, TTTF, TTFT, TTFF, TFTT,TFTF, TFFT, TFFF, FTTT, FTTF, FTFT, FTFF, FFTT, FFTF, FFFT, FFFF.

(b) There are 16 equally likely outcomes. The answers are allthe same in two of them, TTTT and FFFF. Thereforethe probability is 2/16 or 1/8.

(c) There are 16 equally likely outcomes. There are four of them, TFFF, FTFF, FFTF, and FFFT, for which exactlyone answer is “True.” Therefore the probability is 4/16 or 1/4.

(d) There are 16 equally likely outcomes. There are five of them, TFFF, FTFF, FFTF, FFFT, and FFFF, for whichat most one answer is “True.” Therefore the probability is 5/16.

4. (a) The outcomes are the 27 different strings of 3 chosen from red, yellow and green. These areRRR, RRY, RRG,RYR, RYY, RYG, RGR, RGY, RGG, YRR, YRY, YRG, YYR, YYY, YYG, YGR, YGY, YGG, GRR, GRY,GRG, GYR, GYY, GYG, GGR, GGY, GGG.

(b) A = RRR, YYY, GGG

(c) B = RYG, RGY, YRG, YGR, GRY, GYR

(d) C = RGG, GRG, GGR, YGG, GYG, GGY, GGG

(e) The only outcome common toA andC is GGG. ThereforeA∩C = GGG.


SECTION 2.1 27

(f) The setA∪B contains the outcomes that are either inA, in B, or in both. ThereforeA∪B = RRR, YYY, GGG,RYG, RGY, YRG, YGR, GRY, GYR.

(g) Cc contains the outcomes that are not inC. A∩Cc contains the outcomes that are inA but not inC. ThereforeA∩Cc = RRR, YYY.

(h) Ac contains the outcomes that are not inA. Ac∩C contains the outcomes that are inC but not inA. ThereforeAc∩C = RGG, GRG, GGR, YGG, GYG, GGY.

(i) No. They both contain the outcome GGG.

(j) Yes. They have no outcomes in common.

5. (a) The outcomes are the sequences of bolts that end with either #1 or #2. These are1, 2, 31, 32, 41, 42, 341,342, 431, 432.

(b) A = 1, 2

(c) B = 341, 342, 431, 432

(d) C = 2, 32, 42, 342, 432

(e) D = 1, 2, 31, 32

(f) A and E are not mutually exclusive because they both contain the outcome 1. B and E are not mutuallyexclusive because they both contain the events 341 and 431. Cand E are mutually exclusive because they haveno outcomes in common. D and E are not mutually exclusive because they have the events 1 and 31 in common.

6. (a) The equally likely outcomes are the sequences of two distinct bolts. These are12, 13, 14, 21, 23, 24, 31, 32,34, 41, 42, 43.

(b) Of the 12 equally likely outcomes, there are 2 (34 and 43) for which both bolts are 7 mm. The probability istherefore 2/12 or 1/6.

(c) Of the 12 equally likely outcomes, there are 8 (13, 14, 23,24, 31, 32, 41, and 42) for which one bolt is 5 mmand the other is 7 mm. The probability is therefore 8/12 or 2/3.


28 CHAPTER 2

7. (a) 0.6

(b) P(personal computer or laptop computer) = P(personal computer)+P(laptop computer)

= 0.6+0.3

= 0.9

8. (a) False

(b) True

(c) True. This is the definition of probability.

9. (a) False

(b) True

10. (a) P(E∪T) = P(E)+P(T)−P(E∩T)

= 0.10+0.02−0.01

= 0.11

(b) From part (a), the probability that the car needs work on either the engine or the transmission is 0.11. Thereforethe probability that the car needs no work is 1−0.11= 0.89.

(c) We need to findP(E∩Tc). Now P(E) = P(E∩T)+ P(E∩Tc) (this can be seen from a Venn diagram). Weknow thatP(E) = 0.10 andP(E∩T) = 0.01. ThereforeP(E∩Tc) = 0.09.

11. (a) P(J∪A) = P(J)+P(A)−P(J∩A)

= 0.6+0.7−0.5

= 0.8

(b) From part (a), the probability that the household gets the rate in at least one month is 0.8. Therefore theprobability that the household does not get the rate in either month is 1−0.8= 0.2.


SECTION 2.1 29

(c) We need to findP(J∩Ac). NowP(J) = P(J∩A)+P(J∩Ac) (this can be seen from a Venn diagram). We knowthatP(J) = 0.6 andP(J∩A) = 0.5. ThereforeP(J∩Ac) = 0.1.

12. (a) Since 948 bowls were found flawless by both inspectors, 52 bowls were found to have flaws by one or bothinspectors. The probability is therefore 52/1000= 0.052.

(b) Let A be the event that a flaw is found by inspector A, and letB be the event that a flaw is found by inspectorB. We need to findP(A∩B). We know thatP(A) = 0.037 andP(B) = 0.043. From part (a) we know thatP(A∪B) = 0.052.

Now P(A∪B) = P(A)+ P(B)−P(A∩B). Substituting, we find that 0.052= 0.037+ 0.043−P(A∩B). Itfollows thatP(A∩B) = 0.028.

(c) We need to findP(A∩Bc). Now P(A) = P(A∩B)+ P(A∩Bc) (this can be seen from a Venn diagram). Weknow thatP(A) = 0.037 andP(A∩B) = 0.028. ThereforeP(A∩Bc) = 0.009.

13. (a) LetC be the event that a student gets an A in calculus, and letPh be the event that a student gets an A inphysics. Then

P(C∪Ph) = P(C)+P(Ph)−P(C∩Ph)

= 0.164+0.146−0.084

= 0.226

(b) The probability that the student got an A in both courses is 0.042. Therefore the probability that the student gotless than an A in one or both courses is 1−0.042= 0.958.

(c) We need to findP(C∩Pc). Now P(C) = P(C∩P)+ P(C∩Pc) (this can be seen from a Venn diagram). Weknow thatP(C) = 0.164 andP(C∩P) = 0.084. ThereforeP(C∩Pc) = 0.08.

(d) We need to findP(P∩Cc). Now P(P) = P(P∩C)+ P(P∩Cc) (this can be seen from a Venn diagram). Weknow thatP(P) = 0.146 andP(C∩P) = 0.084. ThereforeP(C∩Pc) = 0.062.

14. P(A∪B) = P(A)+P(B)−P(A∩B)

= 0.95+0.90−0.88

= 0.97


30 CHAPTER 2

15. P(A∩B) = P(A)+P(B)−P(A∪B)

= 0.98+0.95−0.99

= 0.94

16. (a)P(O) = 1−P(notO)

= 1− [P(A)+P(B)+P(AB)]

= 1− [0.35+0.10+0.05]

= 0.50

(b) P(does not containB) = 1−P(containsB)

= 1− [P(B)+P(AB)]

= 1− [0.10+0.05]

= 0.85

17. (a) The events of having a major flaw and of having only minor flaws are mutually exclusive. Therefore

P(major flaw or minor flaw) = P(major flaw)+P(only minor flaws) = 0.15+0.05= 0.20.

(b) P(no major flaw) = 1−P(major flaw) = 1−0.05= 0.95.

18. (a)A andB are mutually exclusive, since it is impossible for both events to occur.

(b) If bolts #5 and #8 are torqued correctly, but bolt #3 is nottorqued correctly, then eventsB andD both occur.ThereforeB andD are not mutually exclusive.

(c) If bolts #5 and #8 are torqued correctly, but exactly one of the other bolts is not torqued correctly, then eventsC andD both occur. ThereforeC andD are not mutually exclusive.


SECTION 2.2 31

(d) If the #3 bolt is the only one not torqued correctly, then eventsB andC both occur. ThereforeB andC are notmutually exclusive.

19. (a) False

(b) True

(c) False

(d) True

Section 2.2

1. (a)(4)(4)(4) = 64

(b) (2)(2)(2) = 8

(c) (4)(3)(2) = 24

2. (5)(2)(4) = 40

3. (10)(9)(8)(7)(6)(5) = 151,200

4.

(105

)=

10!5!5!

= 252


32 CHAPTER 2

5.

(104

)=

10!4!6!

= 210

6. (8)(7)(6) = 336

7. (210)(45) = 1,048,576

8. (a)(263)(103) = 17,576,000

(b) (26)(25)(24)(10)(9)(8) = 11,232,000

(c)11,232,00017,576,000

= 0.6391

9. (a) 368 = 2.8211×1012

(b) 368−268 = 2.6123×1012

(c)368−268

368 = 0.9260

10.

(15

6,5,4

)=

15!6!5!4!

= 630,630

11. P(match) = P(BB)+P(WW)


SECTION 2.3 33

= (8/14)(4/6)+ (6/14)(2/6)

= 44/84= 0.5238

12. P(match) = P(RR)+P(GG)+P(BB)

= (6/12)(5/11)+ (4/12)(3/11)+(2/12)(1/11)

= 1/3

Section 2.3

1. (a) 2/10

(b) Given that the first fuse is 10 amps, there are 9 fuses remaining of which 2 are 15 amps. ThereforeP(2nd is 15 amps|1st is 10 amps) = 2/9.

(c) Given that the first fuse is 15 amps, there are 9 fuses remaining of which 1 is 15 amps. ThereforeP(2nd is 15 amps|1st is 15 amps) = 1/9.

2. (a)(8/10)(7/9) = 56/90

(b) P(2 fuses selected) = P(1st is 10 amps and 2nd is 15 amps)

= (8/10)(2/9)

= 16/90

(c) P(more than 3 fuses selected) = P(1st 3 fuses are all 10 amps)

= (8/10)(7/9)(6/8)

= 336/720


34 CHAPTER 2

3. Given that a student is an engineering major, it is almost certain that the student took a calculus course. There-fore P(B|A) is close to 1. Given that a student took a calculus course, it is much less certain that the student isan engineering major, since many non-engineering majors take calculus. ThereforeP(A|B) is much less than1, soP(B|A) > P(A|B).

4. (0.056)(0.027) = 0.001512

5. (a)P(A∩B) = P(A)P(B) = (0.2)(0.09) = 0.018

(b) P(Ac∩Bc) = P(Ac)P(Bc) = (1−0.2)(1−0.09)= 0.728

(c) P(A∪B) = P(A)+P(B)−P(A∩B)

= P(A)+P(B)−P(A)P(B)

= 0.2+0.09− (0.2)(0.09)

= 0.272

6. Let A denote the event that the allocation sector is damaged, and let N denote the event that a non-allocationsector is damaged. ThenP(A∩Nc) = 0.20,P(Ac∩N) = 0.7, andP(A∩N) = 0.10.

(a) P(A) = P(A∩Nc)+P(A∩N) = 0.3

(b) P(N) = P(Ac∩N)+P(A∩N) = 0.8

(c) P(N|A) =P(A∩N)

P(A)

=P(A∩N)

P(A∩N)+P(A∩Nc)

=0.10

0.10+0.20= 1/3


SECTION 2.3 35

(d) P(A|N) =P(A∩N)

P(N)

=P(A∩N)

P(A∩N)+P(Ac∩N)

=0.10

0.10+0.70= 1/8

(e) P(Nc|A) =P(A∩Nc)

P(A)

=P(A∩Nc)

P(A∩Nc)+P(A∩N)

=0.20

0.20+0.10= 2/3

Equivalently, one can computeP(Nc|A) = 1−P(N|A) = 1−1/3= 2/3

(f) P(Ac|N) =P(Ac∩N)

P(N)

=P(Ac∩N)

P(Ac∩N)+P(A∩N)

=0.70

0.70+0.10= 7/8

Equivalently, one can computeP(Ac|N) = 1−P(A|N) = 1−1/8= 7/8

7. Let OK denote the event that a valve meets the specification, letR denote the event that a valve is reground,and letS denote the event that a valve is scrapped. ThenP(OK∩Rc) = 0.7, P(R) = 0.2, P(S∩Rc) = 0.1,P(OK|R) = 0.9, P(S|R) = 0.1.

(a) P(Rc) = 1−P(R) = 1−0.2= 0.8

(b) P(S|Rc) =P(S∩Rc)

P(Rc)=

0.10.8

= 0.125


36 CHAPTER 2

(c) P(S) = P(S∩Rc)+P(S∩R)

= P(S∩Rc)+P(S|R)P(R)

= 0.1+(0.1)(0.2)

= 0.12

(d) P(R|S) =P(S∩R)

P(S)

=P(S|R)P(R)

P(S)

=(0.1)(0.2)

0.12= 0.167

(e) P(OK) = P(OK∩Rc)+P(OK∩R)

= P(OK∩Rc)+P(OK|R)P(R)

= 0.7+(0.9)(0.2)

= 0.88

(f) P(R|OK) =P(R∩OK)

P(OK)

=P(OK|R)P(R)

P(OK)

=(0.9)(0.2)

0.88= 0.205

(g) P(Rc|OK) =P(Rc∩OK)

P(OK)

=0.70.88

= 0.795


SECTION 2.3 37

8. LetSdenote the number rolled by Sarah, and letT denote the number rolled by Thomas.

(a) There are 36 equally likely values for the ordered pair(S,T). The ones for whichT > Sare(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6), and(5,6).ThereforeP(T > S) = 15/36.

(b) P(S> T|S= 3) = P(T < 3|S= 3)

= P(T < 3)

= 1/3

(c) P(S< T|S= 3) = P(T > 3|S= 3)

= P(T > 3)

= 1/2

(d) There are 15 equally likely outcomes(S,T) for which S> T. Of these there are three, (4,3), (5,3), and (6,3),for whichT = 3. ThereforeP(T = 3|S> T) = 3/15.

(e) There are 15 equally likely outcomes(S,T) for whichS> T. Of these there are two, (3,2) and (3,1), for whichS= 3. ThereforeP(S= 3|S> T) = 2/15.

9. LetT1 denote the event that the first device is triggered, and letT2 denote the event that the second device istriggered. ThenP(T1) = 0.9 andP(T2) = 0.8.

(a) P(T1∪T2) = P(T1)+P(T2)−P(T1∩T2)

= P(T1)+P(T2)−P(T1)P(T2)

= 0.9+0.8− (0.9)(0.8)

= 0.98

(b) P(T1c∩T2c) = P(T1c)P(T2c) = (1−0.9)(1−0.8)= 0.02

(c) P(T1∩T2) = P(T1)P(T2) = (0.9)(0.8) = 0.72


38 CHAPTER 2

(d) P(T1∩T2c) = P(T1)P(T2c) = (0.9)(1−0.8) = 0.18

10. Let L denote the event that Laura hits the target, and letPh be the event that Philip hits the target. ThenP(L) = 0.5 andP(Ph) = 0.3.

(a) P(L∪Ph) = P(L)+P(Ph)−P(L∩Ph)

= P(L)+P(Ph)−P(L)P(Ph)

= 0.5+0.3− (0.5)(0.3)

= 0.65

(b) P(exactly one hit) = P(L∩Phc)+P(Lc∩Ph)

= P(L)P(Phc)+P(Lc)P(Ph)

= (0.5)(1−0.3)− (1−0.5)(0.3)

= 0.5

(c) P(L|exactly one hit) =P(L∩exactly one hit)

P(exactly one hit)

=P(L∩Phc)

P(exactly one hit)

=P(L)P(Phc)

P(exactly one hit)

=(0.5)(1−0.3)

0.5= 0.7

11. LetR1 andR2 denote the events that the first and second lights, respectively, are red, letY1 andY2 denote theevents that the first and second lights, respectively, are yellow, and letG1 andG2 denote the events that the firstand second lights, respectively, are green.

(a) P(R1) = P(R1∩R2)+P(R1∩Y2)+P(R1∩G2)

= 0.30+0.04+0.16

= 0.50


SECTION 2.3 39

(b) P(G2) = P(R1∩G2)+P(Y1∩G2)+P(G1∩G2)

= 0.16+0.04+0.20

= 0.40

(c) P(same color) = P(R1∩R2)+P(Y1∩Y2)+P(G1∩G2)

= 0.30+0.01+0.20

= 0.51

(d) P(G2|R1) =P(R1∩G2)

P(R1)

=0.160.50

= 0.32

(e) P(R1|Y2) =P(R1∩Y2)

P(Y2)

=0.04

0.04+0.01+0.05= 0.40

12. (a)150

150+25+30+60=

3053

(b)35

35+90+25=

730

(c)15+70

15+70+30=

1723

(d)15+70

30+35+15+5+60+90+70+30=

1767


40 CHAPTER 2

(e)90+70+25+30

35+15+90+70+25+30=

4353

13. (a) That the gauges fail independently.

(b) One cause of failure, a fire, will cause both gauges to fail. Therefore, they do not fail independently.

(c) Too low. The correct calculation would useP(second gauge fails|first gauge fails) in place ofP(second gauge fails).Because there is a chance that both gauges fail together in a fire, the condition that the first gauge fails makesit more likely that the second gauge fails as well.ThereforeP(second gauge fails|first gauge fails) > P(second gauge fails).

14. No.P(both gauges fail) = P(first gauge fails)P(second gauge fails|first gauge fails).

SinceP(second gauge fails|first gauge fails) ≤ 1, P(both gauges fail) ≤ P(first gauge fails) = 0.01.

15. (a)P(A) = 3/10

(b) Given thatA occurs, there are 9 components remaining, of which 2 are defective.

ThereforeP(B|A) = 2/9.

(c) P(A∩B) = P(A)P(B|A) = (3/10)(2/9) = 1/15

(d) Given thatAc occurs, there are 9 components remaining, of which 3 are defective.

ThereforeP(B|A) = 3/9. NowP(Ac∩B) = P(Ac)P(B|Ac) = (7/10)(3/9) = 7/30.

(e) P(B) = P(A∩B)+P(Ac∩B) = 1/15+7/30= 3/10


SECTION 2.3 41

(f) No. P(B) 6= P(B|A) [or equivalently,P(A∩B) 6= P(A)P(B)].

16. (a)P(A) = 300/1000= 3/10

(b) Given thatA occurs, there are 999 components remaining, of which 299 aredefective.

ThereforeP(B|A) = 299/999.

(c) P(A∩B) = P(A)P(B|A) = (3/10)(299/999)= 299/3330

(d) Given thatAc occurs, there are 999 components remaining, of which 300 aredefective.

ThereforeP(B|A) = 300/999. NowP(Ac∩B) = P(Ac)P(B|Ac) = (7/10)(300/999)= 70/333.

(e) P(B) = P(A∩B)+P(Ac∩B) = 299/3330+70/333= 3/10

(f) P(A|B) =P(A∩B)

P(B)=

299/33303/10

=299999

(g) A andB are not independent, but they are very nearly independent. To see this note thatP(B) = 0.3, whileP(B|A) = 0.2993. SoP(B) is very nearly equal toP(B|A), but not exactly equal. Alternatively, note thatP(A∩B) = 0.0898, whileP(A)P(B) = 0.09. Therefore in most situations it would be reasonable to treatA andB as though they were independent.

17. n = 10,000. The two components are a simple random sample from the population. When the population islarge, the items in a simple random sample are nearly independent.

18. LetY denote the event that the Yankees win the first three games, and let R denote the probability that the RedSox win the last four games.


42 CHAPTER 2

(a) We need to findP(Y∩R). Now the Yankees had probability 0.4 of winning each game, and the games wereindependent, soP(Y) = (0.4)3 = 0.064. Similarly,P(R) = (0.6)4 = 0.1296. Since the games were independent,Y andR were independent, soP(Y∩R) = P(Y)P(R) = (0.064)(0.1296) = 0.008294.

(b) SinceY andRare independent,P(R|Y) = P(R) = 0.1296.

19. (a) On each of the 24 inspections, the probability that the process will not be shut down is 1− 0.05 = 0.95.ThereforeP(not shut down for 24 hours) = (0.95)24 = 0.2920. It follows thatP(shut down at least once) =1−0.2920= 0.7080.

(b) P(not shut down for 24 hours) = (1− p)24 = 0.80. Solving forp yields p = 0.009255.

20. P(black SUV|SUV) =P(black SUV∩SUV)

P(SUV)=

P(black SUV)P(SUV)

=0.050.20

= 0.25.

21. LetR denote the event of a rainy day, and letC denote the event that the forecast is correct. ThenP(R) = 0.1,P(C|R) = 0.8, andP(C|Rc) = 0.9.

(a) P(C) = P(C|R)P(R)+P(C|Rc)P(Rc)

= (0.8)(0.1)+ (0.9)(1−0.1)

= 0.89

(b) A forecast of no rain will be correct on every non-rainy day. Therefore the probability is 0.9.

22. LetE denote the event that a parcel is sent express (soEc denotes the event that a parcel is sent standard), and letN denote the event that a parcel arrives the next day. ThenP(E) = 0.25,P(N|E) = 0.95, andP(N|Ec) = 0.80).

(a) P(E∩N) = P(E)P(N|E) = (0.25)(0.95) = 0.2375.


SECTION 2.3 43

(b) P(N) = P(N|E)P(E)+P(N|Ec)P(Ec)

= (0.95)(0.25)+ (0.80)(1−0.25)

= 0.8375

(c) P(E|N) =P(N|E)P(E)

P(N|E)P(E)+P(N|Ec)P(Ec)

=(0.95)(0.25)

(0.95)(0.25)+ (0.80)(1−0.25)= 0.2836

23. LetE denote the event that a board is rated excellent, letA denote the event that a board is rated acceptable,let U denote the event that a board is rated unacceptable, and letF denote the event that a board fails. ThenP(E) = 0.3, P(A) = 0.6, P(U) = 0.1, P(F|E) = 0.1, P(F |A) = 0.2, andP(F |U) = 1.

(a) P(E∩F) = P(E)P(F|E) = (0.3)(0.1) = 0.03.

(b) P(F) = P(F |E)P(E)+P(F|A)P(A)+P(F|U)P(U)

= (0.1)(0.3)+ (0.2)(0.6)+ (1)(0.1)

= 0.25

(c) P(E|F) =P(F |E)P(E)

P(F |E)P(E)+P(F|A)P(A)+P(F|U)P(U)

=(0.1)(0.3)

(0.1)(0.3)+ (0.2)(0.6)+ (1)(0.1)

= 0.12

24. LetA denote the event that the flaw is found by the first inspector, and letB denote the event that the flaw isfound by the second inspector.

(a) P(A∩B) = P(A)P(B) = (0.9)(0.7) = 0.63


44 CHAPTER 2

(b) P(A∪B) = P(A)+P(B)−P(A∩B)= 0.9+0.7−0.63= 0.97

(c) P(Ac∩B) = P(Ac)P(B) = (1−0.9)(0.7) = 0.07

25. Let F denote the event that an item has a flaw. LetA denote the event that a flaw is detected by the firstinspector, and letB denote the event that the flaw is detected by the second inspector.

(a) P(F |Ac) =P(Ac|F)P(F)

P(Ac|F)P(F)+P(Ac|Fc)P(Fc)

=(0.1)(0.1)

(0.1)(0.1)+ (1)(0.9)

= 0.011

(b) P(F |Ac∩Bc) =P(Ac∩Bc|F)P(F)

P(Ac∩Bc|F)P(F)+P(Ac∩Bc|Fc)P(Fc)

=P(Ac|F)P(Bc|F)P(F)

P(Ac|F)P(Bc|F)P(F)+P(Ac|Fc)P(Bc|Fc)P(Fc)

=(0.1)(0.3)(0.1)

(0.1)(0.3)(0.1)+ (1)(1)(0.9)

= 0.0033

26. LetD denote the event that a person has the disease, and let+ denote the event that the test is positive. ThenP(D) = 0.05,P(+|D) = 0.99, andP(+|Dc) = 0.01.

(a) P(D|+) =P(+|D)P(D)

P(+|D)P(D)+P(+|Dc)P(Dc)

=(0.99)(0.05)

(0.99)(0.05)+ (0.01)(0.95)= 0.8390

(b) P(Dc|−) =P(−|Dc)P(Dc)

P(−|Dc)P(Dc)+P(−|D)P(D)


SECTION 2.3 45

=(0.99)(0.95)

(0.99)(0.95)+ (0.01)(0.05)= 0.9995

27. (a) Each son has probability 0.5 of having the disease. Since the sons are independent, the probability that bothare disease-free is 0.52 = 0.25.

(b) Let C denote the event that the woman is a carrier, and letD be the probability that the son has the disease.ThenP(C) = 0.5, P(D|C) = 0.5, andP(D|Cc) = 0. We need to findP(D).

P(D) = P(D|C)P(C)+P(D|Cc)P(Cc)

= (0.5)(0.5)+ (0)(0.5)

= 0.25

(c) P(C|Dc) =P(Dc|C)P(C)

P(Dc|C)P(C)+P(Dc|Cc)P(Cc)

=(0.5)(0.5)

(0.5)(0.5)+ (1)(0.5)

= 0.3333

28. LetFl denote the event that a bottle has a flaw. LetF denote the event that a bottle fails inspection. We aregivenP(Fl) = 0.0002,P(F |Fl) = 0.995, andP(Fc|Flc) = 0.99.

(a) P(Fl |F) =P(F |Fl)P(Fl)

P(F|Fl)P(Fl)+P(F|Flc)P(Flc)

=P(F|Fl)P(Fl)

P(F|Fl)P(Fl)+ [1−P(Fc|Flc)]P(Flc)

=(0.995)(0.0002)

(0.995)(0.0002)+ (1−0.99)(0.9998)= 0.01952

(b) i. Given that a bottle failed inspection, the probability that it had a flaw is only 0.01952.


46 CHAPTER 2

(c) P(Flc|Fc) =P(Fc|Flc)P(Flc)

P(Fc|Flc)P(Flc)+P(Fc|Fl)P(Fl)

=P(Fc|Flc)P(Flc)

P(Fc|Flc)P(Flc)+ [1−P(F|Fl)]P(Fl)

=(0.99)(0.9998)

(0.99)(0.9998)+ (1−0.995)(0.0002)= 0.999999

(d) ii. Given that a bottle passes inspection, the probability that is has no flaw is 0.999999.

(e) The small probability in part (a) indicates that some good bottles will be scrapped. This is not so serious. Theimportant thing is that of the bottles that pass inspection,very few should have flaws. The large probability inpart (c) indicates that this is the case.

29. LetD represent the event that the man actually has the disease, and let+ represent the event that the test givesa positive signal.

We are given thatP(D) = 0.005,P(+|D) = 0.99, andP(+|Dc) = 0.01.

It follows thatP(Dc) = 0.995,P(−|D) = 0.01, andP(−|Dc) = 0.99.

(a) P(D|−) =P(−|D)P(D)

P(−|D)P(D)+P(−|Dc)P(Dc)

=(0.01)(0.005)

(0.01)(0.005)+ (0.99)(0.995)

= 5.08×10−5

(b) P(++ |D) = 0.992 = 0.9801

(c) P(++ |Dc) = 0.012 = 0.0001

(d) P(D|++) =P(++ |D)P(D)

P(++ |D)P(D)+P(++ |Dc)P(Dc)

=(0.9801)(0.005)

(0.9801)(0.005)+ (0.0001)(0.995)= 0.9801


SECTION 2.3 47

30. P(system functions) = P[(A∩B)∪ (C∪D)]. Now P(A∩B) = P(A)P(B) = (1−0.10)(1−0.05) = 0.855, andP(C∪D) = P(C)+P(D)−P(C∩D) = (1−0.10)+ (1−0.20)− (1−0.10)(1−0.20)= 0.98.

Therefore

P[(A∩B)∪ (C∪D)] = P(A∩B)+P(C∪D)−P[(A∩B)∩ (C∪D)]

= P(A∩B)+P(C∪D)−P(A∩B)P(C∪D)

= 0.855+0.98− (0.855)(0.98)

= 0.9971

31. P(system functions) = P[(A∩B)∩ (C∪D)]. Now P(A∩B) = P(A)P(B) = (1−0.05)(1−0.03)= 0.9215, andP(C∪D) = P(C)+P(D)−P(C∩D) = (1−0.07)+ (1−0.14)− (1−0.07)(1−0.14)= 0.9902.

Therefore

P[(A∩B)∩ (C∪D)] = P(A∩B)P(C∪D)

= (0.9215)(0.9902)

= 0.9125

32. (a)P(A∩B) = P(A)P(B) = (1−0.05)(1−0.03)= 0.9215

(b) P(A∩B) = (1− p)2 = 0.9. Thereforep = 1−√

0.9 = 0.0513.

(c) P(three components all function) = (1− p)3 = 0.90. Thereforep = 1− (0.9)1/3 = 0.0345.

33. LetC denote the event that component C functions, and letD denote the event that component D functions.

(a) P(system functions) = P(C∪D)

= P(C)+P(D)−P(C∩D)

= (1−0.08)+ (1−0.12)− (1−0.08)(1−0.12)

= 0.9904

Alternatively,

P(system functions) = 1−P(system fails)

= 1−P(Cc∩Dc)


48 CHAPTER 2

= 1−P(Cc)P(Dc)

= 1− (0.08)(0.12)

= 0.9904

(b) P(system functions) = 1−P(Cc∩Dc) = 1− p2 = 0.99. Thereforep =√

1−0.99= 0.1.

(c) P(system functions) = 1− p3 = 0.99. Thereforep = (1−0.99)1/3 = 0.2154.

(d) Letn be the required number of components. Thenn is the smallest integer such that 1−0.5n≥ 0.99. It followsthatnln(0.5) ≤ ln0.01, son≥ (ln0.01)(ln0.5) = 6.64. Sincen must be an integer,n = 7.

34. To show thatAc andB are independent, we show thatP(Ac∩B) = P(Ac)P(B). Now B = (Ac∩B)∪ (A∩B),and (Ac ∩B) and (A∩B) are mutually exclusive. ThereforeP(B) = P(Ac ∩B) + P(A∩B), from which itfollows thatP(Ac∩B) = P(B)−P(A∩B). SinceA andB are independent,P(A∩B) = P(A)P(B). ThereforeP(Ac ∩B) = P(B)−P(A)P(B) = P(B)[1−P(A)] = P(Ac)P(B). To show thatA andBc are independent, itsuffices to interchangeA andB in the argument above. To show thatAc andBc are independent, replaceB withBc in the argument above, and use the fact thatA andBc are independent.

Section 2.4

1. (a) Discrete

(b) Continuous

(c) Discrete

(d) Continuous

(e) Discrete


SECTION 2.4 49

2. (a)P(X < 2) = P(X = 0)+P(X = 1) = p(0)+ p(1) = 0.5+0.3= 0.8

(b) P(X ≥ 1) = 1−P(X < 1) = 1−P(X = 0) = 1− p(0) = 1−0.5= 0.5

(c) µX = 0(0.5)+1(0.3)+2(0.1)+3(0.1)= 0.8

(d) σ2X = (0−0.8)2(0.5)+ (1−0.8)2(0.3)+ (2−0.8)2(0.1)+ (3−0.8)2(0.1) = 0.96

Alternatively,σ2X = 02(0.5)+12(0.3)+22(0.1)+32(0.1)−0.82 = 0.96

3. (a)µX = 1(0.4)+2(0.2)+3(0.2)+4(0.1)+5(0.1)= 2.3

(b) σ2X = (1−2.3)2(0.4)+ (2−2.3)2(0.2)+ (3−2.3)2(0.2)+ (4−2.3)2(0.1)+ (5−2.3)2(0.1) = 1.81

Alternatively,σ2X = 12(0.4)+22(0.2)+32(0.2)+42(0.1)+52(0.1)−2.32 = 1.81

(c) σX =√

1.81= 1.345

(d) Y = 10X. Therefore the probability density function is as follows.

y 10 20 30 40 50p(y) 0.4 0.2 0.2 0.1 0.1

(e) µY = 10(0.4)+20(0.2)+30(0.2)+40(0.1)+50(0.1)= 23

(f) σ2Y = (10−23)2(0.4)+ (20−23)2(0.2)+ (30−23)2(0.2)+ (40−23)2(0.1)+ (50−23)2(0.1) = 181

Alternatively,σ2Y = 102(0.4)+202(0.2)+302(0.2)+402(0.1)+502(0.1)−232 = 181

(g) σY =√

181= 13.45

4. (a) p3(x) is the only probability mass function, because it is the onlyone whose probabilities sum to 1.

(b) µX = 0(0.1)+1(0.2)+2(0.4)+3(0.2)+4(0.1)= 2.0,

σ2X = (0−2)2(0.1)+ (1−2)2(0.2)+ (2−2)2(0.4)+ (3−2)2(0.2)+ (4−2)2(0.1) = 1.2


50 CHAPTER 2

5. (a)x 1 2 3 4 5

p(x) 0.70 0.15 0.10 0.03 0.02

(b) P(X ≤ 2) = P(X = 1)+P(X = 2) = 0.70+0.15= 0.85

(c) P(X > 3) = P(X = 4)+P(X = 5) = 0.03+0.02= 0.05

(d) µX = 1(0.70)+2(0.15)+3(0.10)+4(0.03)+5(0.02)= 1.52

(e) σX =√

12(0.70)+22(0.15)+32(0.10)+42(0.03)+52(0.02)−1.522 = 0.9325

6. (a) There are two 5 mm bolts and two 7 mm bolts in the box. NowX = 1 if the first bolt selected is 5 mm.ThereforeP(X = 1) = 1/2. If the first bolt selected is 7 mm and the second is 5 mm thenX = 2.ThereforeP(X = 2) = P(first is 7 mm)P(second is 5 mm|first is 7 mm) = (1/2)(2/3) = 1/3.Finally, X = 3 if the first two bolts are both 7 mm.ThereforeP(X = 3) = P(first is 7 mm)P(second is 7 mm|first is 7 mm) = (1/2)(1/3) = 1/6.

The probability mass function ofX is thereforep(1) = 1/2, p(2) = 1/3, p(3) = 1/6, andp(x) = 0 for valuesof x other than 1, 2, and 3.

(b) P(X > 2) = P(X = 3) = 1/6

(c) µX = 1(1/2)+2(1/3)+3(1/6)= 5/3

(d) σ2X = 12(1/2)+22(1/3)+32(1/6)− (5/3)2 = 5/9

7. (a)∑4x=1cx= 1, soc(1+2+3+4)= 1, soc = 0.1.

(b) P(X = 2) = c(2) = 0.1(2) = 0.2

(c) µX = ∑4x=1xP(X = x) = ∑4

x=1 0.1x2 = (0.1)(12+22+32+42) = 3.0

(d) σ2X = ∑4

x=1(x−µX)2P(X = x) = ∑4x=1(x−3)2(0.1x) = 4(0.1)+1(0.2)+0(0.3)+1(0.4)= 1

Alternatively,σ2X = ∑4

x=1 x2P(X = x)−µ2X = ∑4

x=10.1x3−32 = 0.1(13+23+33+43)−32 = 1

(e) σX =√

1 = 1


SECTION 2.4 51

8. (a)P(X < 2) = P(X ≤ 1) = F(1) = 0.53

(b) P(X > 3) = 1−P(X ≤ 3) = 1−F(3) = 1−0.97= 0.03

(c) P(X = 1) = P(X ≤ 1)−P(X ≤ 0) = F(1)−F(0) = 0.53−0.17= 0.36

(d) P(X = 0) = P(X ≤ 0) = F(0) = 0.17

(e) For any integerx, P(X = x) = F(x)−F(x−1). The value ofx for which this quantity is greatest isx = 1.

9. (a)

x p1(x)0 0.21 0.162 0.1283 0.10244 0.08195 0.0655

(b)

x p2(x)0 0.41 0.242 0.1443 0.08644 0.05185 0.0311

(c) p2(x) appears to be the better model. Its probabilities are all fairly close to the proportions of days observed inthe data. In contrast, the probabilities of 0 and 1 forp1(x) are much smaller than the observed proportions.

(d) No, this is not right. The data are a simple random sample,and the model represents the population. Simplerandom samples generally do not reflect the population exactly.

10. LetA denote an acceptable chip, andU an unacceptable one.


52 CHAPTER 2

(a) P(A) = 0.9

(b) P(UA) = P(U)P(A) = (0.1)(0.9) = 0.09

(c) P(X = 3) = P(UUA) = P(U)P(U)P(A) = (0.1)(0.1)(0.9) = 0.009

(d) p(x) =

(0.9)(0.1)x−1 x = 1,2,3, ...

0 otherwise

11. LetA denote an acceptable chip, andU an unacceptable one.

(a) If the first two chips are both acceptable, thenY = 2. This is the smallest possible value.

(b) P(Y = 2) = P(AA) = (0.9)2 = 0.81

(c) P(Y = 3|X = 1) =P(Y = 3 andX = 1)

P(X = 1).

Now P(Y = 3 andX = 1) = P(AUA) = (0.9)(0.1)(0.9) = 0.081, andP(X = 1) = P(A) = 0.9.

ThereforeP(Y = 3|X = 1) = 0.081/0.9= 0.09.

(d) P(Y = 3|X = 2) =P(Y = 3 andX = 2)

P(X = 2).

Now P(Y = 3 andX = 2) = P(UAA) = (0.1)(0.9)(0.9) = 0.081, and

P(X = 2) = P(UA) = (0.1)(0.9) = 0.09.

ThereforeP(Y = 3|X = 2) = 0.081/0.09= 0.9.

(e) If Y = 3 the only possible values forX areX = 1 andX = 2.

Therefore

P(Y = 3) = P(Y = 3|X = 1)P(X = 1)+P(Y = 3|X = 2)P(X = 2)

= (0.09)(0.9)+ (0.9)(0.09)

= 0.162


SECTION 2.4 53

12. (a) 0, 1, 2, 3

(b) P(X = 3) = P(SSS) = (0.8)3 = 0.512

(c) P(FSS) = (0.2)(0.8)2 = 0.128

(d) P(SFS) = P(SSF) = (0.8)2(0.2) = 0.128

(e) P(X = 2) = P(FSS)+P(SFS)+P(SSF) = 0.384

(f) P(X = 1) = P(SFF)+P(FSF)+P(FFS)

= (0.8)(0.2)2+(0.8)(0.2)2+(0.8)(0.2)2

= 0.096

(g) P(X = 0) = P(FFF) = (0.2)3 = 0.008

(h) µX = 0(0.08)+1(0.096)+2(0.384)+3(0.512)= 2.4

(i) σ2X = (0−2.4)2(0.08)+ (1−2.4)2(0.096)+ (2−2.4)2(0.384)+ (3−2.4)2(0.512) = 0.48

Alternatively,σ2X = 02(0.08)+12(0.096)+22(0.384)+32(0.512)−2.42 = 0.48

(j) P(Y = 3) = P(SSSF)+P(SSFS)+P(SFSS)+P(FSSS)

= (0.8)3(0.2)+ (0.8)3(0.2)+ (0.8)3(0.2)+ (0.8)3(0.2)

= 0.4096

13. (a)Z 50

10

x−101800

dx=x2−20x

3600

50

10

= 0.444

(b)Z 70

10x

x−101800

dx=x3−15x2

5400

70

10

= 50


54 CHAPTER 2

(c) σ2X =

Z 70

10x2 x−10

1800dx−502 =

3x4−40x3

21600

70

10

−502 = 200

σX =√

200= 14.142

(d) F(x) =

Z x

−∞f (t)dt

If x < 10,F(x) =

Z x

−∞0dt = 0

If 10 ≤ x < 70,F(x) =

Z 10

−∞0dt+

Z x

10

t −101800

dt =x2/2−10x+50

1800.

If X ≥ 70,F(x) =

Z 10

−∞0dt+

Z 70

10

t −101800

dt+Z x

700dt = 1.

(e) The medianxm solvesF(xm) = 0.5. Thereforex2

m/2−10xm+501800

= 0.5, soxm = 52.426.

14. (a)Z 11

8

x54

dx=x2

108

11

8

= 0.528

(b) µX =

Z 12

6x

x54

dx=x3

162

12

6

= 9.333

(c) σ2X =

Z 12

6x2 x

54dx−µ2

X

=x4

216

12

6

−87.111

= 2.889

(d) σX =√

2.889= 1.70

(e) F(x) =

Z x

−∞f (t)dt

If x < 6, F(x) =

Z x

−∞0dt = 0.


SECTION 2.4 55

If 6 ≤ x < 12,F(x) =

Z 6

−∞0dt+

Z x

6

t54

dt =x2−36

108.

If x≥ 12,F(x) =

Z 6

−∞0dt+

Z 12

6

t54

dt+Z x

120dt = 1.

(f) The medianxm solvesF(xm) = 0.5. Thereforex2

m−36108

= 0.5, soxm = 9.4868.

(g) The 10th percentilex10 solvesF(x10) = 0.1. Thereforex2

10−36108

= 0.1, sox10 = 6.8411.

(h)Z 12

10

x54

dx=x2

108

12

10

= 0.4074

15. (a)µ =

Z ∞

00.2te−0.2t dt

= −te−0.2t∞

0

−Z ∞

0−e−0.2t dt

= 0−5e−0.2t∞

0

= 5

(b) σ2 =

Z ∞

00.2t2e−0.2t dt−µ2

= −t2e−0.2t∞

0

−Z ∞

0−2te−0.2t dt−25

= 0+10Z ∞

00.2te−0.2t dt−25

= 0+10(5)−25

= 25

σX =√

25= 5

(c) F(x) =

Z x

−∞f (t)dt.


56 CHAPTER 2

If x≤ 0, F(x) =

Z x

−∞0dt = 0.

If x > 0, F(x) =

Z 0

−∞0dt+

Z x

00.2e−0.2t dt = 1−e−0.2x.

(d) P(T < 10) = P(T ≤ 10) = F(10) = 1−e−2 = 0.8647

(e) The medianxm solvesF(xm) = 0.5. Therefore 1−e−0.2xm = 0.5, soxm = 3.466.

(f) The 90th percentilex90 solvesF(x90) = 0.9. Therefore 1−e−0.2x90 = 0.9, sox90 = 11.5129.

16. (a)µ=

Z 10.25

9.753x[1−16(x−10)2]dx= −12x4+320x3−2398.5x2

10.25

9.75

= 10

(b) σ2 =

Z 10.25

9.753(x−10)2[1−16(x−10)2]dx= (x−10)3−9.6(x−10)5

10.25

9.75

= 0.0125.

σ =√

0.0125= 0.1118

(c) F(x) =

Z x

−∞f (t)dt.

If x≤ 9.75,F(x) =Z x

−∞0dt = 0.

If 9.75< x < 10.25,

F(x) =

Z x

−∞0dt+

Z x

9.753[1−16(t−10)2]dt = 3t−16(t−10)3

x

9.75

= 3x−16(x−10)3−29.5

If x≥ 10.25,F(x) = 1.

(d) None of them.F(9.75) = 0.

(e) All of them.F(10.25) = 1, so all of the rings have diameters less than or equal to 10.25 cm. Since none of therings have diameters less than 9.75 cm, all of them have diameters between 9.75 and 10.25 cm.


SECTION 2.4 57

17. With this process, the probability that a ring meets the specification isZ 10.1

9.915[1−25(x−10.05)2]/4dx=

Z 0.05

−0.1515[1−25x2]/4dx= 0.25(15x−125x3)

0.05

−0.15

= 0.641.

With the process in Exercise 16, the probability isZ 10.1

9.93[1−16(x−10)2]dx=

Z 0.1

−0.13[1−16x2]dx= 3x−16x3

0.1

−0.1

= 0.568.

Therefore this process is better than the one in Exercise 16.

18. (a)P(X > 3) =

Z ∞

3

81(x+3)4 dx= − 27

(x+3)3

∞

3

= 1/8

(b) P(1 < X < 3) =

Z 3

1

81(x+3)4 dx= − 27

(x+3)3

3

1

= 19/64

(c) µ=Z ∞

0x

81(x+3)4 dx=

Z ∞

3(u−3)

81u4 du= 81

Z ∞

3(u−3−3u−4)du= 81

(−1

2u−2+u−3

) ∞

3

= 3/2

(d) σ2 =

Z ∞

0x2 81

(x+3)4 dx−µ2

=

Z ∞

3(u−3)281

u4 du− (3/2)2

= 81Z ∞

3(u−2−6u−3+9u−4)du − 9/4

= 81(−u−1+3u−2−3u−3)

∞

3

− 9/4

= 27/4

(e) F(x) =

Z x

−∞f (t)dt.

If x < 0, F(x) =Z x

−∞0dt = 0.

If x≥ 0, F(x) =Z x

0

81(t +3)4 dt = − 27

(t +3)3

x

0

= 1−27(x+3)−3


58 CHAPTER 2

(f) The medianxm solvesF(xm) = 0.5. Therefore 1−27(xm+3)−3 = 0.5, soxm = 0.7798.

(g) The 30th percentilex30 solvesF(x30) = 0.3. Therefore 1−27(x30+3)−3 = 0.3, sox30 = 0.3787.

19. (a)P(X > 1) =

Z 3

1(4/27)x2(3−x)dx=

4x3

27− x4

27

3

1

= 8/9

(b) P(1 < X < 2) =

Z 2

1(4/27)x2(3−x)dx=

4x3

27− x4

27

2

1

= 13/27

(c) µ=Z 3

0(4/27)x3(3−x)dx=

427

(3x4

4− x5

5

) 3

0

= 9/5

(d) σ2 =

Z 3

0(4/27)x4(3−x)dx−µ2 =

427

(3x5

5− x6

6

) 3

0

− (9/5)2 = 9/25

(e) F(x) =

Z x

−∞f (t)dt

If x≤ 0, F(x) =

Z x

−∞0dt = 0

If 0 < x < 3, F(x) =Z x

0(4/27)t2(3− t)dt = (4x3−x4)/27

If x≥ 3, F(x) =

Z 3

0(4/27)t2(3− t)dt = 1

20. (a)P(X < 0.02) =

Z 0.02

0625xdx=

625x2

2

0.02

0

= 0.125

(b) µ=

Z 0.04

0625x2dx+

Z 0.08

0.04(50x−625x2)dx=

625x3

3

0.04

0

+

(25x2− 625x3

3

) 0.08

0.04

= 0.04


SECTION 2.4 59

(c) The variance is

σ2 =

Z 0.04

0625x3dx+

Z 0.08

0.04(50x2−625x3)dx−µ2

=625x4

4

0.04

0

+

(50x3

3− 625x4

4

) 0.08

0.04

−0.042

= 0.0002667

The standard deviation isσ =√

0.0002667= 0.01633.

(d) F(x) =Z x

−∞f (t)dt

If x≤ 0, F(x) =

Z x

−∞0dt = 0

If 0 < x≤ 0.04,F(x) =

Z x

0625t dt = 625x2/2

If 0.04< x≤ 0.08,F(x) =Z 0.04

0625t dt+

Z x

0.04(50−625t)dt = 50x− 625

2x2−1

If x > 0.08,F(x) =

Z 0.04

0625t dt+

Z 0.08

0.04(50−625t)dt = 1

(e) The medianxm solvesF(xm) = 0.5. SinceF(x) is described by different expressions forx≤ 0.04 andx> 0.04,we computeF(0.04). F(0.04) = 625(0.04)2/2 = 0.5 soxm = 0.04.

(f) P(0.015< X < 0.063)=

Z 0.04

0.015625xdx+

Z 0.063

0.04(50−625x)dx=

625x2

2

0.04

0

+

(50x− 625x2

2

) 0.063

0.04

= 0.9097

21. (a)P(X > 0.5) =Z 1

0.51.2(x+x2)dx= 0.6x2 +0.4x3

1

0.5

= 0.8

(b) µ=

Z 1

01.2x(x+x2)dx= 0.4x3+0.3x4

1

0

= 0.7

(c) X is within±0.1 of the mean if 0.6 < X < 0.8.

P(0.6 < X < 0.8) =

Z 0.8

0.61.2(x+x2)dx= 0.6x2+0.4x3

0.8

0.6

= 0.2864


60 CHAPTER 2

(d) The variance is

σ2 =

Z 1

01.2x2(x+x2)dx−µ2

= 0.3x4+0.24x51

0

−0.72

= 0.05


0.05= 0.2236.

(e) X is within ±2σ of the mean if 0.2528< X < 1.1472. SinceP(X > 1) = 0, X is within ±2σ of the mean if0.2528< X < 1.

P(0.2528< X > 1) =

Z 1

0.25281.2(x+x2)dx= 0.6x2+0.4x3

1

0.2528

= 0.9552

(f) F(x) =

Z x

−∞f (t)dt

If x≤ 0, F(x) =Z x

−∞0dt = 0

If 0 < x < 1, F(x) =

Z x

01.2(t + t2)dt = 0.6x2 +0.4x3

If x > 1, F(x) =

Z 1

01.2(t + t2)dt = 1

22. (a)P(X < 0.2) =Z 0.2

0

e1−x

e−1dx=

(e

e−1

)(−e−x)

0.2

0

= 0.2868

(b) µ =

Z 1

0

xe1−x

e−1dx

=

(e

e−1

)Z 1

0xe−x dx

=

(e

e−1

)(−xe−x

1

0

+

Z 1

0e−x dx

)

=

(e

e−1

)(−e−1−e−x

1

0

)

=e−2e−1

= 0.4180


SECTION 2.4 61

(c) σ2 =

Z 1

0

x2e1−x

e−1dx−µ2

=

(e

e−1

)(−x2e−x

1

0

+

Z 1

02xe−xdx

)−(

e−2e−1

)2

=

(e

e−1

)[−e−1 +2

(−xe−x

1

0

+

Z 1

0e−xdx

)]−(

e−2e−1

)2

=

(e

e−1

)[−e−1 +2

(−e−1−e−x

1

0

)]−(

e−2e−1

)2

=

(e

e−1

)[−e−1+2

(1−2e−1)]−

(e−2e−1

)2

= 1− e(e−1)2

= 0.07933

(d) F(x) =

Z x

−∞f (t)dt

If x≤ 0, F(x) =

Z x

−∞0dt = 0

If 0 < x < 1, F(x) =Z x

0

e1−t

e−1dt =

1−e−x

1−e−1

If x≥ 1, F(x) =Z 1

0

e1−t

e−1dt = 1

(e) The medianxm solvesF(xm) = 0.5. Therefore1−e−xm

1−e−1 = 0.5.

Solving, we obtaine−xm =e+12e

, soxm = 1+ ln2− ln(e+1) = 0.3799.

(f) P(0 < X < 0.3) =

Z 0.3

0

e1−x

e−1dx=

(e

e−1

)(−e−x)

0.3

0

= 0.4100

23. (a)P(X < 0.4) =Z 0.4

06x(1−x)dx= 3x2−2x3

0.4

0

= 0.352


62 CHAPTER 2

(b) P(0.2 < X < 0.6) =

Z 0.6

0.26x(1−x)dx= 3x2−2x3

0.6

0.2

= 0.544

(c) µ=

Z 1

06x2(1−x)dx= 2x3−1.5x4

1

0

= 0.5

(d) The variance is

σ2 =

Z 1

06x3(1−x)dx−µ2

= 1.5x4−1.2x51

0

−0.52

= 0.05


0.05= 0.2236.

(e) X is within±σ of the mean if 0.2764< X < 0.7236.

P(0.2764< X < 0.7236) =

Z 0.7236

0.27646x(1−x)dx= 3x2−2x3

0.7236

0.2764

= 0.6261

(f) F(x) =

Z x

−∞f (t)dt

If x≤ 0, F(x) =Z x

−∞0dt = 0

If 0 < x < 1, F(x) =

Z x

06t(1− t)dt = 3x2−2x3

If x≥ 1, F(x) =

Z 1

06t(1− t)dt = 1

24. (a)c solves the equationZ ∞

1c/x3dx= 1. Therefore−0.5c/x2

∞

1

= 1, soc = 2.

(b) µX =

Z ∞

1cx/x3dx=

Z ∞

12/x2dx= −2

x

∞

1

= 2

(c) F(x) =

Z x

−∞f (t)dt


SECTION 2.4 63

If x < 1, F(x) =

Z x

−∞0dt = 0.

If x≥ 1, F(x) =

Z 1

−∞0dt+

Z x

12/t3dt = − 1

t2

x

1

= 1−1/x2.

(d) The medianxm solvesF(xm) = 0.5. Therefore 1−1/x2m = 0.5, soxm = 1.414.

(e) P(X ≤ 10) = F(10) = 1−1/102 = 0.99

(f) P(X ≤ 2.5) = F(2.5) = 1−1/2.52 = 0.84

(g) P(X ≤ 2.5|X ≤ 10) =P(X ≤ 2.5 andX ≤ 10)

P(X ≤ 10)=

P(X ≤ 2.5)

P(X ≤ 10)=

0.840.99

= 0.85

25. (a)P(X < 2) =

Z 2

0xe−x dx=

(−xe−x

2

0

+

Z 2

0e−x dx

)=

(−2e−2−e−x

2

0

)= 1−3e−2 = 0.5940

(b) P(1.5 < X < 3) =

Z 3

1.5xe−x dx=

(−xe−x

3

1.5

+

Z 3

1.5e−xdx

)=

(−3e−3+1.5e−1.5−e−x

3

1.5

)

= 2.5e−1.5−4e−3 = 0.3587

(c) µ=Z ∞

0x2e−x dx= −x2e−x

∞

0

+Z ∞

02xe−x dx= 0+2xe−x

∞

0

= 2

(d) F(x) =Z x

−∞f (t)dt

If x < 0, F(x) =

Z x

−∞0dt = 0

If x > 0, F(x) =

Z x

0te−t dt = 1− (x+1)e−x

26. (a)µX =Z ∞

−∞(2√

2π)−1xe−(x−10)2/8dx

=

Z ∞

−∞(2√

2π)−1(x−10)e−(x−10)2/8dx+

Z ∞

−∞(2√

2π)−110e−(x−10)2/8dx


64 CHAPTER 2

Making the substitutionu = (x−10)/√

8,Z ∞

−∞(2√

2π)−1(x−10)e−(x−10)2/8dx=

Z ∞

−∞(2√

2π)−18ue−u2du= 0.

Also,Z ∞

−∞(2√

2π)−110e−(x−10)2/8dx= 10Z ∞

−∞(2√

2π)−1e−(x−10)2/8dx= 10(1) = 1.

SoµX = 10.

(b) σ2X =

Z ∞

−∞(2√

2π)−1(x−10)2e−(x−10)2/8dx. Let u = −4(2√

2π)−1(x−10), dv= −x−104

e−(x−10)2/8dx.

Then

σ2X = −(x−10)2e−(x−10)2/8

∞

−∞

−Z ∞

−∞−4(2

√2π)−1e−(x−10)2/8dx

= 0+4Z ∞

−∞(2√

2π)−1e−(x−10)2/8dx

= 0+4(1)

= 4


4 = 2.

Section 2.5

1. (a)µ2X = 2µX = 2(10.5) = 21.0

σ2X = 2σX = 2(0.5) = 1.0

(b) µX−Y = µX −µY = 10.5−5.7= 4.8

σX−Y =√

σ2X + σ2

Y =√

0.52+0.32 = 0.583

(c) µ3X+2Y = 3µX +2µY = 3(10.5)+2(5.7) = 42.9

σ3X+2Y =√

32σ2X +22σ2

Y =√

9(0.52)+4(0.32) = 1.62

2. (a)µV = µ3R = 3µR = 3(100) = 300


SECTION 2.5 65

(b) σV = σ3R = 3σR = 3(10) = 30

3. LetX1, ...,X5 be the lifetimes of the five bulbs. LetS= X1 + · · ·+X5 be the total lifetime.

µS = ∑µXi = 5(700) = 3500

σS =√

∑σ2Xi

=√

5(202) = 44.7

4. (a)µR = µR1 +µR2 = 50+100= 150

(b) σR =√

σ2R1

+ σ2R2

=√

52 +102 = 11.2

5. LetX1, ...,X5 be the thicknesses of the five layers. LetS= X1 + · · ·+X5 be the total thickness.

(a) µS = ∑µXi = 5(0.125) = 0.625

(b) σS =√

∑σ2Xi

=√

5(0.0052) = 0.0112

6. LetX denote the time on the first machine and letY denote the time on the second machine.

(a) µX+Y = µX +µY = 10+15= 25

(b) σX+Y =√

σ2X + σ2

Y =√

22 +32 = 3.606


66 CHAPTER 2

7. (a)µM = µX+1.5Y = µX +1.5µY = 0.125+1.5(0.350)= 0.650

(b) σM = σX+1.5Y =√

σ2X +1.52σ2

Y =√

0.052+1.52(0.12) = 0.158

8. LetX1, ...,X12 be the weights of the 12 boxes. LetS= X1 + · · ·+X12 be the total weight. The average weightper box isX.

(a) µS = ∑µXi = 12(12.02) = 144.24

(b) σS =√

∑σ2Xi

=√

12(0.032) = 0.104

(c) µX = µXi = 12.02

(d) σX = σXi /√

12= 0.03/√

12= 0.00866

(e) Letn be the required number of boxes. Then 0.03/√

n = 0.005, son = 36.

9. LetX1 andX2 denote the lengths of the pieces chosen from the population with mean 30 and standard deviation0.1, and letY1 andY2 denote the lengths of the pieces chosen from the population with mean 45 and standarddeviation 0.3.

(a) µX1+X2+Y1+Y2 = µX1 +µX2 +µY1 +µY2 = 30+30+45+45= 150

(b) σX1+X2+Y1+Y2 =√

σ2X1

+ σ2X2

+ σ2Y1

+ σ2Y2

=√

0.12+0.12+0.32+0.32 = 0.447

10. The daily revenue isR= 1.60X1+1.75X2+1.90X3.

(a) µR = 1.60µ1+1.75µ2+1.90µ3 = 1.60(1500)+1.75(500)+1.90(300)= 3845


SECTION 2.5 67

(b) σR =√

1.602σ21 +1.752σ2

2 +1.902σ23 =

√1.602(1802)+1.752(902)+1.902(402) = 336.94

11. The tank holds 20 gallons of gas. LetY1 be the number of miles traveled on the first gallon, letY2 be thenumber of miles traveled on the second gallon, and so on, withY20 being the number of miles traveled on the20th gallon. ThenµYi = 25 miles andσYi = 2 miles. LetX = Y1 +Y2 + · · ·Y20 denote the number of milestraveled on one tank of gas.

(a) µX = µY1 + · · ·+µY20 = 20(25) = 500 miles.

(b) σ2X = σ2

Y1+ · · ·+ σ2

Y20= 20(22) = 80. SoσX =

√80= 8.944.

(c) µX/20 = (1/20)µX = (1/20)(500) = 25.

(d) σX/20 = (1/20)σX = (1/20)(8.944) = 0.4472

12. LetX denote the number of mismatches and letY denote the number of gaps. ThenµX = 5, σX = 2, µY = 2,andσY = 1. LetSdenote the Needleman-Wunsch score. ThenS= X +3Y.

(a) µS = µX+3Y = µX +3µY = 5+3(2) = 11

(b) σ2S = σ2

X+3Y = σ2X +9σ2

Y = 22 +9(12) = 13

13. (a)µ= 0.0695+1.0477

20+

0.864920

+0.7356

20+

0.217130

+2.8146

60+

0.591315

+0.0079

10+5(0.0006) = 0.2993

(b) σ =√

0.00182 +(0.026920 )2 +(0.0225

20 )2 +(0.011320 )2 +(0.0185

30 )2 +(0.028460 )2 +(0.0031

15 )2 +(0.000610 )2 + 52(0.0002)2 = 0.00288

14. (a)µX = µO+2N+2C/3 = µO +2µN +(2/3)µC = 0.1668+2(0.0255)+(2/3)(0.0247)= 0.2343

(b) σX = σO+2N+2C/3 =√

σ2O +4σ2

N +(2/3)2σ2C =

√0.03402+4(0.0194)2+(2/3)2(0.0131)2 = 0.05232


68 CHAPTER 2

Section 2.6

1. (a) 0.08

(b) P(X > 0 andY ≤ 1) = P(1,0)+P(1,1)+P(2,0)+P(2,1)= 0.15+0.08+0.10+0.03= 0.36

(c) P(X ≤ 1) = 1−P(X = 2) = 1−P(2,0)−P(2,1)−P(2,2)= 1−0.10−0.03−0.01= 0.86

(d) P(Y > 0) = 1−P(Y = 0) = 1−P(0,0)−P(1,0)−P(2,0)= 1−0.40−0.15−0.10= 0.35

(e) P(X = 0) = P(0,0)+P(0,1)+P(0,2)= 0.40+0.12+0.08= 0.60

(f) P(Y = 0) = P(0,0)+P(1,0)+P(2,0)= 0.40+0.15+0.10= 0.65

(g) P(X = 0 andY = 0) = 0.40

2. (a) The marginal probability mass functionpX(x) is found by summing along the rows of the joint probabilitymass function.

yx 0 1 2 pX(x)0 0.40 0.12 0.08 0.601 0.15 0.08 0.03 0.262 0.10 0.03 0.01 0.14

pY(y) 0.65 0.23 0.12

pX(0) = 0.60, pX(1) = 0.26, pX(2) = 0.14, pX(x) = 0 if x 6= 0,1, or 2

(b) The marginal probability mass functionpY(y) is found by summing down the columns of the joint probabilitymass function. SopY(0) = 0.65, pY(1) = 0.23, pY(2) = 0.12, pY(y) = 0 if y 6= 0,1, or 2

(c) µX = 0pX(0)+1pX(1)+2pX(2) = 0(0.60)+1(0.26)+2(0.14)= 0.54

(d) µY = 0pY(0)+1pY(1)+2pY(2) = 0(0.65)+1(0.23)+2(0.12)= 0.47


SECTION 2.6 69

(e) σ2X = 02pX(0)+12pX(1)+22pX(2)−µ2

X = 02(0.60)+12(0.26)+22(0.14)−0.542 = 0.5284

σX =√

0.5284= 0.7269

(f) σ2Y = 02pY(0)+12pY(1)+22pY(2)−µ2

Y = 02(0.65)+12(0.23)+22(0.12)−0.472 = 0.4891

σY =√

0.4891= 0.6994

(g) Cov(X,Y) = µXY−µXµY

µXY = (0)(0)pX,Y(0,0)+ (0)(1)pX,Y(0,1)+ (0)(2)pX,Y(0,2)+ (1)(0)pX,Y(1,0)+ (1)(1)pX,Y(1,1)

+ (1)(2)pX,Y(1,2)+ (2)(0)pX,Y(2,0)+ (2)(1)pX,Y(2,1)+ (2)(2)pX,Y(2,2)

= (0)(0)(0.40)+ (0)(1)(0.12)+ (0)(2)(0.08)+(1)(0)(0.15)+(1)(1)(0.08)

+ (1)(2)(0.03)+ (2)(0)(0.10)+ (2)(1)(0.03)+(2)(2)(0.01)

= 0.24

µX = 0.54, µY = 0.47

Cov(X,Y) = 0.24− (0.54)(0.47)= −0.0138

(h) ρX,Y =Cov(X,Y)

σXσY=

−0.0138(0.7269)(0.6994)

= −0.0271

(i) No. The joint probability density function is not equal to the product of the marginals. For example,P(X = 1 andY = 1) = 0.08, butP(X = 1)P(Y = 1) = (0.26)(0.23) = 0.0598.

3. (a) pY|X(0|1) =pX,Y(1,0)

pX(1)=

0.150.26

= 0.577

pY|X(1|1) =pX,Y(1,1)

pX(1)=

0.080.26

= 0.308

pY|X(2|1) =pX,Y(1,2)

pX(1)=

0.030.26

= 0.115

(b) pX|Y(0|1) =pX,Y(0,1)

pY(1)=

0.120.23

= 0.522

pX|Y(1|1) =pX,Y(1,1)

pY(1)=

0.080.23

= 0.348


70 CHAPTER 2

pX|Y(2|1) =pX,Y(2,1)

pY(1)=

0.030.23

= 0.130

(c) E(Y |X = 1) = 0pY|X(0|1)+1pY|X(1|1)+2pY|X(2|1) = 0.538

(d) E(X |Y = 1) = 0pX|Y(0|1)+1pX|Y(1|1)+2pX|Y(2|1) = 0.609


yx 1 2 3 pX(x)1 0.15 0.10 0.10 0.352 0.10 0.20 0.15 0.453 0.05 0.05 0.10 0.20

pY(y) 0.30 0.35 0.35

pX(1) = 0.35, pX(2) = 0.45, pX(3) = 0.20, pX(x) = 0 if x 6= 1,2, or 3

(b) The marginal probability mass functionpY(y) is found by summing down the columns of the joint probabilitymass function. SopY(1) = 0.30, pY(2) = 0.35, pY(3) = 0.35, pY(y) = 0 if y 6= 1,2, or 3

(c) No,X andY are not independent. For example,P(X = 1 andY = 1) = 0.15, butP(X = 1)P(Y = 1) = (0.35)(0.30) = 0.105 6= 0.15.

(d) µX = 1pX(1)+2pX(2)+3pX(3) = 1(0.35)+2(0.45)+3(0.20)= 1.85

µY = 1pY(1)+2pY(2)+3pY(3) = 1(0.30)+2(0.35)+3(0.35)= 2.05

(e) σ2X = 12pX(1)+22pX(2)+32pX(3)−µ2

X = 12(0.35)+22(0.45)+32(0.20)−1.852 = 0.5275

σX =√

0.5275= 0.7263

σ2Y = 12pY(1)+22pY(2)+32pY(3)−µ2

Y = 12(0.30)+22(0.35)+32(0.35)−2.052 = 0.6475

σY =√

0.6475= 0.8047

(f) Cov(X,Y) = µXY−µXµY



SECTION 2.6 71

+ (2)(3)pX,Y(2,3)+ (3)(1)pX,Y(3,1)+ (3)(2)pX,Y(3,2)+ (3)(3)pX,Y(3,3)

= (1)(1)(0.15)+ (1)(2)(0.10)+ (1)(3)(0.10)+(2)(1)(0.10)+(2)(2)(0.20)

+ (2)(3)(0.15)+ (3)(1)(0.05)+ (3)(2)(0.05)+(3)(3)(0.10)

= 3.90

µX = 1.85,µY = 2.05

Cov(X,Y) = 3.90− (1.85)(2.05)= 0.1075

(g) ρX,Y =Cov(X,Y)

σXσY=

0.1075(0.7263)(0.8047)

= 0.1839

5. (a)µX+Y = µX +µY = 1.85+2.05= 3.90

(b) σX+Y =√

σ2X + σ2

Y +2Cov(X,Y) =√

0.5275+0.6475+2(0.1075)= 1.179

(c) P(X +Y = 4) = P(1,3)+P(2,2)+P(3,1)= 0.10+0.20+0.05= 0.35

6. (a) pY|X(1|2) =pX,Y(2,1)

pX(2)=

0.100.45

= 2/9

pY|X(2|2) =pX,Y(2,2)

pX(2)=

0.200.45

= 4/9

pY|X(3|2) =pX,Y(2,3)

pX(2)=

0.150.45

= 1/3

(b) pX|Y(1|3) =pX,Y(1,3)

pY(3)=

0.100.35

= 2/7

pX|Y(2|3) =pX,Y(2,3)

pY(3)=

0.150.35

= 3/7

pX|Y(3|3) =pX,Y(3,3)

pY(3)=

0.100.35

= 2/7


72 CHAPTER 2

(c) E(Y |X = 2) = 1pY|X(1|2)+2pY|X(2|2)+3pY|X(3|2) = 19/9

(d) E(X |Y = 3) = 1pX|Y(1|3)+2pX|Y(2|3)+3pX|Y(3|3) = 2

7. (a) 100X +200Y

(b) µ100X+200Y = 100µX +200µY = 100(1.85)+200(2.05)= 595 ms

(c) σ100X+200Y =√

1002σ2X +2002σ2

Y +2(100)(200)Cov(X,Y)

=√

1002(0.5275)+2002(0.6475)+2(100)(200)(0.1075)

= 188.35ms

8. (a)P(X = 2 andY = 2) = P(X = 2)P(Y = 2|X = 2). Now P(X = 2) = 0.30. Given thatX = 2, Y = 2 if andonly if each of the two customers purchases one item. Therefore P(Y = 2|X = 2) = 0.052 = 0.0025, soP(X = 2 andY = 2) = (0.30)(0.0025) = 0.00075.

(b) P(X = 2 andY = 6) = P(X = 2)P(Y = 6|X = 2). Now P(X = 2) = 0.30. Given thatX = 2, LetY1 be thenumber of items purchased by the first customer, and letY2 be the number of items purchased by the secondcustomer. Then the eventY = 6 is equivalent to the eventY1 = 5 andY2 = 1 or Y1 = 4 andY2 = 2 orY1 = 3 andY2 = 3 or Y1 = 2 andY2 = 4 or Y1 = 1 andY2 = 5.

Therefore

P(Y = 6|X = 2) = P(Y1 = 5 andY2 = 1)+P(Y1 = 4 andY2 = 2)+P(Y1 = 3 andY2 = 3)

+ P(Y1 = 2 andY2 = 4)+P(Y1 = 1 andY2 = 5)

= (0.15)(0.05)+ (0.30)(0.15)+(0.25)(0.25)+(0.15)(0.30)+(0.05)(0.15)

= 0.1675

P(X = 2 andY = 6) = (0.30)(0.1675) = 0.05025

(c) P(Y = 2) = P(X = 1 andY = 2)+P(X = 2 andY = 2).

P(X = 1 andY = 2) = P(X = 1)P(Y = 2|X = 1) = (0.25)(0.15) = 0.0375.

From part (a),P(X = 2 andY = 2) = 0.00075.

P(Y = 2) = 0.0375+0.00075= 0.03825.


SECTION 2.6 73


yx 0 1 2 3 4 pX(x)0 0.05 0.04 0.01 0.00 0.00 0.101 0.05 0.10 0.03 0.02 0.00 0.202 0.03 0.05 0.15 0.05 0.02 0.303 0.00 0.02 0.08 0.10 0.05 0.254 0.00 0.00 0.02 0.05 0.08 0.15

pY(y) 0.13 0.21 0.29 0.22 0.15

pX(0) = 0.10, pX(1) = 0.20, pX(2) = 0.30, pX(3) = 0.25, pX(4) = 0.15, pX(x) = 0 if x 6= 0,1,2,3, or 4.

(b) The marginal probability mass functionpY(y) is found by summing down the columns of the joint probabilitymass function. SopY(0) = 0.13, pY(1) = 0.21, pY(2) = 0.29, pY(3) = 0.22, pY(4) = 0.15, pY(y) = 0 ify 6= 0,1,2,3, or 4.

(c) No. The joint probability mass function is not equal to the product of the marginals. For example,pX,Y(0,0) =0.05 6= pX(0)pY(0).

(d) µX = 0pX(0)+1pX(1)+2pX(2)+3pX(3)+4pX(4) = 0(0.10)+1(0.20)+2(0.30)+3(0.25)+4(0.15)= 2.15

µY = 0pY(0)+1pY(1)+2pY(2)+3pY(3)+4pY(4) = 0(0.13)+1(0.21)+2(0.29)+3(0.22)+4(0.15)= 2.05

(e) σ2X = 02pX(0)+12pX(1)+22pX(2)+32pX(3)+42pX(4)−µ2

X

= 02(0.10)+12(0.20)+22(0.30)+32(0.25)+42(0.15)−2.152

= 1.4275

σX =√

1.4275= 1.1948

σ2Y = 02pY(0)+12pY(1)+22pY(2)+32pY(3)+42pY(4)−µ2

Y

= 02(0.13)+12(0.21)+22(0.29)+32(0.22)+42(0.15)−2.052

= 1.5475

σY =√

1.5475= 1.2440

(f) Cov(X,Y) = µXY−µXµY.


+ (1)(0)pX,Y(1,0)+ (1)(1)pX,Y(1,1)+ (1)(2)pX,Y(1,2)+ (1)(3)pX,Y(1,3)+ (1)(4)pX,Y(1,4)



74 CHAPTER 2



= (0)(0)(0.05)+ (0)(1)(0.04)+ (0)(2)(0.01)+(0)(3)(0.00)+(0)(4)(0.00)

+ (1)(0)(0.05)+ (1)(1)(0.10)+ (1)(2)(0.03)+(1)(3)(0.02)+(1)(4)(0.00)

+ (2)(0)(0.03)+ (2)(1)(0.05)+ (2)(2)(0.15)+(2)(3)(0.05)+(2)(4)(0.02)

+ (3)(0)(0.00)+ (3)(1)(0.02)+ (3)(2)(0.08)+(3)(3)(0.10)+(3)(4)(0.05)

+ (4)(0)(0.00)+ (4)(1)(0.00)+ (4)(2)(0.02)+(4)(3)(0.05)+(4)(4)(0.08)

= 5.46

µX = 2.15,µY = 2.05

Cov(X,Y) = 5.46− (2.15)(2.05)= 1.0525

(g) ρX,Y =Cov(X,Y)

σXσY=

1.0525(1.1948)(1.2440)

= 0.7081

10. (a)µX+Y = µX +µY = 2.15+2.05= 4.20

(b) σ2X+Y = σ2

X + σ2Y +2Cov(X,Y) = 1.4275+1.5475+2(1.0525)= 5.08

(c) P(X +Y = 6) = P(X = 2 andY = 4)+P(X = 3 andY = 3)+P(X = 4 andY = 2) = 0.14

11. (a)pY|X(0|3) =pX,Y(3,0)

pX(3)=

0.000.25

= 0

pY|X(1|3) =pX,Y(3,1)

pX(3)=

0.020.25

= 0.08

pY|X(2|3) =pX,Y(3,2)

pX(3)=

0.080.25

= 0.32

pY|X(3|3) =pX,Y(3,3)

pX(3)=

0.100.25

= 0.40

pY|X(4|3) =pX,Y(4,3)

pX(3)=

0.050.25

= 0.20

(b) pX|Y(0|4) =pX,Y(0,4)

pY(4)=

0.000.15

= 0


SECTION 2.6 75

pX|Y(1|4) =pX,Y(1,4)

pY(4)=

0.000.15

= 0

pX|Y(2|4) =pX,Y(2,4)

pY(4)=

0.020.15

= 2/15

pX|Y(3|4) =pX,Y(3,4)

pY(4)=

0.050.15

= 1/3

pX|Y(4|4) =pX,Y(4,4)

pY(4)=

0.080.15

= 8/15

(c) E(Y |X = 3) = 0pY|X(0|3)+1pY|X(1|3)+2pY|X(2|3)+3pY|X(3|3)+4pY|X(4|3) = 2.72

(d) E(X |Y = 4) = 0pX|Y(0|4)+1pX|Y(1|4)+2pX|Y(2|4)+3pX|Y(3|4)+4pX|Y(4|4) = 3.4


yx 0 1 2 pX(x)0 0.10 0.05 0.05 0.201 0.10 0.10 0.05 0.252 0.05 0.20 0.10 0.353 0.05 0.05 0.10 0.20

pY(y) 0.30 0.40 0.30

pX(0) = 0.20, pX(1) = 0.25, pX(2) = 0.35, pX(3) = 0.20, pX(x) = 0 if x 6= 0,1,2, or 3

(b) The marginal probability mass functionpY(y) is found by summing down the columns of the joint probabilitymass function. SopY(0) = 0.30, pY(1) = 0.40, pY(2) = 0.30, pY(y) = 0 if y 6= 0,1, or 2.

(c) µX = 0pX(0)+1pX(1)+2pX(2)+3pX(3) = 0(0.20)+1(0.25)+2(0.35)+3(0.20)= 1.55

(d) µY = 0pY(0)+1pY(1)+2pY(2) = 0(0.30)+1(0.40)+2(0.30)= 1.00

(e) σ2X = 02pX(0)+12pX(1)+22pX(2)+32pX(3)−µ2

X

= 02(0.20)+12(0.25)+22(0.35)+32(0.20)−1.552

= 1.0475


76 CHAPTER 2

σX =√

1.0475= 1.0235

(f) σ2Y = 02pY(0)+12pY(1)+22pY(2)−µ2

Y = 02(0.30)+12(0.40)+22(0.30)−1.002 = 0.6000.

σY =√

0.6000= 0.7746

(g) Cov(X,Y) = µXY−µXµY

µXY = (0)(0)pX,Y(0,0)+ (0)(1)pX,Y(0,1)+ (0)(2)pX,Y(0,2)+ (1)(0)pX,Y(1,0)

+ (1)(1)pX,Y(1,1)+ (1)(2)pX,Y(1,2)+ (2)(0)pX,Y(2,0)+ (2)(1)pX,Y(2,1)

+ (2)(2)pX,Y(2,2)+ (3)(0)pX,Y(3,0)+ (3)(1)pX,Y(3,1)+ (3)(2)pX,Y(3,2)

= (0)(0)(0.10)+ (0)(1)(0.05)+ (0)(2)(0.05)+(1)(0)(0.10)

+ (1)(1)(0.10)+ (1)(2)(0.05)+ (2)(0)(0.05)+(2)(1)(0.20)

+ (2)(2)(0.10)+ (3)(0)(0.05)+ (3)(1)(0.05)+(3)(2)(0.10)

= 1.75

µX = 1.55,µY = 1.00

Cov(X,Y) = 1.75− (1.55)(1.00)= 0.20

(h) ρX,Y =Cov(X,Y)

σXσY=

0.20(1.0235)(0.7746)

= 0.2523

13. (a)µZ = µX+Y = µX +µY = 1.55+1.00= 2.55

(b) σZ = σX+Y =√

σ2X + σ2

Y +2Cov(X,Y) =√

1.0475+0.6000+2(0.20)= 1.4309

(c) P(Z = 2) = P(X +Y = 2)

= P(X = 0 andY = 2)+P(X = 1 andY = 1)+P(X = 2 andY = 0)

= 0.05+0.10+0.05

= 0.2

14. (a)T = 2X +5Y, soµT = µ2X+5Y = 2µX +5µY = 2(1.55)+5(1.00)= 8.10.


SECTION 2.6 77

(b) σT = σ2X+5Y

=√

22σ2X +52σ2

Y +2(2)(5)Cov(X,Y)

=√

22(1.0475)+52(0.6000)+2(2)(5)(0.20)

= 4.82

(c) P(T = 9) = P(X = 2 andY = 1) = 0.20

15. (a)pY|X(0|3) =pX,Y(3,0)

pX(3)=

0.050.20

= 0.25

pY|X(1|3) =pX,Y(3,1)

pX(3)=

0.050.20

= 0.25

pY|X(2|3) =pX,Y(3,2)

pX(3)=

0.100.20

= 0.50

(b) pX|Y(0|1) =pX,Y(0,1)

pY(1)=

0.050.40

= 0.125

pX|Y(1|1) =pX,Y(1,1)

pY(1)=

0.100.40

= 0.25

pX|Y(2|1) =pX,Y(2,1)

pY(1)=

0.200.40

= 0.50

pX|Y(3|1) =pX,Y(3,1)

pY(1)=

0.050.40

= 0.125

(c) E(Y |X = 3) = 0pY|X(0|3)+1pY|X(1|3)+2pY|X(2|3) = 1.25.

(d) E(X |Y = 1) = 0pX|Y(0|1)+1pX|Y(1|1)+2pX|Y(2|1)+3pX|Y(3|1) = 1.625

16. (a)P(X < 0.5 andY > 0.75) =Z 1

0.75

Z 0.5

04xydxdy

=

Z 1

0.75

(2x2y

0.5

0

)dy


78 CHAPTER 2

=Z 1

0.75

y2

dy

=y2

4

1

0.75

= 0.1094

(b) fX(x) =

Z 1

0f (x,y)dy.

If x≤ 0 orx≥ 1 then f (x,y) = 0 for all y so fX(x) = 0.

If 0 < x < 1 then fX(x) =

Z 1

04xydy= 2xy2

1

0

= 2x.

fY(y) =Z 1

0f (x,y)dx.

If y≤ 0 ory≥ 1 then f (x,y) = 0 for all x so fY(y) = 0.

If 0 < y < 1 then fY(y) =Z 1

04xydx= 2x2y

1

0

= 2y.

(c) Yes, f (x,y) = fX(x) fY(y).

17. (a) Cov(X,Y) = µXY−µXµY

µXY =

Z 2

1

Z 5

4

16

xy(x+y)dydx

=

Z 2

1

16

(x2y2

2+

xy3

3

) 5

4

dx

=

Z 2

1

16

(9x2

2+

61x3

)dx

=16

(3x3

2+

61x2

6

) 2

1

=416

fX(x) =16

(x+

92

)for 1≤ x≤ 2 (see Example 2.55).

µX =

Z 2

1

16

x

(x+

92

)dx=

16

(x3

3+

9x2

4

) 2

1

=10972

.


SECTION 2.6 79

fY(y) =16

(y+

32

)for 4≤ y≤ 5 (see Example 2.55).

µY =

Z 5

4

16

y

(y+

32

)dy=

16

(y3

3+

3y2

4

) 5

4

=32572

.

Cov(X,Y) = µXY−µXµY =416

−(

10972

)(32572

)= −0.000193.

(b) ρX,Y =Cov(X,Y)

σXσY.

σ2X =

Z 2

1

16

x2(

x+92

)dx−µ2

X =16

(x4

4+

3x3

2

) 2

1

−(

10972

)2

= 0.08314.

σ2Y =

Z 5

4

16

y2(

y+32

)dx−µ2

Y =16

(y4

4+

y3

2

) 5

4

−(

32572

)2

= 0.08314.

ρX,Y =−0.000193√

(0.08314)(0.08314)= −0.00232.

18. (a)P(X ≤ 0.25 andY ≤ 0.85) =

Z 0.25

0

Z 0.85

0(x+y)dydx

=Z 0.25

0

(xy+

y2

2

) 0.85

0

dx

=

Z 0.25

0(0.36125+0.85x)dx

= 0.36125x+0.425x20.25

0

= 0.1169

(b) Cov(X,Y) = µXY−µXµY.

µXY =

Z 1

0

Z 1

0xy(x+y)dxdy

=

Z 1

0

(x3y3

+x2y2

2

) 1

0

dy


80 CHAPTER 2

=

Z 1

0

(y3

+y2

2

)dy

=

(y2

6+

y3

6

) 1

0

dy

=13

For 0< x < 1, fX(x) =

Z 1

0(x+y)dy= x+0.5. Forx≤ 0 andx≥ 1, fX(x) = 0.

µX =

Z 1

0x

(x+

12

)dx=

(x3

3+

x2

4

) 1

0

=712

.

For 0< y < 1, fY(y) =Z 1

0(x+y)dx= y+0.5. Fory≤ 0 andy≥ 1, fY(y) = 0.

µY =

Z 1

0y

(y+

12

)dy=

(y3

3+

y2

4

) 1

0

=712

.

Cov(X,Y) =13−(

712

)2

= − 1144

.

(c) ρX,Y =Cov(X,Y)

σXσY.

σ2X =

Z 1

0x2(

x+12

)dx−µ2

X =

(x4

4+

x3

6

) 1

0

−(

712

)2

=11144

.

σ2Y =

Z 1

0y2(

y+12

)dy−µ2

Y =

(y4

4+

y3

6

) 1

0

−(

712

)2

=11144

.

ρX,Y =−1/144√

11/144√

11/144= − 1

11.

(d) No,ρX,Y 6= 0.

19. (a) fX(x) =

x+1/2 0< x < 1

0 for other values ofxfY(y) =

y+1/2 0< y < 1

0 for other values ofyThese were computed in the solution to Exercise 18.

(b) fY|X(y|0.75) =f (0.75, y)fX(0.75)

.


SECTION 2.6 81

fX(x) = x+0.5 for 0< x < 1. This was computed in the solution to Exercise 18.

If 0 < y < 1, thenf (0.75, y)fX(0.75)

=y+0.75

1.25=

4y+35

.

If y≤ 0 ory≥ 1, thenf (0.75, y) = 0, so fY|X(y|0.75) = 0.

ThereforefY|X(y|0.75) =

4y+35

0 < y < 1

0 for other values ofy

(c) E(Y |X = 0.75) =

Z ∞

−∞y fY|X(y|0.75)dy=

Z 1

0y

(4y+3

5

)dy=

(4y3

15+

3y2

10

) 1

0

= 0.5667

20. (a)P(X < 3.25 andY > 0.8) =

Z 3.25

3

Z 1

0.8

48xy2

49dydx

=

Z 3.25

3

(16xy3

49

1

0.8

)dx

=

Z 3.25

3

976x6125

dx

=488x2

6125

3.25

3

= 0.12449

(b) For 3< x < 4, fX(x) =

Z 1

0.5

48xy2

49dy=

2x7

. Forx≤ 3 andx≥ 4, f (x,y) = 0, so fX(x) = 0.

For 0.5 < y < 1, fX(x) =

Z 4

3

48xy2

49dx=

24y2

7. Fory≤ 0.5 andy≥ 1, f (x,y) = 0, so fY(y) = 0.

ThereforefX(x) =

2x/7 3< x < 4

0 for other values ofxfY(y) =

24y2/7 0.5 < y < 1

0 for other values ofy

(c) Yes, f (x,y) = fX(x) fY(y).


82 CHAPTER 2

21. (a)P(X < 9.98) =

Z 9.98

9.9510dx= 10x

9.98

9.95

= 0.3

(b) P(Y > 5.01) =

Z 5.1

5.015dy= 5y

5.1

5.01

= 0.45

(c) SinceX andY are independent,P(X < 9.98 andY > 5.01) = P(X < 9.98)P(Y > 5.01) = (0.3)(0.45) = 0.135

(d) µX =Z 10.05

9.9510xdx= 5x2

10.05

9.95

= 10

(e) µY =

Z 5.1

4.95ydy= 2.5y2

5.1

4.9

= 5

(f) µA = µXY = µXµY = (10)(5) = 50

22. (a)µX =

Z 6

4

(34

x− 3x(x−5)2

4

)dx=

(−9x2+

5x3

2− 3x4

16

) 6

4

= 5

(b) σ2X =

Z 6

4

(3(x−5)2

4− 3(x−5)4

4

)dx=

((x−5)3

4− 3(x−5)5

20

) 6

4

= 0.2

(c) µY = µ0.0394X = 0.0394µX = 0.0394(5) = 0.197

σ2Y = σ2

0.0394X = (0.0394)2σ2X = (0.0394)2(0.2) = 0.00031

(d) LetX1, X2, andX3 be the three thicknesses, in millimeters.

ThenS= X1 +X2 +X3 is the total thickness.

µS = µX1 +µX2 +µX3 = 3(5) = 15.

σ2S = σ2

X1+ σ2

X2+ σ2

X3= 3(0.2) = 0.6.


SECTION 2.6 83

23. (a) The probability mass function ofY is the same as that ofX, so fY(y) = e−y if y > 0 and fY(y) = 0 if y ≤ 0.SinceX andY are independent,f (x,y) = fX(x) fY(y).

Thereforef (x,y) =

e−x−y x > 0 andy > 0

0 otherwise

(b) P(X ≤ 1 andY > 1) = P(X ≤ 1)P(Y > 1)

=

(Z 1

0e−xdx

)(Z ∞

1e−y dy

)

=

(−e−x

1

0

)(−e−y

∞

1

)

= (1−e−1)(e−1)

= e−1−e−2

= 0.2325

(c) µX =

Z ∞

0xe−xdx= −xe−x

∞

0

−Z ∞

0e−x dx= 0− (−e−x)

∞

0

= 0+1= 1

(d) SinceX andY have the same probability mass function,µY = µX = 1.

ThereforeµX+Y = µX +µY = 1+1= 2.

(e) P(X +Y ≤ 2) =

Z ∞

−∞

Z 2−x

−∞f (x,y)dydx

=

Z 2

0

Z 2−x

0e−x−y dydx

=

Z 2

0e−x

(−e−y

2−x

0

)dx

=

Z 2

0e−x(1−ex−2)dx

=

Z 2

0(e−x−e−2)dx

= (−e−x−xe−2)

2

0

= 1−3e−2

= 0.5940


84 CHAPTER 2

24. (a) Yes. LetXi denote the number of bytes downloaded in theith second. The total number of bytes isX1 +X2 +X3 +X4+X5. The mean of the total number is the sum of the five means, each of which is equal to 105.

(b) No. X1, ...,X5 are not independent, so the standard deviation of the sum depends on the covariances betweentheXi .

25. (a) Letµ = 40.25 be the mean SiO2 content, and letσ = 0.36 be the standard deviation of the SiO2 content, in arandomly chosen rock. LetX be the average content in a random sample of 10 rocks.

ThenµX = µ= 40.25, andσX =σ√10

=0.36√

10= 0.11.

(b) Letn be the required number of rocks. Thenσ√n

=0.36√

n= 0.05.

Solving forn yieldsn = 51.84. Sincen must be an integer, taken = 52.

26. (a)P(X = 1) = 1/3, P(Y = 1) = 2/3, P(X = 1 andY = 1) = 1/3 6= P(X = 1)P(Y = 1).

(b) µXY = (−1)(1/3)+ (0)(1/3)+ (1)(1/3)= 0.

Now µX = (−1)(1/3)+ (0)(1/3)+ (1)(1/3)= 0, andµY = (0)(1/3)+ (1)(2/3) = 2/3.

So Cov(X,Y) = µXY−µXµY = 0− (0)(2/3) = 0.

Since Cov(X,Y) = 0, ρX,Y = 0.

27. (a)R= 0.3X +0.7Y

(b) µR = µ0.3X+0.7Y = 0.3µX +0.7µY = (0.3)(6)+ (0.7)(6) = 6.

The risk isσR = σ0.3X+0.7Y =√

0.32σ2X +0.72σ2

Y +2(0.3)(0.7)Cov(X,Y).

Cov(X,Y) = ρX,YσXσY = (0.3)(3)(3) = 2.7.

ThereforeσR =√

0.32(32)+0.72(32)+2(0.3)(0.7)(2.7) = 2.52.

(c) µR = µ(0.01K)X+(1−0.01K)Y = (0.01K)µX +(1−0.01K)µY = (0.01K)(6)+ (1−0.01K)(6)= 6.


SECTION 2.6 85

σR =√

(0.01K)2σ2X +(1−0.01K)2σ2

Y +2(0.01K)(1−0.01K)Cov(X,Y).

ThereforeσR=√

(0.01K)2(32)+ (1−0.01K)2(32)+2(0.01K)(1−0.01K)(2.7)= 0.03√

1.4K2−140K +10,000.

(d) σR is minimized when 1.4K2−140K +10000 is minimized.

Nowd

dK(1.4K2−140K +10000) = 2.8K−140, so

ddK

(1.4K2−140K +10000) = 0 if K = 50.

σR is minimized whenK = 50.

(e) For any correlationρ, the risk is 0.03√

K2 +(100−K)2+2ρK(100−K).

If ρ 6= 1 this quantity is minimized whenK = 50.

28. µV =Z 21

19

Z 6

53πr2h(h−20)2(r −5)drdh

= 3πZ 21

19h(h−20)2dh

Z 6

5r2(r −5)dr

= (3π)

(h4

4− 40h3

3+200h2

) 21

19

(r4

4− 5r3

3

) 6

5

= 2021.0913cm3

29. (a)σM1 =√

σ2M1 =

√σ2

R+E1=√

σ2R+ σ2

E1 =√

22 +12 = 2.2361. Similarly,σM2 = 2.2361.

(b) µM1M2 = µR2+E1R+E2R+E1E2= µR2 +µE1µR+µE2µR+µE1µE2 = µR2

(c) µM1µM2 = µR+E1µR+E2 = (µR+µE1)(µR+µE2) = µRµR = µ2R

(d) Cov(M1,M2) = µM1M2 −µM1µM2 = µR2 −µ2R = σ2

R

(e) ρM1,M2 =Cov(M1,M2)

σM1σM2

=σ2

R

σM1σM2

=4

(2.2361)(2.2361)= 0.8

30. Cov(X,X) = µX·X −µXµX = µX2 − (µX)2 = σ2X .


86 CHAPTER 2

31. (a) Cov(aX,bY) = µaX·bY−µaXµbY = µabXY−aµXbµY = abµXY−abµXµY

= ab(µXY−µXµY) = abCov(X,Y).

(b) ρaX,bY = Cov(aX,bY)/(σaXσbY) = abCov(X,Y)/(abσXσY) = Cov(X,Y)/(σXσY) = ρX,Y.

32. Cov(X +Y,Z) = µ(X+Y)Z −µX+YµZ

= µXZ+YZ− (µX +µY)µZ

= µXZ +µYZ−µXµZ−µYµZ

= µXZ−µXµZ +µYZ−µYµZ

= Cov(X,Z)+Cov(Y,Z)

33. (a)V(X− (σX/σY)Y) = σ2X +(σX/σY)2σ2

Y −2(σX/σY)Cov(X,Y)

= 2σ2X −2(σX/σY)Cov(X,Y)

(b) V(X− (σX/σY)Y) ≥ 0

2σ2X −2(σX/σY)Cov(X,Y) ≥ 0

2σ2X −2(σX/σY)ρX,YσXσY ≥ 0

2σ2X −2ρX,Yσ2

X ≥ 0

1−ρX,Y ≥ 0

ρX,Y ≤ 1

(c) V(X +(σX/σY)Y) ≥ 0

2σ2X +2(σX/σY)Cov(X,Y) ≥ 0

2σ2X +2(σX/σY)ρX,YσXσY ≥ 0

2σ2X +2ρX,Yσ2

X ≥ 0

1+ ρX,Y ≥ 0

ρX,Y ≥ −1


SECTION 2.6 87

34. (a)µX = µ1.12C+2.69N+O−0.21Fe

= 1.12µC +2.69µN +µO−0.21µFe

= 1.12(0.0247)+2.69(0.0255)+0.1668−0.21(0.0597)

= 0.2505

(b) Cov(C,N) = ρC,NσCσN = −0.44(0.0131)(0.0194)= −1.118×10−4

Cov(C,O) = ρC,OσCσO = 0.58(0.0131)(0.0340)= 2.583×10−4

Cov(C,Fe) = ρC,FeσCσFe = 0.39(0.0131)(0.0413)= 2.110×10−4

Cov(N,O) = ρN,OσNσO = −0.32(0.0194)(0.0340)= −2.111×10−4

Cov(N,Fe) = ρN,FeσNσFe = 0.09(0.0194)(0.0413)= 7.211×10−5

Cov(O,Fe) = ρO,FeσOσFe = −0.35(0.0340)(0.0413)= −4.915×10−4

(c) σ2X = σ2

1.12C+2.69N+O−0.21Fe

= 1.122σ2C +2.692σ2

N + σ2O +0.212σ2

Fe+2(1.12)(2.69)Cov(C,N)+2(1.12)Cov(C,O)−2(1.12)(0.21)Cov(C,Fe)

+2(2.69)Cov(N,O)−2(2.69)(0.21)Cov(N,Fe)−2(0.21)Cov(O,Fe)

= 1.122(0.0131)2+2.692(0.0194)2+0.03402+0.212(0.0413)2+2(1.12)(2.69)(−0.0001118)

+2(1.12)(0.0002583)−2(1.12)(0.21)(0.0002110)+2(2.69)(−0.0002111)−2(2.69)(0.21)(0.00007211)

−2(0.21)(0.0004915)

= 0.0029648

σ =√

0.0029648= 0.05445

35. µY = µ7.84C+11.44N+O−1.58Fe

= 7.84µC +11.44µN +µO−1.58µFe

= 7.84(0.0247)+11.44(0.0255)+0.1668−1.58(0.0597)

= 0.5578

σ2Y = σ2

7.84C+11.44N+O−1.58Fe

= 7.842σ2C +11.442σ2

N + σ2O +1.582σ2

Fe+2(7.84)(11.44)Cov(C,N)+2(7.84)Cov(C,O)−2(7.84)(1.58)Cov(C,Fe)

+2(11.44)Cov(N,O)−2(11.44)(1.58)Cov(N,Fe)−2(1.58)Cov(O,Fe)

= 7.842(0.0131)2+11.442(0.0194)2+0.03402+1.582(0.0413)2+2(7.84)(11.44)(−0.0001118)

+2(7.84)(0.0002583)−2(7.84)(1.58)(0.0002110)+2(11.44)(−0.0002111)−2(11.44)(1.58)(0.00007211)

−2(1.58)(0.0004915)

= 0.038100


88 CHAPTER 2

σ =√

0.038100= 0.1952

36. (a) Letc =R ∞−∞ h(y)dy.

Now fX(x) =R ∞−∞ f (x,y)dy=

R ∞−∞ g(x)h(y)dy= g(x)

R ∞−∞ h(y)dy= cg(x).

fY(y) =R ∞−∞ f (x,y)dx=

R ∞−∞ g(x)h(y)dx= h(y)

R ∞−∞ g(x)dx= (1/c)h(y)

R ∞−∞ cg(x)dx

= (1/c)h(y)R ∞−∞ fX(x)dx= (1/c)h(y)(1) = (1/c)h(y).

(b) f (x,y) = g(x)h(y) = cg(x)(1/c)h(y) = fX(x)gY(y). ThereforeX andY are independent.

37. (a)R ∞−∞

R ∞−∞ f (x,y)dxdy=

R dc

R ba kdxdy= k

R dc

R ba dxdy= k(d−c)(b−a) = 1.

Thereforek =1

(b−a)(d−c).

(b) fX(x) =R d

c kdy=d−c

(b−a)(d−c)=

1b−a

(c) fY(y) =R b

a kdx=b−a

(b−a)(d−c)=

1d−c

(d) f (x,y) =1

(b−a)(d−c)=

(1

b−a

)(1

d−c

)= fX(x) fY(y)


1. LetA be the event that component A functions, letB be the event that component B functions, letC be the eventthat component C functions, and letD be the event that component D functions. ThenP(A) = 1−0.1 = 0.9,P(B) = 1−0.2 = 0.8, P(C) = 1−0.05= 0.95, andP(D) = 1−0.3= 0.7. The event that the system functionsis (A∪B)∪ (C∪D).

P(A∪B) = P(A)+P(B)−P(A∩B)= P(A)+P(B)−P(A)P(B) = 0.9+0.8− (0.9)(0.8)= 0.98.

P(C∪D) = P(C)+P(D)−P(C∩D) = P(C)+P(D)−P(C)P(D) = 0.95+0.7− (0.95)(0.7)= 0.985.

P[(A∪B)∪ (C∪D)] = P(A∪B)+P(C∪D)−P(A∪B)P(C∪D) = 0.98+0.985− (0.98)(0.985)= 0.9997.



2. P(more than 5 rolls necessary) = P(first 5 rolls are not 6) =

(56

)5

= 0.4019.

3. LetA denote the event that the resistance is above specification,and letB denote the event that the resistance isbelow specification. ThenA andB are mutually exclusive.

(a) P(doesn’t meet specification) =P(A∪B) = P(A)+P(B) = 0.05+0.10= 0.15

(b) P[B|(A∪B)] =P[(B∩ (A∪B)]

P(A∪B)=

P(B)

P(A∪B)=

0.100.15

= 0.6667

4. LetD denote the event that the bag is defective, letL1 denote the event that the bag came from line 1, and letL2 denote them event that the bag came from line 2. ThenP(D |L1) = 1/100 andP(D |L2) = 3/100.

(a) P(L1) =23

(b) P(D) = P(D |L1)P(L1)+P(D |L2)P(L2) = (1/100)(2/3)+ (3/100)(1/3)=160

(c) P(L1 |D) =P(D |L1)P(L1)

P(D)=

(1/100)(2/3)

1/60=

25

(d) P(L1 |Dc) =P(Dc |L1)P(L1)

P(Dc)=

[1−P(D |L1)]P(L1)

P(Dc)=

(1−1/100)(2/3)

59/60=

198295

= 0.6712

5. LetR be the event that the shipment is returned. LetD1 be the event that the first fuse chosen is defective, letD2 be the event that the second fuse chosen is defective, and letD3 be the event that the third fuse chosen isdefective. Since the sample size of 3 is a small proportion ofthe population size of 10,000, it is reasonable totreat these events as independent, each with probability 0.1.

P(R) = 1−P(Rc) = 1−P(Dc1∩Dc

2∩Dc3) = 1− (0.9)3 = 0.271.

(If dependence is taken into account, the answer to six significant digits is 0.271024.)


90 CHAPTER 2

6. (a)(0.99)10 = 0.904

(b) Let p be the required probability. Thenp10 = 0.95. Solving forp, p = 0.9949.

7. Let A be the event that the bit is reversed at the first relay, and letB be the event that the bit is reversed atthe second relay. ThenP(bit received is the same as the bit sent) = P(Ac ∩Bc) + P(A∩B) = P(Ac)P(Bc) +P(A)P(B) = 0.92+0.12 = 0.82.

8. (a)Z 1

−1k(1−x2)dx= k

Z 1

−1(1−x2)dx= 1. Since

Z 1

−1(1−x2)dx=

(x− x3

3

) 1

−1

=43

, k =34

= 0.75.

(b)Z 1

00.75(1−x2)dx= 0.75

(x− x3

3

) 1

0

= 0.5

(c)Z 0.25

−0.250.75(1−x2)dx= 0.75

(x− x3

3

) 0.25

−0.25

= 0.3672

(d) µ=

Z 1

−10.75x(1−x2)dx= 0.75

(x2

2− x4

4

) 1

−1

= 0

(e) The medianxm solvesZ xm

−10.75(1−x2)dx= 0.5. Therefore 0.75

(x− x3

3

) xm

−1

= 0.5, soxm = 0

(f) σ2 =

Z 1

−10.75x2(1−x2)dx−µ2

= 0.75

(x3

3− x5

5

) 1

−1

−02

= 0.2

σ =√

0.2 = 0.4472



9. Let A be the event that two different numbers come up, and letB be the event that one of the dice comes up6. ThenA contains 30 equally likely outcomes (6 ways to choose the number for the first die times 5 ways tochoose the number for the second die). Of these 30 outcomes, 10 belong toB, specifically (1,6), (2,6), (3,6),(4,6), (5,6), (6,1), (6,2), (6,3), (6,4), and (6,5). ThereforeP(B|A) = 10/30= 1/3.

10. LetA be the event that the first component is defective and letB be the event that the second component isdefective.

(a) P(X = 0) = P(Ac∩Bc) = P(Ac)P(Bc|Ac) =

(810

)(79

)= 0.6222

(b) P(X = 1) = P(A∩Bc)+P(Ac∩B)

= P(A)P(Bc|A)+P(Ac)P(B|Ac)

=

(210

)(89

)+

(810

)(29

)

= 0.3556

(c) P(X = 2) = P(A∩B) = P(A)P(B|A) =

(210

)(19

)= 0.0222

(d) pX(0) = 0.6222,pX(1) = 0.3556,pX(2) = 0.0222,pX(x) = 0 if x 6= 0,1, or 2.

(e) µX = 0pX(0)+1pX(1)+2pX(2) = 0(0.6222)+1(0.3556)+2(0.0222)= 0.4

(f) σX =√

02pX(0)+12pX(1)+22pX(2)−µ2X =

√02(0.6222)+12(0.3556)+22(0.0222)−0.42 = 0.5333

11. (a)P(X ≤ 2 andY ≤ 3) =

Z 2

0

Z 3

0

16

e−x/2−y/3dydx

=

Z 2

0

12

e−x/2

(−e−y/3

3

0

)dx

=Z 2

0

12

e−x/2(1−e−1)dx

= (e−1−1)e−x/22

0


92 CHAPTER 2

= (1−e−1)2

= 0.3996

(b) P(X ≥ 3 andY ≥ 3) =

Z ∞

3

Z ∞

3

16

e−x/2−y/3dydx

=

Z ∞

3

12

e−x/2

(−e−y/3

∞

3

)dx

=

Z ∞

3

12

e−x/2e−1dx

= −e−1e−x/2∞

3

= e−5/2

= 0.0821

(c) If x≤ 0, f (x,y) = 0 for all y so fX(x) = 0.

If x > 0, fX(x) =

Z ∞

0

16

e−x/2−y/3dy=12

e−x/2

(−e−y/3

∞

0

)=

12

e−x/2.

ThereforefX(x) =

12

e−x/2 x > 0

0 x≤ 0

(d) If y≤ 0, f (x,y) = 0 for all x so fY(y) = 0.

If y > 0, fY(y) =

Z ∞

3

16

e−x/2−y/3dx=13

e−y/3

(−e−x/2

∞

0

)=

13

e−y/3.

ThereforefY(y) =

13

e−y/3 y > 0

0 y≤ 0

(e) Yes, f (x,y) = fX(x) fY(y).

12. (a)A andB are mutually exclusive ifP(A∩B) = 0, or equivalently, ifP(A∪B) = P(A)+P(B).

So if P(B) = P(A∪B)−P(A) = 0.7−0.3= 0.4, thenA andB are mutually exclusive.



(b) A andB are independent ifP(A∩B) = P(A)P(B). Now P(A∪B) = P(A)+P(B)−P(A∩B).

SoA andB are independent ifP(A∪B) = P(A)+P(B)−P(A)P(B), that is, if 0.7= 0.3+P(B)−0.3P(B). Thisequation is satisfied ifP(B) = 4/7.

13. Let D denote the event that a snowboard is defective, letE denote the event that a snowboard is made inthe eastern United States, letW denote the event that a snowboard is made in the western United States, andlet C denote the event that a snowboard is made in Canada. ThenP(E) = P(W) = 10/28, P(C) = 8/28,P(D|E) = 3/100,P(D|W) = 6/100, andP(D|C) = 4/100.

(a) P(D) = P(D|E)P(E)+P(D|W)P(W)+P(D|C)P(C)

=

(1028

)(3

100

)+

(1028

)(6

100

)+

(828

)(4

100

)

=1222800

= 0.0436

(b) P(D∩C) = P(D|C)P(C) =

(828

)(4

100

)=

322800

= 0.0114

(c) LetU be the event that a snowboard was made in the United States.

ThenP(D∩U) = P(D)−P(D∩C) =1222800

− 322800

=90

2800.

P(U |D) =P(D∩U)

P(D)=

90/2800122/2800

=90122

= 0.7377.

14. (a) Discrete. The possible values are 10, 60, and 80.

(b) µX = 10pX(10)+60pX(60)+80pX(80) = 10(0.40)+60(0.50)+80(0.10)= 42

(c) σ2X = 102pX(10)+602pX(60)+802pX(80)−µ2

X = 102(0.40)+602(0.50)+802(0.10)−422 = 716

σX =√

716= 26.76

(d) P(X > 50) = P(X = 60)+P(X = 80) = 0.5+0.1= 0.6


94 CHAPTER 2

15. The total number of pairs of cubicles is(6

2

)= 6!

2!4! = 15. Each is equally likely to be chosen. Of these pairs,five are adjacent (1 and 2, 2 and 3, 3 and 4, 4 and 5, 5 and 6). Therefore the probability that an adjacent pair ofcubicles is selected is 5/15, or 1/3.

16. The total number of combinations of four shoes that can beselected from eight is(8

4

)= 8!

4!4! = 70. The fourshoes will contain no pair if exactly one shoe is selected from each pair. Since each pair contains two shoes,the number of ways to select exactly one shoe from each pair is24 = 16. Therefore the probability that the fourshoes contain no pair is 16/70, or 8/35.

17. (a)µ3X = 3µX = 3(2) = 6, σ23X = 32σ2

X = (32)(12) = 9

(b) µX+Y = µX +µY = 2+2= 4, σ2X+Y = σ2

X + σ2Y = 12 +32 = 10

(c) µX−Y = µX −µY = 2−2= 0, σ2X−Y = σ2

X + σ2Y = 12 +32 = 10

(d) µ2X+6Y = 2µX +6µY = 2(2)+6(2) = 16, σ22X+6Y = 22σ2

X +62σ2Y = (22)(12)+ (62)(32) = 328

18. Cov(X,Y) = ρX,YσXσY = (0.5)(2)(1) = 1

(a) µX+Y = µX +µY = 1+3= 4, σ2X+Y = σ2

X + σ2Y +2Cov(X,Y) = 22 +12+2(1) = 7

(b) µX−Y = µX −µY = 1−3= −2, σ2X−Y = σ2

X + σ2Y −2Cov(X,Y) = 22 +12−2(1) = 3

(c) µ3X+2Y = 3µX +2µY = 3(1)+2(3) = 9,

σ23X+2Y = 32σ2

X +22σ2Y +2(3)(2)Cov(X,Y) = (32)(22)+ (22)(12)+2(3)(2)(1) = 52

(d) µ5Y−2X = 5µY −2µX = 5(3)−2(1) = 13,

σ25Y−2X = 52σ2

Y +(−2)2σ2X +2(5)(−2)Cov(X,Y) = 52(12)+ (−2)2(22)+2(5)(−2)(1) = 21



19. The marginal probability mass functionpX(x) is found by summing along the rows of the joint probabilitymass function.

yx 100 150 200 pX(x)

0.02 0.05 0.06 0.11 0.220.04 0.01 0.08 0.10 0.190.06 0.04 0.08 0.17 0.290.08 0.04 0.14 0.12 0.30pY(y) 0.14 0.36 0.50

(a) For additive concentration(X): pX(0.02) = 0.22, pX(0.04) = 0.19, pX(0.06) = 0.29, pX(0.08) = 0.30, andpX(x) = 0 for x 6= 0.02, 0.04, 0.06, or 0.08.

For tensile strength(Y): The marginal probability mass functionpY(y) is found by summing down the columnsof the joint probability mass function. ThereforepY(100) = 0.14, pY(150) = 0.36, pY(200) = 0.50, andpY(y) = 0 for y 6= 100, 150, or 200.

(b) No,X andY are not independent. For exampleP(X = 0.02∩Y = 100) = 0.05, butP(X = 0.02)P(Y = 100) = (0.22)(0.14) = 0.0308.

(c) P(Y ≥ 150|X = 0.04) =P(Y ≥ 150 andX = 0.04)

P(X = 0.04)

=P(Y = 150 andX = 0.04)+P(Y = 200 andX = 0.04)

P(X = 0.04)

=0.08+0.10

0.19= 0.947

(d) P(Y > 125|X = 0.08) =P(Y > 125 andX = 0.08)

P(X = 0.08)

=P(Y = 150 andX = 0.08)+P(Y = 200 andX = 0.08)

P(X = 0.08)

=0.14+0.12

0.30= 0.867

(e) The tensile strength is greater than 175 ifY = 200. Now

P(Y = 200|X = 0.02) =P(Y = 200 andX = 0.02)

P(X = 0.02)=

0.110.22

= 0.500,


96 CHAPTER 2

P(Y = 200|X = 0.04) =P(Y = 200 andX = 0.04)

P(X = 0.04)=

0.100.19

= 0.526,

P(Y = 200|X = 0.06) =P(Y = 200 andX = 0.06)

P(X = 0.06)=

0.170.29

= 0.586,

P(Y = 200|X = 0.08) =P(Y = 200 andX = 0.08)

P(X = 0.08)=

0.120.30

= 0.400.

The additive concentration should be 0.06.

20. (a)µX = 0.02pX(0.02)+0.04pX(0.04)+0.06pX(0.06)+0.08pX(0.08)

= 0.02(0.22)+0.04(0.19)+0.06(0.29)+0.08(0.30)

= 0.0534

(b) µY = 100pY(100)+150pY(150)+200pY(200) = 100(0.14)+150(0.36)+200(0.50)= 168

(c) σ2X = 0.022pX(0.02)+0.042pX(0.04)+0.062pX(0.06)+0.082pX(0.08)−µ2

X

= 0.022(0.22)+0.042(0.19)+0.062(0.29)+0.082(0.30)−0.05342

= 0.00050444

σX =√

0.00050444= 0.02246

(d) σ2Y = 1002pY(100)+1502pY(150)+2002pY(200)−µ2

Y

= 1002(0.14)+1502(0.36)+2002(0.50)−1682

= 1276

σY =√

1276= 35.721

(e) Cov(X,Y) = µXY−µXµY.

µXY = (0.02)(100)P(X = 0.02 andY = 100)+ (0.02)(150)P(X = 0.02andY = 150)

+ (0.02)(200)P(X = 0.02 andY = 200)+ (0.04)(100)P(X = 0.04 andY = 100)

+ (0.04)(150)P(X = 0.04 andY = 150)+ (0.04)(200)P(X = 0.04 andY = 200)

+ (0.06)(100)P(X = 0.06 andY = 100)+ (0.06)(150)P(X = 0.06 andY = 150)

+ (0.06)(200)P(X = 0.06 andY = 200)+ (0.08)(100)P(X = 0.08 andY = 100)

+ (0.08)(150)P(X = 0.08 andY = 150)+ (0.08)(200)P(X = 0.08 andY = 200)

= (0.02)(100)(0.05)+ (0.02)(150)(0.06)+(0.02)(200)(0.11)+(0.04)(100)(0.01)



+ (0.04)(150)(0.08)+ (0.04)(200)(0.10)+(0.06)(100)(0.04)+(0.06)(150)(0.08)

+ (0.06)(200)(0.17)+ (0.08)(100)(0.04)+(0.08)(150)(0.14)+(0.08)(200)(0.12)

= 8.96

Cov(X,Y) = 8.96− (0.0534)(168)= −0.0112

(f) ρX,Y =Cov(X,Y)

σXσY=

−0.0112(0.02246)(35.721)

= −0.01396

21. (a)pY|X(100|0.06) =p(0.06,100)

pX(0.06)=

0.040.29

=429

= 0.138

pY|X(150|0.06) =p(0.06,150)

pX(0.06)=

0.080.29

=829

= 0.276

pY|X(200|0.06) =p(0.06,200)

pX(0.06)=

0.170.29

=1729

= 0.586

(b) pX|Y(0.02|100) =p(0.02,100)

pY(100)=

0.050.14

=514

= 0.357

pX|Y(0.04|100) =p(0.04,100)

pY(100)=

0.010.14

=114

= 0.071

pX|Y(0.06|100) =p(0.06,100)

pY(100)=

0.040.14

=414

= 0.286

pX|Y(0.08|100) =p(0.08,100)

pY(100)=

0.040.14

=414

= 0.286

(c) E(Y |X = 0.06) = 100pY|X(100|0.06)+150pY|X(150|0.06)+200pY|X(200|0.06)

= 100(4/29)+150(8/29)+200(17/29)

= 172.4

(d) E(X |Y = 100) = 0.02pX|Y(0.02|100)+0.04pX|Y(0.04|100)+0.06pX|Y(0.06|100)+0.08pX|Y(0.08|100)

= 0.02(5/14)+0.04(1/14)+0.06(4/14)+0.08(4/14)

= 0.0500


98 CHAPTER 2

22. LetD denote the event that an item is defective, letS1 denote the event that an item is produced on the first shift,let S2 denote the event that an item is produced on the second shift,and letS3 denote the event that an item isproduced on the third shift. ThenP(S1) = 0.50,P(S2) = 0.30,P(S3) = 0.20,P(D|S1) = 0.01,P(D|S2) = 0.02,andP(D|S3) = 0.03.

(a) P(S1|D) =P(D|S1)P(S1)

P(D|S1)P(S1)+P(D|S2)P(S2)+P(D|S3)P(S3)

=(0.01)(0.50)

(0.01)(0.50)+ (0.02)(0.30)+(0.03)(0.20)= 0.294

(b) P(S3|Dc) =P(Dc|S3)P(S3)

P(Dc|S1)P(S1)+P(Dc|S2)P(S2)+P(Dc|S3)P(S3)

=[1−P(D|S3)]P(S3)

[1−P(D|S1)]P(S1)+ [1−P(D|S2)]P(S2)+ [1−P(D|S3)]P(S3)

=(1−0.03)(0.20)

(1−0.01)(0.50)+ (1−0.02)(0.30)+(1−0.03)(0.20)= 0.197

23. (a) Under scenario A:

µ= 0(0.65)+5(0.2)+15(0.1)+25(0.05)= 3.75

σ =√

02(0.65)+52(0.2)+152(0.1)+252(0.05)−3.752 = 6.68

(b) Under scenario B:

µ= 0(0.65)+5(0.24)+15(0.1)+20(0.01)= 2.90

σ =√

02(0.65)+52(0.24)+152(0.1)+202(0.01)−2.902 = 4.91

(c) Under scenario C:

µ= 0(0.65)+2(0.24)+5(0.1)+10(0.01)= 1.08

σ =√

02(0.65)+22(0.24)+52(0.1)+102(0.01)−1.082 = 1.81

(d) LetL denote the loss.

Under scenario A,P(L < 10) = P(L = 0)+P(L = 5) = 0.65+0.2= 0.85.

Under scenario B,P(L < 10) = P(L = 0)+P(L = 5) = 0.65+0.24= 0.89.



Under scenario C,P(L < 10) = P(L = 0)+P(L = 2)+P(L = 5) = 0.65+0.24+0.1= 0.99.

24. LetL denote the loss.

(a) P(A∩L = 5) = P(L = 5|A)P(A) = (0.20)(0.20) = 0.040

(b) P(L = 5) = P(L = 5|A)P(A)+P(L = 5|B)P(B)+P(L = 5|C)P(C)

= (0.20)(0.20)+ (0.30)(0.24)+(0.50)(0.1)

= 0.162

(c) P(A|L = 5) =P(A∩L = 5)

P(L = 5)=

0.0400.162

= 0.247

25. (a)p(0,0) = P(X = 0 andY = 0) =

(310

)(29

)=

115

= 0.0667

p(1,0) = P(X = 1 andY = 0) =

(410

)(39

)+

(310

)(49

)=

415

= 0.2667

p(2,0) = P(X = 2 andY = 0) =

(410

)(39

)=

215

= 0.1333

p(0,1) = P(X = 0 andY = 1) =

(310

)(39

)+

(310

)(39

)=

315

= 0.2000

p(1,1) = P(X = 1 andY = 1) =

(410

)(39

)+

(310

)(49

)=

415

= 0.2667

p(0,2) = P(X = 0 andY = 2) =

(310

)(29

)=

115

= 0.0667

p(x,y) = 0 for all other pairs(x,y).

The joint probability mass function is

yx 0 1 20 0.0667 0.2000 0.06671 0.2667 0.2667 02 0.1333 0 0

(b) The marginal probability density function ofX is:


100 CHAPTER 2

pX(0) = p(0,0)+ p(0,1)+ p(0,2)=115

+315

+115

=13

pX(1) = p(1,0)+ p(1,1) =415

+415

=815

.

pX(2) = p(2,0) =215

µX = 0pX(0)+1pX(1)+2pX(2) = 0

(13

)+1

(815

)+2

(215

)=

1215

= 0.8

(c) The marginal probability density function ofY is:

pY(0) = p(0,0)+ p(1,0)+ p(2,0)=115

+415

+215

=715

pY(1) = p(0,1)+ p(1,1) =315

+415

=715

.

pY(2) = p(0,2) =115

µY = 0pY(0)+1pY(1)+2pY(2) = 0

(715

)+1

(715

)+2

(115

)=

915

= 0.6

(d) σX =√

02pX(0)+12pX(1)+22pX(2)−µ2X

=√

02(1/3)+12(8/15)+22(2/15)− (12/15)2

=√

96/225= 0.6532

(e) σY =√

02pY(0)+12pY(1)+22pY(2)−µ2Y

=√

02(7/15)+12(7/15)+22(1/15)− (9/15)2

=√

84/225= 0.6110

(f) Cov(X,Y) = µXY−µXµY.

µXY = (0)(0)p(0,0)+ (1)(0)p(1,0)+ (2)(0)p(2,0)+(0)(1)p(0,1)+(1)(1)p(1,1)+(0)(2)p(0,2)

= (1)(1)415

=415

Cov(X,Y) =415

−(

1215

)(915

)= − 48

225= −0.2133



(g) ρX,Y =Cov(X,Y)

σXσY=

−48/225√96/225

√84/225

= −0.5345

26. (a) The constantc solvesZ 1

0

Z 1

0c(x+y)2dxdy= 1.

SinceZ 1

0

Z 1

0(x+y)2dxdy=

76

, c =67

.

(b) For 0< x < 1, fX(x) =Z 1

0

67(x+y)2dy=

2(x+y)3

7

1

0

=6x2 +6x+2

7.

Forx≤ 0 orx≥ 1 f (x,y) = 0 for all y, so fX(x) = 0.

ThereforefX(x) =

6x2 +6x+27

0 < x < 1

0 otherwise

(c) fY|X(y|x) =f (x,y)fX(x)

. If x≤ 0 orx≥ 1 fX(x) = 0, so fY|X(y|x) is undefined.

Now assume 0< x < 1. If y≤ 0 ory≥ 1 then f (x,y) = 0 for all x so fY|X(y|x) = 0.

If 0 < y < 1 then

fY|X(y|x) =(6/7)(x+y)2

(6x2 +6x+2)/7=

3(x+y)2

3x2 +3x+1.

ThereforefY|X(y|x) =

3(x+y)2

3x2 +3x+10 < y < 1

0 otherwise

(d) E(Y |X = 0.4) =

Z ∞

−∞y fY|X(y|0.4)dy

=Z 1

0

3y(0.4+y)2

3(0.4)2 +3(0.4)+1dy

=0.24y2+0.8y3+0.75y4

2.68

1

0

= 0.6679

(e) No, fY|X(y|x) 6= fY(y).


102 CHAPTER 2

27. (a)µX =

Z ∞

−∞x fX(x)dx=

Z 1

0x

6x2 +6x+27

dx=114

(3x4 +4x3+2x2)

1

0

=914

= 0.6429

(b) σ2X =

Z 1

0x2 6x2 +6x+2

7dx−µ2

X =17

(6x5

5+

3x4

2+

2x3

3

) 1

0

−(

914

)2

=1992940

= 0.06769

(c) Cov(X,Y) = µXY−µXµY.

µXY =Z 1

0

Z 1

0xy

(67

)(x+y)2dxdy

=

Z 1

0

67

y

(x4

4+

2x3y3

+x2y2

2

1

0

)dy

=Z 1

0

67

(y3

2+

2y2

3+

y4

)dy

=67

(y2

8+

2y3

9+

y4

8

) 1

0

=1742

µX =914

, computed in part (a). To computeµY, note that the joint density is symmetric inx andy, so the

marginal density ofY is the same as that ofX. It follows thatµY = µX =914

.

Cov(X,Y) =1742

−(

914

)(914

)=

−5588

= −0.008503.

(d) Since the marginal density ofY is the same as that ofX, σ2Y = σ2

X =1992940

.

ThereforeρX,Y =Cov(X,Y)

σXσY=

−5/588√199/2940

√199/2940

=−25199

= −0.1256.

28. Since the coin is fair,P(H) = P(T) = 1/2. The tosses are independent.SoP(HTTHH) = P(H)P(T)P(T)P(H)P(H) = (1/2)5 = 1/32, andP(HHHHH) = P(H)P(H)P(H)P(H)P(H) = (1/2)5 = 1/32 as well.



29. (a)pX(0) = 0.6, pX(1) = 0.4, pX(x) = 0 if x 6= 0 or 1.

(b) pY(0) = 0.4, pY(1) = 0.6, pY(y) = 0 if y 6= 0 or 1.

(c) Yes. It is reasonable to assume that knowledge of the outcome of one coin will not help predict the outcome ofthe other.

(d) p(0,0) = pX(0)pY(0) = (0.6)(0.4) = 0.24, p(0,1) = pX(0)pY(1) = (0.6)(0.6) = 0.36,

p(1,0) = pX(1)pY(0) = (0.4)(0.4) = 0.16, p(1,1) = pX(1)pY(1) = (0.4)(0.6) = 0.24,

p(x,y) = 0 for other values of(x,y).

30. The probability mass function ofX is pX(x) = 1/6 for x= 1,2,3,4,5, or 6, andpX(x) = 0 for other values ofx.ThereforeµX = 1(1/6)+2(1/6)+3(1/6)+4(1/6)+5(1/6)+6(1/6)= 3.5. The probability mass function ofY is the same as that ofX, soµY = µX = 3.5. SinceX andY are independent,µXY = µXµY = (3.5)(3.5)= 12.25.

31. (a) The possible values of the pair(X,Y) are the ordered pairs(x,y) where each ofx andy is equal to 1, 2, or 3.There are nine such ordered pairs, and each is equally likely. ThereforepX,Y(x,y) = 1/9 for x = 1,2,3 andy = 1,2,3, andpX,Y(x,y) = 0 for other values of(x,y).

(b) BothX andY are sampled from the numbers1,2,3, with each number being equally likely.

ThereforepX(1) = pX(2) = pX(3) = 1/3, andpX(x) = 0 for other values ofx. pY is the same.

(c) µX = µY = 1(1/3)+2(1/3)+3(1/3)= 2

(d) µXY = ∑3x=1 ∑3

y=1 xypX,Y(x,y) =19

3

∑x=1

3

∑y=1

xy=19(1+2+3)(1+2+3)= 4.

Another way to computeµXY is to note thatX andY are independent, soµXY = µXµY = (2)(2) = 4.

(e) Cov(X, Y) = µXY−µXµY = 4− (2)(2) = 0


104 CHAPTER 2

32. (a) The values ofX andY must both be integers between 1 and 3 inclusive, and may not beequal. There are sixpossible values for the ordered pair(X,Y), specifically(1,2),(1,3),(2,1),(2,3),(3,1),(3,2). Each of these sixordered pairs is equally likely.

ThereforepX,Y(x,y) = 1/6 for (x, y) = (1,2),(1,3),(2,1),(2,3),(3,1),(3,2), and pX,Y(x,y) = 0 for othervalues of(x,y).

(b) The value ofX is chosen from the integers 1, 2, 3 with each integer being equally likely.

ThereforepX(1) = pX(2) = pX(3) = 1/3. The marginal probability mass functionpY is the same.

To see this, compute

pY(1) = pX,Y(2,1)+ pX,Y(3,1) = 1/6+1/6= 1/3

pY(2) = pX,Y(1,2)+ pX,Y(3,2) = 1/6+1/6= 1/3

pY(3) = pX,Y(1,3)+ pX,Y(2,3) = 1/6+1/6= 1/3

(c) µX = µY = 1(1/3)+2(1/3)+3(1/3)= 2

(d) µXY = (1)(2)pX,Y(1,2)+ (1)(3)pX,Y(1,3)+ (2)(1)pX,Y(2,1)

+ (2)(3)pX,Y(2,3)+ (3)(1)pX,Y(3,1)+ (3)(2)pX,Y(3,2)

= [(1)(2)+ (1)(3)+ (2)(1)+ (2)(3)+ (3)(1)+ (3)(2)](1/6)

=113

(e) Cov(X, Y) = µXY−µXµY =113

− (2)(2) = −13

33. (a)µX =R ∞−∞ x f(x)dx. Sincef (x) = 0 for x≤ 0, µX =

R ∞0 x f(x)dx.

(b) µX =R ∞

0 x f(x)dx≥ R ∞k x f(x)dx≥ R ∞

k k f(x)dx= kP(X ≥ k)

(c) µX/k≥ kP(X ≥ k)/k = P(X ≥ k)

(d) µX = µ(Y−µY)2 = σ2Y

(e) P(|Y−µY| ≥ kσY) = P((Y−µY)2 ≥ k2σ2Y) = P(X ≥ k2σ2

Y)

(f) P(|Y−µY| ≥ kσY) = P(X ≥ k2σ2Y) ≤ µX/(k2σ2

Y) = σ2Y/(k2σ2

Y) = 1/k2



34. µA = πµR2. We now findµ2R.

σ2R = µR2 −µ2

R. Substituting, we obtain 1= µR2 −102. ThereforeµR2 = 101, andµA = 101π.

35. (a) If the pooled test is negative, it is the only test performed, soX = 1. If the pooled test is positive, thennadditional tests are carried out, one for each individual, so X = n+1. The possible values ofX are therefore 1andn+1.

(b) The possible values ofX are 1 and 5. NowX = 1 if the pooled test is negative. This occurs if none of theindividuals has the disease. The probability that this occurs is (1− 0.1)4 = 0.6561. ThereforeP(X = 1) =0.6561. It follows thatP(X = 5) = 0.3439.SoµX = 1(0.6561)+5(0.3439)= 2.3756.

(c) The possible values ofX are 1 and 7. NowX = 1 if the pooled test is negative. This occurs if none of theindividuals has the disease. The probability that this occurs is (1−0.2)6 = 0.262144. ThereforeP(X = 1) =0.262144. It follows thatP(X = 7) = 0.737856.SoµX = 1(0.262144)+7(0.737856)= 5.4271.

(d) The possible values ofX are 1 andn+1. NowX = 1 if the pooled test is negative. This occurs if none of theindividuals has the disease. The probability that this occurs is (1− p)n. ThereforeP(X = 1) = (1− p)n. Itfollows thatP(X = n+1) = 1− (1− p)n.SoµX = 1(1− p)n+(n+1)(1− (1− p)n) = n+1−n(1− p)n.

(e) The pooled method is more economical if 11−10(1− p)10< 10. Solving forp yields p < 0.2057.


106 CHAPTER 3

Chapter 3

Section 3.1

1. (ii). The measurements are close together, and thus precise, but they are far from the true value of 100C, andthus not accurate.

2. (a) No. They both measured 90C, so we cannot tell which one is on the average closer to the true value.

(b) Yes, the first one is more precise because its uncertaintyis smaller.

3. (a) True

(b) False

(c) False

(d) True

4. (a) The uncertainty is 0.02(100) = 2 g

(b) The uncertainty is 0.02(50) = 1 g

5. (a) No, we cannot determine the standard deviation of the process from a single measurement.

(b) Yes, the bias can be estimated to be 2 pounds, because the reading is 2 pounds when the true weight is 0.

6. (a) Yes, the uncertainty can be estimated with the standard deviation of the four measurements, which is 1.71.


SECTION 3.2 107

(b) Yes, the bias can be estimated to be 2 pounds, because the reading is 2 pounds when the true weight is 0.

7. (a) Yes, the uncertainty can be estimated with the standard deviation of the five measurements, which is 21.3µg.

(b) No, the bias cannot be estimated, since we do not know the true value.

8. (a) Yes, the uncertainty can be estimated with the standard deviation of the five measurements, which is 0.65µg.

(b) Yes, the bias can be estimated with the difference between the average measurement and the true value. Thisdifference is 26.2µg.

9. We can get a more accurate estimate by subtracting the biasof 26.2µg, obtaining 100.8µg above 1 kg.

10. (a) Yes, the uncertainty can be estimated with the standard deviation of the five measurements, which is 0.0109 ppm.

(b) Yes, the bias can be estimated to be the average measurement minus the true value. This difference is−0.0624 ppm.

11. (a) No, they are in increasing order, which would be highly unusual for a simple random sample.

(b) No, since they are not a simple random sample from a population of possible measurements, we cannot estimatethe uncertainty.

Section 3.2

1. (a)σ3X = 3σX = 3(0.2) = 0.6


108 CHAPTER 3

(b) σX−Y =√

σ2X + σ2

Y =√

0.22+0.42 = 0.45

(c) σ2X+3Y =√

22σ2X +32σ2

Y =√

(22)(0.22)+ (32)(0.42) = 1.3

2. Letn be the number of measurements required. Then 2/√

n = 0.5, son = 16.

3. Letn be the necessary number of measurements. ThenσX = 3/√

n = 1. Solving forn yieldsn = 9.

4. LetA represent the measured area,H the measured height, andV the calculated volume. ThenA = 20,H = 5,andσH = 0.1. The volume isV = AH = 100 cm3.

The uncertainty isσV = 20σH = 20(0.1) = 2.0 cm3.

5. LetX represent the estimated mean annual level of land uplift forthe years 1774–1884, and letY represent theestimated mean annual level of land uplift for the years 1885–1984. ThenX = 4.93,Y = 3.92,σX = 0.23, andσY = 0.19. The difference isX−Y = 1.01 mm.

The uncertainty isσX−Y =√

σ2X + σ2

Y =√

0.232+0.192 = 0.30 mm.

6. Let X represent the measured diameter of the hole, and letY represent the measured diameter of the piston.ThenX = 20.00,Y = 19.90,σX = 0.01, andσY = 0.02. The clearance is 0.5X−0.5Y = 0.05 cm.

The uncertainty isσ0.5X−0.5Y =√

0.52σ2X +(−0.5)2σ2

Y =√

(0.52)(0.012)+ (0.52)(0.022) = 0.011 cm.

7. LetX represent the measured length. ThenX = 72 andσX = 0.1. LetV represent the estimated volume. ThenV = (1.5)(3.5)X = 5.25X = 5.25(72) = 378.0 in3.

The uncertainty isσV = σ5.25X = 5.25σX = 5.25(0.1) = 0.5 in3.

8. T = 1.5, L = 0.559,σL = 0.005

g = (4π2/T2)L = 17.546L = 17.546(0.559)= 9.81 m/s2

σg = σ17.546L = 17.546σL = 17.546(0.005)= 0.088 m/s2


SECTION 3.2 109

9. C = 1.25,L = 1, A = 1.30,σA = 0.05

M = (1/LC)A = 0.8A = 0.8(1.30) = 1.04 cm2/mol

σM = 0.8σA = (0.8)(0.05) = 0.04 cm2/mol

10. (a)µ= 1.49,h = 10,τ = 30.0, στ = 0.1

V = τh/µ= 6.7114τ = 6.7114(30) = 201.34 mm/s

σV = σ6.7114τ = 6.7114στ = 6.7114(0.1) = 0.67 mm/s

(b) σV = 6.7114στ = 0.2, soστ = 0.0298 mm/s

11. (a)Ta = 36,k = 0.025,t = 10,T0 = 72,σT0 = 0.5

T = Ta +(T0−Ta)e−kt = 36+0.7788(T0−36) = 36+0.7788(72−36)= 64.04F

σT = σTa+(T0−Ta)e−kt = σ36+0.7788(T0−36) = σ0.7788(T0−36) = 0.7788σ(T0−36) = 0.7788σT0 = 0.7788(0.5) = 0.39F

(b) T0 = 72,k = 0.025,t = 10,Ta = 36,σTa = 0.5

T = Ta +(T0−Ta)e−kt = Ta +0.7788(72−Ta) = 56.074+0.2212Ta = 56.074+0.2212(36)= 64.04F

σT = σ56.074+0.2212Ta = σ0.2212Ta = 0.2212σTa = 0.2212(0.5) = 0.11F

12. τ0 = 50,w = 1.0, k = 0.29,σk = 0.05

τ = τ0(1−kw) = 50(1−k) = 50−50k= 50−50(0.29)= 35.5 MPa

στ = σ50−50k = σ−50k = σ50k = 50σk = 50(0.05) = 2.5 MPa

13. (a) The uncertainty in the average of 9 measurements is approximately equal tos/√

9 = 0.081/3= 0.027 cm.

(b) The uncertainty in a single measurement is approximately equal tos, which is 0.081 cm.

14. (a) The bias is 2 g and the uncertainty isσ = 3 g.

(b) The bias is 2 g and the uncertainty isσ/√

4 = 1.5 g.


110 CHAPTER 3

(c) The bias is 2 g and the uncertainty isσ/√

400= 0.15 g.

(d) The uncertainty gets smaller; it is proportional to 1/√

n wheren is the number of measurements.

(e) The bias stays the same. The mean of the sample average is the same no matter what the sample size, so thebias is the same as well.

15. (a) LetX = 87.0 denote the average of the eight measurements, lets= 2.0 denote the sample standard deviation,and letσ denote the unknown population standard deviation. The volume is estimated withX. The uncertaintyis σX = σ/

√8≈ s/

√8 = 2.0/

√8 = 0.7 mL.

(b) σX ≈ s/√

16= 2.0/√

16= 0.5 mL

(c) Letn be the required number of measurements. Thenσ/√

n = 0.4.Approximatingσ with s= 2.0 yields 2.0/

√n≈ 0.4, son≈ 25.

16. The average of the five measurements isX = 37.0, and the sample standard deviation iss= 1.28841.

(a) k = X±s/√

5 = 37.0±1.28841/√

5 = 37.0±0.6 N/m.

(b) The uncertainty is approximatelys/√

10= 0.41 N/m.

(c) Letn be the required number of measurements. Thenσ/√

n = 0.3.Approximatingσ with s= 1.28841 yields 1.28841/

√n = 0.3. Solving forn yieldsn = 18.44.

Sincen must be an integer,n = 18 or 19.

(d) Since the second spring is similar to the first, it is reasonable to assume that the uncertainty in a measurementof its spring constant will be similar to that of the first. Therefore the uncertainty in a single measurement isapproximated withs= 1.29 N/m.

17. LetX denote the average of the 10 yields at 65C, and letY denote the average of the 10 yields at 80C. LetsX

denote the sample standard deviation of the 10 yields at 65C, and letsY denote the sample standard deviationof the 10 yields at 80C. ThenX = 70.14,sX = 0.897156,Y = 90.50,sY = 0.794425.

(a) At 65C, the yield isX±sX/√

10= 70.14±0.897156/√

10= 70.14±0.28.

At 80C, the yield isY±sY/√

10= 90.50±0.794425/√

10= 90.50±0.25.


SECTION 3.2 111

(b) The difference is estimated withY−X = 90.50−70.14= 20.36.

The uncertainty isσY−X =√

σ2Y

+ σ2X

=√

0.252+0.282 = 0.38.

18. No, the 8 measurements are not a simple random sample froma single population. Instead, they consist of tworandom samples, each of size 4, from two different populations.

19. (a) LetσX = 0.05 denote the uncertainty in the instrument. ThenσX = σX/√

10= 0.05/√

10= 0.016.

(b) Let σY = 0.02 denote the uncertainty in the instrument. ThenσY = σY/√

5 = 0.02/√

5 = 0.0089.

(c) The uncertainty in12X + 12Y is

σ0.5X+0.5Y =√

0.52σ2X

+0.52σ2Y

=√

0.52(0.052/10)+0.52(0.022/5) = 0.0091.

The uncertainty in1015X + 5

15Y is

σ(2/3)X+(1/3)Y =√

(2/3)2σ2X

+(1/3)2σ2Y

=√

(2/3)2(0.052/10)+ (1/3)2(0.022/5) = 0.011.

The uncertainty in12X + 1

2Y is smaller.

(d) cbest=σ2

Y

σ2X

+ σ2Y

=0.022/5

0.052/10+0.022/5= 0.24.

The minimum uncertainty is√

c2bestσ2

X+(1−cbest)2σ2

Y= 0.0078.

20. (a)σX+Y =√

σ2X

+ σ2Y

=√

σ21/4+ σ2

2/12=√

0.022/4+0.082/12= 0.025.

(b) σX+Y

√σ2

1/n+ σ22/(16−n) =

√0.022/n+0.082/(16−n) =

√0.0004/n+0.0064/(16−n)

(c) The integer value ofn than minimizesσX+Y is n = 3. Therefore measure the first component 3 times and thesecond component 13 times.


112 CHAPTER 3

Section 3.3

1. X = 4.0, σX = 0.4.

(a)dYdX

= 2X = 8, σY =

∣∣∣∣dYdX

∣∣∣∣σX = 3.2

(b)dYdX

=1

2√

X= 0.25, σY =

∣∣∣∣dYdX

∣∣∣∣σX = 0.1

(c)dYdX

=−1X2 = − 1

16, σY =

∣∣∣∣dYdX

∣∣∣∣σX = 0.025

(d)dYdX

=1X

= 0.25, σY =

∣∣∣∣dYdX

∣∣∣∣σX = 0.1

(e)dYdX

= eX = e4 = 54.598, σY =

∣∣∣∣dYdX

∣∣∣∣σX = 21.8

(f)dYdX

= cosX = cos4= −0.653644, σY =

∣∣∣∣dYdX

∣∣∣∣σX = 0.26

2. X = 2.0, σX = 0.2

(a)Y =1X

,dYdX

=−1X2 = −0.25, σY =

∣∣∣∣dYdX

∣∣∣∣σX = 0.05, Y = 0.5±0.05

(b) Y = 0.5X,dYdX

= 0.5, σY =

∣∣∣∣dYdX

∣∣∣∣σX = 0.1, Y = 1.0±0.1.

(c) Y =9

X2 ,dYdX

= − 18X3 = −2.25, σY =

∣∣∣∣dYdX

∣∣∣∣σX = 0.45, Y = 2.25±0.45.

(d) Y =4√X

,dYdX

= −2X−3/2 = −0.707107, σY =

∣∣∣∣dYdX

∣∣∣∣σX = 0.14, Y = 2.83±0.14.


SECTION 3.3 113

3. s= 5, t = 1.01,σt = 0.02,g = 2st−2 = 9.80

dgdt

= −4st−3 = −19.4118 σg =

∣∣∣∣dgdt

∣∣∣∣σt = 0.39

g = 9.80±0.39 m/s2

4. T = 300,σT = 0.4,V = 20.04√

T = 347.10dVdT

= 10.02T−1/2 = 0.578505 σV =

∣∣∣∣dVdT

∣∣∣∣σT = 0.23

V = 347.10±0.23 m/s

5. (a)g = 9.80,L = 0.742,σL = 0.005,T = 2π√

L/g = 2.00709√

L = 1.7289

dTdL

= 1.003545L−1/2 = 1.165024 σT =

∣∣∣∣dTdL

∣∣∣∣σL = 0.0058

T = 1.7289±0.0058 s

(b) L = 0.742,T = 1.73,σT = 0.01,g = 4π2LT−2 = 29.292986T−2 = 9.79

dgdT

= −58.5860T−3 = −11.315 σT =

∣∣∣∣dgdT

∣∣∣∣σT = 0.11

g = 9.79±0.11 m/s2

6. k = 7.57,r = 2.0, σr = 0.1, h = k/r = 7.57/r = 3.79

dkdr

= −7.57r−2 = −1.8925 σh =

∣∣∣∣dkdr

∣∣∣∣σr = 0.19

h = 3.79±0.19 mm

7. (a)g = 9.80,d = 0.15, l = 30.0, h = 5.33,σh = 0.02,F =√

gdh/4l = 0.110680√

h = 0.2555


114 CHAPTER 3

dFdh

= 0.055340h−1/2 = 0.023970 σF =

∣∣∣∣dFdh

∣∣∣∣σh = 0.0005

F = 0.2555±0.0005 m/s

(b) g = 9.80, l = 30.0, h = 5.33,d = 0.15,σd = 0.03,F =√

gdh/4l = 0.659760√

d = 0.2555

dFdh

= 0.329880d−1/2 = 0.851747 σF =

∣∣∣∣dFdd

∣∣∣∣σd = 0.026

F = 0.256±0.026 m/s

Note thatF is given to only three significant digits in the final answer, in order to keep the uncertainty to nomore than two significant digits.

(c) g = 9.80,d = 0.15,h = 5.33, l = 30.00,σl = 0.04,F =√

gdh/4l = 1.399562l−1/2 = 0.2555

dFdl

= −0.699781l−3/2 = −0.00425873 σF =

∣∣∣∣dFdl

∣∣∣∣σl = 0.0002

F = 0.2555±0.0002 m/s

8. θ = 0.70,σθ = 0.02,n = 1/sinθ = 1.55227

dndθ

= − cosθsin2 θ

= 1.84292 σn =

∣∣∣∣dndθ

∣∣∣∣σθ = 0.037

n = 1.55±0.037

9. m= 750,V0 = 500.0,V1 = 813.2, σV0 = σV1 = 0.1, D =m

V1−V0=

750V1−V0

= 2.3946.

Let V = V1−V0. ThenV = 813.2−500.0= 313.2, and

σV = σV1−V0 =√

σ2V1

+ σ2V0

=√

0.12+0.12 = 0.141421.

NowdDdV

= −750V2 = −0.007646, σD =

∣∣∣∣dDdV

∣∣∣∣σV = 0.0011.

D = 2.3946±0.0011 g/mL

10. (a)C0 = 0.1, t = 45,C = 0.0811,σC = 0.0005,k = (1/t)(1/C−1/C0) = (1/45)(1/C)−2/9= 0.051788


SECTION 3.3 115

dkdC

= − 145C2 = −3.37867 σk =

∣∣∣∣dkdC

∣∣∣∣σC = 0.0017

k = 0.0518±0.0017 L/mol·min

(b) C0 = 0.1,C = 0.0750,k = 0.0518,σk = 0.0017,t = (1/k)(1/C−1/C0) = (10/3)(1/k) = 64.350064

dtdk

= − 103k2 = −1242.28 σt =

∣∣∣∣dtdk

∣∣∣∣σk = 2.1

t = 64.4±2.1 minutes

11. (a) The relative uncertainty is 0.1/37.2 = 0.27%

(b) The relative uncertainty is 0.003/8.040 = 0.037%

(c) The relative uncertainty is 37/936 = 4.0%

(d) The relative uncertainty is 0.3/54.8 = 0.5%

12. (a) The absolute uncertainty is 0.003(48.41) = 0.15

(b) The absolute uncertainty is 0.006(991.7) = 6.0

(c) The absolute uncertainty is 0.09(0.011) = 0.001

(d) The absolute uncertainty is 0.01(7.86) = 0.08

13. s= 2.2, t = 0.67,σt = 0.02,g = 2s/t2 = 4.4/t2 = 9.802

lng = ln4.4−2lnt,d lng

dt= −2/t = −2.985, σlng =

∣∣∣∣d lng

dt

∣∣∣∣σt = 0.060

g = 9.802 m/s2 ± 6.0%


116 CHAPTER 3

14. T = 298.4, σT = 0.2,V = 20.04√

T = 346.18

lnV = ln20.04+0.5lnT,d lnVdT

= 0.5/T = 0.001676, σlnV =

∣∣∣∣d lnVdT

∣∣∣∣σT = 0.0003351

V = 346.18 m/s± 0.034%

15. (a)g = 9.80,L = 0.855,σL = 0.005,T = 2π√

L/g = 2.00709√

L = 1.85588

lnT = ln2.00709+0.5lnL,d lnT

dL= 0.5/L = 0.584795, σlnT =

∣∣∣∣d lnT

dL

∣∣∣∣σL = 0.0029

T = 1.856 s± 0.29%

(b) L = 0.855,T = 1.856,σT = 0.005,g = 4π2LT−2 = 33.754047T−2 = 9.799

lng = ln33.754047−2lnT,d lngdT

= −2/T = −1.0776, σlng =

∣∣∣∣d lngdT

∣∣∣∣σT = 0.0054

g = 9.799 m/s2 ± 0.54%

16. k = 7.57,r = 2.5, σr = 0.2, h = k/r = 7.57/r = 3.028

lnh = ln7.57− lnr,d lnhdr

= −1/r = −0.4, σlnh =

∣∣∣∣d lnhdr

∣∣∣∣σr = 0.08

h = 3.028± 8.0%

17. (a)g = 9.80,d = 0.20, l = 35.0, h = 4.51,σh = 0.03,F =√

gdh/4l = 0.118332√

h = 0.2513

lnF = ln0.118332+0.5lnh,d lnF

dh= 0.5/h= 0.110865, σlnF =

∣∣∣∣d lnF

dh

∣∣∣∣σh = 0.0033

F = 0.2513 m/s± 0.33%

(b) g = 9.80, l = 35.0, h = 4.51,d = 0.20,σd = 0.008,F =√

gdh/4l = 0.561872√

d = 0.2513

lnF = ln0.561872+0.5lnd,d lnF

dd= 0.5/d = 2.5, σlnF =

∣∣∣∣d lnF

dd

∣∣∣∣σd = 0.02

F = 0.2513 m/s± 2.0%


SECTION 3.3 117

(c) g = 9.80,d = 0.20,h = 4.51, l = 35.00,σl = 0.4, F =√

gdh/4l = 1.486573l−1/2 = 0.2513

lnF = ln1.486573−0.5lnl ,d lnF

dl= −0.5/l = −0.014286, σlnF =

∣∣∣∣d lnF

dl

∣∣∣∣σl = 0.0057

F = 0.2513 m/s± 0.57%

18. θ = 0.90,σθ = 0.01,n = 1/sinθ = 1.276606

lnn = − ln(sinθ),d lnndθ

= −cosθsinθ

= −0.793551 σlnn =

∣∣∣∣d lnndθ

∣∣∣∣σθ = 0.0079

n = 1.28± 0.79%

19. m= 288.2,V0 = 400.0,V1 = 516.0, σV0 = 0.1, σV1 = 0.2, D =m

V1−V0=

288.2V1−V0

= 2.484

Let V = V1−V0. ThenV = 516.0−400.0= 116.0, and

σV = σV1−V0 =√

σ2V1

+ σ2V0

=√

0.12+0.22 = 0.223607

lnD = ln288.2− lnV,d lnDdV

= −1/V = −0.008621, σlnD =

∣∣∣∣d lnDdV

∣∣∣∣σV = 0.0019.

D = 2.484 g/mL± 0.19%

20. (a)C0 = 0.04,t = 30,C = 0.0038,σC = 0.0002,k = (1/t)(1/C−1/C0) = 1/(30C)−5/6= 7.94

lnk = ln(1/(30C)−5/6),d lnkdC

= − 1C−25C2 = −290.782, σlnk =

∣∣∣∣d lnkdC

∣∣∣∣σC = 0.058

k = 7.94 L/mol·s± 5.8%

(b) C0 = 0.04,t = 50,C = 0.0024,σC = 0.0002,k = (1/t)(1/C−1/C0) = 1/(50C)−1/2= 7.83

lnk = ln(1/(50C)−1/2),d lnkdC

= − 1C−25C2 = −443.262, σlnk =

∣∣∣∣d lnkdC

∣∣∣∣σC = 0.089

k = 7.83 L/mol·s± 8.9%

(c) lnk = 0.5lnk1 +0.5lnk2,


118 CHAPTER 3

σlnk =√

0.52σ2ln k1

+0.52σ2ln k2

=√

0.52(0.058)2+0.52(0.089)2 = 0.053.

The relative uncertainty is 5.3%.

Section 3.4

1. (a)X = 10.0, σX = 0.5,Y = 5.0, σY = 0.1,U = XY2 = 250

∂U∂X

= Y2 = 25.0,∂U∂Y

= 2XY = 100.0, σU =

√(∂U∂X

)2

σ2X +

(∂U∂Y

)2

σ2Y = 16.0

U = 250±16

(b) X = 10.0, σX = 0.5,Y = 5.0, σY = 0.1,U = X2 +Y2 = 125

∂U∂X

= 2X = 20.0,∂U∂Y

= 2Y = 10.0, σU =

√(∂U∂X

)2

σ2X +

(∂U∂Y

)2

σ2Y = 10.0

U = 125±10

(c) X = 10.0, σX = 0.5,Y = 5.0, σY = 0.1,U = (X +Y2)/2 = 17.50

∂U∂X

= 1/2,∂U∂Y

= Y = 5.0, σU =

√(∂U∂X

)2

σ2X +

(∂U∂Y

)2

σ2Y = 0.56

U = 17.50±0.56

2. (a)µ= 1.49,τ = 30.0, στ = 0.1, h = 10.0, σh = 0.2,V = τh/µ = τh/1.49= 201.3

∂V∂τ

= h/1.49= 6.71141,∂V∂h

= τ/1.49= 20.1342, σV =

√(∂V∂τ

)2

σ2τ +

(∂V∂h

)2

σ2h = 4.1

V = 201.3±4.1 mm/s

(b) σV =

√(∂V∂τ

)2

σ2τ +

(∂V∂h

)2

σ2h ,

∂V∂τ

= 6.71141,∂V∂h

= 20.1342

If στ = 0.01 andσh = 0.2, thenσV = 4.0.

If στ = 0.1 andσh = 0.1, thenσV = 2.1.

Reducing the uncertainty inh to 0.1 mm provides the greater reduction.


SECTION 3.4 119

3. (a)P1 = 10.1, σP1 = 0.3, P2 = 20.1, σP2 = 0.4, P3 =√

P1P2 = 14.25

∂P3

∂P1= 0.5

√P2/P1 = 0.705354

∂P3

∂P2= 0.5

√P1/P2 = 0.354432

σP3 =

√(∂P3

∂P1

)2

σ2P1

+

(∂P3

∂P2

)2

σ2P2

= 0.25

P3 = 14.25±0.25 MPa

(b) σP3 =

√(∂P3

∂P1

)2

σ2P1

+

(∂P3

∂P2

)2

σ2P2

,∂P3

∂P1= 0.705354,

∂P3

∂P2= 0.354432

If σP1 = 0.2 andσP2 = 0.4, thenσP3 = 0.20.

If σP1 = 0.3 andσP2 = 0.2, thenσP3 = 0.22.

Reducing the uncertainty inP1 to 0.2 MPa provides the greater reduction.

4. (a)M1 = 1.32,σM1 = 0.01,M2 = 1.04,σM2 = 0.01,W = (M1−M2)/M1 = 0.2121

∂W∂M1

= M2/M21 = 0.596878

∂W∂M2

= −1/M1 = −0.757576

σW =

√(∂W∂M1

)2

σ2M1

+

(∂W∂M2

)2

σ2M2

= 0.0096

W = 0.2121±0.0096

(b) σW =

√(∂W∂M1

)2

σ2M1

+

(∂W∂M2

)2

σ2M2

,∂W∂M1

= 0.596878,∂W∂M2

= −0.757576

If σM1 = 0.005 andσM2 = 0.01, thenσW = 0.0081.

If σM1 = 0.01 andσM2 = 0.005, thenσW = 0.0071.

Reducing the uncertainty inM2 to 0.005 kg provides the greater reduction.

5. (a) p = 2.3, σp = 0.2, q = 3.1, σq = 0.2, f = pq/(p+q) = 1.320

∂ f∂p

= q2/(p+q)2 = 0.329561

∂ f∂q

= p2/(p+q)2 = 0.181413


120 CHAPTER 3

σ f =

√(∂ f∂p

)2

σ2p +

(∂ f∂q

)2

σ2q = 0.075

f = 1.320±0.075 cm

(b) σ f =

√(∂ f∂p

)2

σ2p +

(∂ f∂q

)2

σ2q ,

∂ f∂p

= 0.329561,∂ f∂q

= 0.181413

If σp = 0.1 andσq = 0.2, thenσ f = 0.049.

If σp = 0.2 andσq = 0.1, thenσ f = 0.068.

Reducing the uncertainty inp to 0.1 cm provides the greater reduction.

6. (a)P = 242.52, σP = 0.03,V = 10.103, σV = 0.002,T = PV/8.31= 294.847

∂T∂P

= V/8.31= 1.21576

∂T∂V

= P/8.31= 29.1841

σT =

√(∂T∂P

)2

σ2P +

(∂T∂V

)2

σ2V = 0.069

T = 294.847±0.069 K

(b) P = 242.52, σP = 0.03,T = 290.11, σT = 0.02,V = 8.31T/P = 9.9407

∂V∂P

= −8.31T/P2 = −0.0409891

∂V∂T

= 8.31/P= 0.0342652

σV =

√(∂V∂P

)2

σ2P +

(∂V∂T

)2

σ2T = 0.0014

V = 9.9407±0.0014 L

(c) V = 10.103, σV = 0.002,T = 290.11, σT = 0.02,P = 8.31T/V = 238.624

∂P∂V

= −8.31T/V2 = −23.6191

∂P∂T

= 8.31/V = 0.822528

σV =

√(∂V∂P

)2

σ2P +

(∂V∂T

)2

σ2T = 0.050

P = 238.624±0.050 kPa


SECTION 3.4 121

7. g = 9.80,d = 0.18, σd = 0.02,h = 4.86, σh = 0.06, l = 32.04, σl = 0.01,

F =√

gdh/4l = 1.565248√

dh/l = 0.259

∂F∂d

= 0.782624√

h/dl = 0.718437

∂F∂h

= 0.782624√

d/hl = 0.0266088

∂F∂l

= −0.782624√

dh/l3 = 0.00403616

σF =

√(∂F∂d

)2

σ2d +

(∂F∂h

)2

σ2h +

(∂F∂l

)2

σ2l = 0.014

F = 0.259±0.014 m/s

8. Let I1 be the estimated index at a temperature of 166 K, letI2 be the estimated index at a temperature of 196 K,and letR= I1/I2 be the ratio. ThenI1 = 0.00116, σI1 = 0.0001,I2 = 0.00129, σI2 = 0.0001.

∂R∂I1

= 1/I2 = 775.194

∂R∂I2

= −I1/I22 = −697.073

σR =

√(∂R∂I1

)2

σ2I1

+

(∂R∂I2

)2

σ2I2

= 0.10

9. (a)τ0 = 50, στ0 = 1, w = 1.2, σw = 0.1, k = 0.29, σk = 0.05,

τ = τ0(1−kw) = 32.6

∂τ∂τ0

= 1−kw= 0.652

∂τ∂k

= −τ0w = −60

∂τ∂w

= −τ0k = −14.5

στ =

√(∂τ∂τ0

)2

σ2τ0

+

(∂τ∂k

)2

σ2k +

(∂τ∂w

)2

σ2w = 3.4

τ = 32.6±3.4 MPa

(b) στ =

√(∂τ∂τ0

)2

σ2τ0

+

(∂τ∂k

)2

σ2k +

(∂τ∂w

)2

σ2w ,


122 CHAPTER 3

∂τ∂τ0

= 1−kw= 0.652,∂τ∂k

= −τ0w = −60,∂τ∂w

= −τ0k = −14.5

If στ0 = 0.1, σk = 0.05, andσw = 0.1, thenστ = 3.3.



Reducing the uncertainty ink to 0.025 mm−1 provides the greatest reduction.

(c) If στ0 = 0, σk = 0.05, andσw = 0, στ = 3.0. Thus implementing the procedure would reduce the uncertaintyin τ only to 3.0 MPa. It is probably not worthwhile to implement the new procedure for a reduction this small.

10. θ1 = 0.3672, σθ1 = 0.005,θ2 = 0.2943, σθ2 = 0.004,n =sinθ1

sinθ2= 1.238

∂n∂θ1

=cosθ1

sinθ2= 3.21762

∂n∂θ2

= −sinθ1cosθ2

sin2 θ2= −4.08326

σn =

√(∂n∂θ1

)2

σ2θ1

+

(∂n∂θ2

)2

σ2θ2

= 0.023

n = 1.238±0.023

11. (a)g = 3867.4, σg = 0.3, b = 1037.0, σb = 0.2, m= 2650.4, σm = 0.1, y = mb/g = 710.68.

∂y∂g

= −mb/g2 = −0.18376

∂y∂b

= m/g = 0.685318

∂y∂m

= b/g = 0.268139

σy =

√(∂y∂g

)2

σ2g +

(∂y∂m

)2

σ2m+

(∂y∂b

)2

σ2b = 0.15

y = 710.68±0.15 g

(b) σy =

√(∂y∂g

)2

σ2g +

(∂y∂m

)2

σ2m+

(∂y∂b

)2

σ2b ,


SECTION 3.4 123

∂y∂g

= −mb/g2 = −0.18376,∂y∂b

= m/g = 0.685318,∂y∂m

= b/g = 0.268139

If σg = 0.1, σb = 0.2, andσm = 0.1, σy = 0.14.

If σg = 0.3, σb = 0.1, andσm = 0.1, σy = 0.09.

If σg = 0.3, σb = 0.2, andσm = 0, σy = 0.15.

Reducing the uncertainty inb to 0.1 g provides the greatest reduction.

12. (a)l = 14.0, σl = 0.1, d = 4.4, σd = 0.1, R= kl/d2 = 0.723k

∂R∂l

= k/d2 = 0.0516529k

∂R∂d

= −2kl/d3 = −0.3287k

σR =

√(∂R∂l

)2

σ2l +

(∂R∂d

)2

σ2d = 0.033k

R= 0.723k±0.033k Ω

(b) σR =

√(∂R∂l

)2

σ2l +

(∂R∂d

)2

σ2d,

∂R∂l

= k/d2 = 0.0516529k,∂R∂d

= −2kl/d3 = −0.3287k

If σl = 0.05 andσd = 0.1, σR = 0.033k.

If σl = 0.1 andσd = 0.05,σR = 0.017k.

Reducing the uncertainty ind to 0.05 cm provides the greater reduction.

13. (a)F = 800, σF = 1, R= 0.75, σR = 0.1, L0 = 25.0, σL0 = 0.1, L1 = 30.0, σL1 = 0.1,

Y =FL0

πR2(L1−L0)= 2264

∂Y∂F

=L0

πR2(L1−L0)= 2.82942

∂Y∂R

=−2FL0

πR3(L1−L0)= −6036.1

∂Y∂L0

=FL1

πR2(L1−L0)2 = 543.249

∂Y∂L1

=−FL0

πR2(L1−L0)2 = −452.707

σY =

√(∂Y∂F

)2

σ2F +

(∂Y∂R

)2

σ2R+

(∂Y∂L0

)2

σ2L0

+

(∂Y∂L1

)2

σ2L1

= 608

Y = 2264±608 N/mm2


124 CHAPTER 3

(b) σY =

√(∂Y∂F

)2

σ2F +

(∂Y∂R

)2

σ2R+

(∂Y∂L0

)2

σ2L0

+

(∂Y∂L1

)2

σ2L1

,

∂Y∂F

=L0

πR2(L1−L0)= 2.82942,

∂Y∂R

=−2FL0

πR3(L1−L0)= −6036.1,

∂Y∂L0

=FL1

πR2(L1−L0)2 = 543.249,∂Y∂L1

=−FL0

πR2(L1−L0)2 = −452.707

If σF = 0, σR = 0.1, σL0 = 0.1, andσL1 = 0.1, thenσY = 608.

If σF = 1, σR = 0, σL0 = 0.1, andσL1 = 0.1, thenσY = 71.

If σF = 1, σR = 0.1, σL0 = 0, andσL1 = 0.1, thenσY = 605.

If σF = 1, σR = 0.1, σL0 = 0.1, andσL1 = 0, thenσY = 606.

R is the only variable that substantially affects the uncertainty in Y.

14. T = 50,k = 0.025,T0 = 70.1, σT0 = 0.2, Ta = 35.7, σTa = 0.1,

t = ln(T0−Ta)/k− ln(T −Ta)/k = 40ln(T0−Ta)−40ln(50−Ta) = 35.11

∂t∂T0

= 40/(T0−Ta) = 1.16279

∂t∂Ta

= 40/(50−Ta)−40/(T0−Ta) = 1.63441

σt =

√(∂t

∂T0

)2

σ2T0

+

(∂t

∂Ta

)2

σ2Ta

= 0.28

t = 35.11±0.28 min

15. t = 10,T = 54.1, σT = 0.2, T0 = 70.1, σT0 = 0.2, Ta = 35.7, σTa = 0.1,

k = ln(T0−Ta)/t − ln(T −Ta)/t = 0.1ln(T0−Ta)−0.1ln(T −Ta) = 0.0626

∂k∂T

=−1

10(T −Ta)= −0.00543478

∂k∂T0

=1

10(T0−Ta)= 0.00290698

∂k∂Ta

=1

10(T −Ta)− 1

10(T0−Ta)= 0.00252781

σk =

√(∂k∂T

)2

σ2T +

(∂k∂T0

)2

σ2T0

+

(∂k∂Ta

)2

σ2Ta

= 0.0013

k = 0.0626±0.0013 min−1


SECTION 3.4 125

16. (a)a = 2.5, σa = 0.1, b = 0.05, σb = 0.01,w = 1.2, σw = 0.1, v = a+bw= 2.56

∂v∂a

= 1

∂v∂b

= w = 1.2

∂v∂w

= b = 0.05

σv =

√(∂v∂a

)2

σ2a +

(∂v∂b

)2

σ2b +

(∂v∂w

)2

σ2w = 0.10

v = 2.56±0.10 mm

(b) σv =

√(∂v∂a

)2

σ2a +

(∂v∂b

)2

σ2b +

(∂v∂w

)2

σ2w,

∂v∂a

= 1,∂v∂b

= w = 1.2∂v∂w

= b = 0.05

If σa = 0, σb = 0.01, andσw = 0.1, thenσv = 0.013.

If σa = 0.1, σb = 0, andσw = 0.1, thenσv = 0.10.

If σa = 0.1, σb = 0.01, andσw = 0, thenσv = 0.10.

a is the only variable that substantially affects the uncertainty in v.

17. (a) No, they both involve the quantitiesh andr.

(b) r = 0.9, σr = 0.1, h = 1.7, σh = 0.1, R= c2πr(h+2r)

πr2(h+4r/3)= c

2h+4rhr +4r2/3

= 2.68c

∂R∂r

= −c2h2+16rh/3+16r2/3

(4r2/3+ rh)2 = −2.68052c

∂R∂h

= −c4r2/3

(4r2/3+ rh)2 = −0.158541c

σR =

√(∂R∂r

)2

σ2r +

(∂R∂h

)2

σ2h = 0.27c

R= 2.68c±0.27c

18. X = 5.0, σX = 0.2,Y = 10.0, σY = 0.5

(a)U = X√

Y = 15.8, lnU = lnX +0.5lnY

∂ lnU∂X

=1X

= 0.2


126 CHAPTER 3

∂ lnU∂Y

=1

2Y= 0.05

σlnU =

√(∂ lnU

∂X

)2

σ2X +

(∂ lnU

∂Y

)2

σ2Y = 0.047

U = 15.8±4.7%

(b) U = 2Y/√

X = 8.94, lnU = ln2−0.5lnX + lnY

∂ lnU∂X

=−12X

= −0.1

∂ lnU∂Y

=1Y

= 0.1

σlnU =

√(∂ lnU

∂X

)2

σ2X +

(∂ lnU

∂Y

)2

σ2Y = 0.054

U = 8.94±5.4%

(c) U = X2 +Y2 = 125, lnU = ln(X2 +Y2)

∂ lnU∂X

=2X

X2 +Y2 = 0.08

∂ lnU∂Y

=2Y

X2 +Y2 = 0.16

σlnU =

√(∂ lnU

∂X

)2

σ2X +

(∂ lnU

∂Y

)2

σ2Y = 0.082

U = 125±8.2%

Alternatively (and more easily),U = X2 +Y2 = 125,

∂U∂X

= 2X = 10

∂U∂Y

= 2Y = 20

σU =

√(∂U∂X

)2

σ2X +

(∂U∂Y

)2

σ2Y = 10.198

The relative uncertainty isσU/U = 10.198/125= 0.082.

U = 125±8.2%

19. µ= 1.49,τ = 35.2, στ = 0.1, h = 12.0, σh = 0.3,V = τh/µ= τh/1.49= 283.49,

lnV = lnτ+ lnh− ln1.49∂ lnV

∂τ= 1/τ = 0.028409


SECTION 3.4 127

∂ lnV∂h

= 1/h = 0.083333

σlnV =

√(∂ lnV

∂τ

)2

σ2τ +

(∂ lnV

∂h

)2

σ2h = 0.025

V = 283.49 mm/s± 2.5%

20. P1 = 15.3, σP1 = 0.2, P2 = 25.8, σP2 = 0.1, P3 =√

P1P2 = 19.87, lnP3 = 0.5lnP1 +0.5lnP2

∂ lnP3

∂P1=

12P1

= 0.0326797

∂ lnP3

∂P2=

12P2

= 0.0193798

σlnP3 =

√(∂ lnP3

∂P1

)2

σ2P1

+

(∂ lnP3

∂P2

)2

σ2P2

= 0.0068

P3 = 19.87 MPa± 0.68%

21. p = 4.3, σp = 0.1, q = 2.1, σq = 0.2, f = pq/(p+q) = 1.41, ln f = ln p+ lnq− ln(p+q)

∂ ln f∂p

= 1/p−1/(p+q)= 0.0763081

∂ ln f∂q

= 1/q−1/(p+q)= 0.31994

σln f =

√(∂ ln f

∂p

)2

σ2p +

(∂ ln f

∂q

)2

σ2q = 0.064

f = 1.41 cm± 6.4%

22. (a)P = 224.51, σP = 0.04,V = 11.237, σV = 0.002,T = PV/8.31= 303.59, lnT = lnP+ lnV − ln8.31

∂ lnT∂P

= 1/P= 0.00445414

∂ lnT∂V

= 1/V = 0.0889917

σlnT =

√(∂ lnT

∂P

)2

σ2P +

(∂ lnT∂V

)2

σ2V = 0.00025

T = 303.59 K ± 0.025%


128 CHAPTER 3

(b) P = 224.51, σP = 0.04,T = 289.33, σT = 0.02,V = 8.31T/P = 10.71, lnV = ln8.31+ lnT − lnP

∂ lnV∂P

= −1/P= −0.00445414

∂ lnV∂T

= 1/T = 0.00345626

σlnV =

√(∂ lnV

∂P

)2

σ2P +

(∂ lnV

∂T

)2

σ2T = 0.00019

V = 10.71 L± 0.019%

(c) V = 11.203, σV = 0.002,T = 289.33, σT = 0.02,P = 8.31T/V = 214.62, lnV = ln8.31+ lnT − lnV

∂ lnP∂V

= −1/V = −0.0892618

∂ lnP∂T

= 1/T = 0.00345626

σlnP =

√(∂ lnP∂V

)2

σ2V +

(∂ lnP∂T

)2

σ2T = 0.00019

P = 214.62 kPa± 0.019%

23. θ1 = 0.216, σθ1 = 0.003,θ2 = 0.456, σθ2 = 0.005,n =sinθ1

sinθ2= 0.487, lnn = ln(sinθ1)− ln(sinθ2)

∂ lnn∂θ1

= cotθ1 = 4.5574

∂ lnn∂θ2

= −cotθ2 = −2.03883

σlnn =

√(∂ lnn∂θ1

)2

σ2θ1

+

(∂ lnn∂θ2

)2

σ2θ2

= 0.017

n = 0.487±1.7%

24. (a)l = 10.0, σl /l = 0.005,d = 10.4, σd/d = 0.005,R= kl/d2 = 0.0925k, lnR= lnk+ ln l −2lnd

∂ lnR∂l

= 1/l ,∂ lnR

∂d= −2/d

σlnR =

√(∂ lnR

∂l

)2

σ2l +

(∂ lnR

∂d

)2

σ2d =

√(σl

l

)2+4(σd

d

)2= 0.011

R= 0.0925k Ω±1.1%


SECTION 3.4 129

The relative uncertainty does not depend onk.

(b) σlnR =

√(σl

l

)2+4(σd

d

)2

If σl/l = 0.002 andσd/d = 0.005, thenσlnR = 1.0%.

If σl/l = 0.005 andσd/d = 0.002, thenσlnR = 0.64%.

d should be remeasured so as to provide the greater improvement in the relative uncertainty of the resistance.

25. F = 750, σF = 1, R= 0.65, σR = 0.09,L0 = 23.7, σL0 = 0.2, L1 = 27.7, σL1 = 0.2,

Y =FL0

πR2(L1−L0)= 3347.9, lnY = lnF + lnL0− lnπ−2lnR− ln(L1−L0)

∂ lnY∂F

= 1/F = 0.001333

∂ lnY∂L0

= 1/L0 +1/(L1−L0) = 0.292194

∂ lnY∂R

= −2/R= −3.07692

∂ lnY∂L1

= −1/(L1−L0) = −0.25

σlnY =

√(∂ lnY∂F

)2

σ2F +

(∂ lnY∂L0

)2

σ2L0

+

(∂ lnY

∂R

)2

σ2R+

(∂ lnY∂L1

)2

σ2L1

= 0.29

Y = 3347.9 N/mm2 ± 29%

26. T = 50,k = 0.032,T0 = 73.1, σT0 = 0.1, Ta = 37.5, σTa = 0.2,

t = ln(T0−Ta)/k− ln(T −Ta)/k = 31.25ln(T0−Ta)−31.25ln(50−Ta) = 32.71

It is easier to computeσt and then divide byt than to computeσlnt .

∂t∂T0

=31.25

T0−Ta= 0.877809

∂t∂Ta

=31.25

50−Ta− 31.25

T0−Ta= 1.62219

σt =

√(∂t

∂T0

)2

σ2T0

+

(∂t

∂Ta

)2

σ2Ta

= 0.3361

σt

t= 0.3361/32.71= 0.010

t = 32.7 min± 1.0%


130 CHAPTER 3

27. r = 0.8, σr = 0.1, h = 1.9, σh = 0.1

(a) S= 2πr(h+2r) = 17.59, lnS= ln(2π)+ lnr + ln(h+2r)

∂ lnS∂r

=1r

+2

h+2r= 1.82143

∂ lnS∂h

=1

h+2r= 0.285714

σlnS =

√(∂ lnS

∂r

)2

σ2r +

(∂ lnS

∂h

)2

σ2h = 0.18

S= 17.59µm± 18%

(b) V = πr2(h+4r/3) = 5.965, lnV = ln(π)+2lnr + ln(h+4r/3)

∂ lnV∂r

=2r

+4

3h+4r= 2.94944

∂ lnV∂h

=3

3h+4r= 0.337079

σlnV =

√(∂ lnV

∂r

)2

σ2r +

(∂ lnV

∂h

)2

σ2h = 0.30

V = 5.965µm3 ± 30%

(c) R= c2πr(h+2r)

πr2(h+4r/3)= c

2h+4rrh+4r2/3

= 2.95c, lnR= lnc+ ln(2h+4r)− ln(rh+4r2/3)

∂ lnR∂r

=2

2r +h− 8r +3h

4r2 +3rh= −1.12801

∂ lnR∂h

=1

2r +h− 3

4r +3h= −0.0513644

σlnR =

√(∂ lnR

∂r

)2

σ2r +

(∂ lnR

∂h

)2

σ2h = 0.11

R= 2.95c±11%

Note that the uncertainty inR cannot be determined directly from the uncertainties inSandV, becauseSandV are not independent.

(d) No

28. P3 = P0.51 P0.5

2 . The relative uncertainties areσP1

P1= 0.05 and

σP2

P2= 0.02.



σP3

P3=

√(0.5

σP1

P1

)2

+

(0.5

σP2

P2

)2

=√

0.0252+0.012 = 0.027.


29. R= kld−2. The relative uncertainties areσk

k= 0 (sincek is a constant),

σl

l= 0.03, and

σd

d= 0.02.

σR

R=

√(σk

k

)2+(σl

l

)2+(−2

σd

d

)2=√

02 +0.032+(−0.04)2 = 0.05.



1. (a)X = 25.0, σX = 1,Y = 5.0, σY = 0.3, Z = 3.5, σZ = 0.2. LetU = X +YZ.

∂U∂X

= 1,∂U∂Y

= Z = 3.5,∂U∂Z

= Y = 5.0

σU =

√(∂U∂X

)2

σ2X +

(∂U∂Y

)2

σ2Y +

(∂U∂Z

)2

σ2Z = 1.8

(b) X = 25.0, σX = 1,Y = 5.0, σY = 0.3, Z = 3.5, σZ = 0.2. LetU = X/(Y−Z).

∂U∂X

= 1/(Y−Z) = 0.66667,∂U∂Y

= −X/(Y−Z)2 = −11.1111,∂U∂Z

= X/(Y−Z)2 = 11.1111

σU =

√(∂U∂X

)2

σ2X +

(∂U∂Y

)2

σ2Y +

(∂U∂Z

)2

σ2Z = 4.1

(c) X = 25.0, σX = 1,Y = 5.0, σY = 0.3, Z = 3.5, σZ = 0.2. LetU = X√

Y+eZ.

∂U∂X

=√

Y+eZ = 6.17377,∂U∂Y

=X

2√

Y +eZ= 2.02469,

∂U∂Z

=XeZ

2√

Y +eZ= 67.0487

σU =

√(∂U∂X

)2

σ2X +

(∂U∂Y

)2

σ2Y +

(∂U∂Z

)2

σ2Z = 15

(d) X = 25.0, σX = 1,Y = 5.0, σY = 0.3, Z = 3.5, σZ = 0.2. LetU = X ln(Y2 +Z).

∂U∂X

= ln(Y2 +Z) = 3.3499,∂U∂Y

=2XY

Y2 +Z= 8.77193,

∂U∂Z

=X

Y2 +Z= 0.877193


132 CHAPTER 3

σU =

√(∂U∂X

)2

σ2X +

(∂U∂Y

)2

σ2Y +

(∂U∂Z

)2

σ2Z = 4.3

2. The relative uncertainty inX isσX

X= 0.05, the relative uncertainty inY is

σY

Y= 0.10, and the relative

uncertainty inZ isσZ

Z= 0.15.

(a) LetU = XY/Z, so lnU = lnX + lnY− lnZ.

∂ lnU∂X

= 1/X,∂ lnU

∂Y= 1/Y,

∂ lnU∂Z

= −1/Z

σlnU =

√(∂ lnU

∂X

)2

σ2X +

(∂ lnU

∂Y

)2

σ2Y +

(∂ lnU

∂Z

)2

σ2Z

=

√(σX

X

)2+(σY

Y

)2+(σZ

Z

)2

=√

0.052+0.102+0.152

= 0.19

The relative uncertainty inU is 19%.

(b) LetU = X3/2(YZ)1/3, so lnU = (3/2) lnX +(1/3) lnY+(1/3) lnZ.

∂ lnU∂X

=3

2X,

∂ lnU∂Y

=1

3Y,

∂ lnU∂Z

=1

3Z

σlnU =

√(∂ lnU

∂X

)2

σ2X +

(∂ lnU

∂Y

)2

σ2Y +

(∂ lnU

∂Z

)2

σ2Z

=

√(32

)2(σX

X

)2+

(13

)2(σY

Y

)2+

(13

)2(σZ

Z

)2

=√

(3/2)2(0.05)2+(1/3)2(0.10)2+(1/3)2(0.15)2

= 0.096

The relative uncertainty inU is 9.6%.

(c) LetU = X3√

Z/Y, so lnU = 3lnX−0.5lnY+0.5lnZ.

∂ lnU∂X

=3X

,∂ lnU

∂Y= − 1

2Y,

∂ lnU∂Z

=1

2Z



σlnU =

√(∂ lnU

∂X

)2

σ2X +

(∂ lnU

∂Y

)2

σ2Y +

(∂ lnU

∂Z

)2

σ2Z

=

√

32(σX

X

)2+

(12

)2(σY

Y

)2+

(12

)2(σZ

Z

)2

=√

32(0.05)2 +(1/2)2(0.10)2 +(1/2)2(0.15)2

= 0.175

The relative uncertainty inU is 17.5%.

3. (a) LetX andY represent the measured lengths of the components, and letT = X +Y be the combined length.ThenσX = σY = 0.1.

σT =√

σ2X + σ2

Y = 0.14 mm

(b) Letσ be the required uncertainty inX andY. ThenσT =√

σ2 + σ2 = 0.05. Solving forσ yieldsσ = 0.035 mm.

4. t1 = 20,t2 = 40,m1 = 17.7, σm1 = 1.7, m2 = 35.9, σm2 = 5.8,

r = (m2−m1)/(t2− t1) = 0.05m2−0.05m1 = 0.91

(a) σr =√

0.052σ2m1

+0.052σ2m2

= 0.30

r = 0.91±0.30

(b)σr

r= 0.30/0.91= 33%

5. (a)γ = 9800,η = 0.85, ση = 0.02,Q = 60, σQ = 1, H = 3.71, σH = 0.10

P = ηγQH = 9800ηQH = 1.854×106

∂P∂η

= 9800QH = 2.18148×106

∂P∂Q

= 9800ηH = 30904.3


134 CHAPTER 3

∂P∂H

= 9800ηQ= 499800

σP =

√(∂P∂η

)2

σ2η +

(∂P∂Q

)2

σ2Q +

(∂P∂H

)2

σ2H = 73188.9

P = (1.854±0.073)×106 W

(b) The relative uncertainty isσP

P=

0.073×106

1.854×106 = 0.039= 3.9%.

(c) σP =

√(∂P∂η

)2

σ2η +

(∂P∂Q

)2

σ2Q +

(∂P∂H

)2

σ2H ,

∂P∂η

= 9800QH = 2.18148×106,∂P∂Q

= 9800ηH = 30904.3,∂P∂H

= 9800ηQ= 499800

If ση = 0.01,σQ = 1, andσH = 0.1, thenσP = 6.3×104.

If ση = 0.02,σQ = 0.5, andσH = 0.1, thenσP = 6.8×104.

If ση = 0.02,σQ = 1, andσH = 0.05, thenσP = 5.9×104.

Reducing the uncertainty inH to 0.05 m provides the greatest reduction.

6. (a) p = 0.360, σp = 0.048,q = 0.250, σq = 0.043,PAB = pq= 0.090

∂PAB

∂p= q = 0.25

∂PAB

∂q= p = 0.36

σPAB =

√(∂PAB

∂p

)2

σ2p +

(∂PAB

∂q

)2

σ2q = 0.020

PAB = 0.090±0.020

(b) The relative uncertainty isσPAB

PAB=

0.0200.090

= 0.22= 22%.

Alternatively, lnPAB = ln p+ lnq, so∂ lnPAB

∂q= 1/q = 4,

∂ lnPAB

∂p= 1/p = 2.77778,

σPAB

PAB= σlnPAB =

√(∂ lnPAB

∂p

)2

σ2p +

(∂ lnPAB

∂q

)2

σ2q = 0.22= 22%



(c) σPAB =

√(∂PAB

∂p

)2

σ2p +

(∂PAB

∂q

)2

σ2q,

∂PAB

∂p= q = 0.25

∂PAB

∂q= p = 0.36

If σp = 0.02 andσq = 0.043, thenσPAB = 0.016.

If σp = 0.048 andσq = 0.02, thenσPAB = 0.014.

Reducing the uncertainty in the type B proportion to 0.02 provides the greater reduction.

7. (a)H = 4, c = 0.2390,∆T = 2.75, σ∆T = 0.02,m= 0.40, σm = 0.01

C =cH(∆T)

m=

0.956(∆T)

m= 6.57

∂C∂∆T

=0.956

m= 2.39

∂C∂m

=−0.956(∆T)

m2 = −16.4312

σC =

√(∂C

∂∆T

)2

σ2∆T +

(∂C∂m

)2

σ2m = 0.17

C = 6.57±0.17 kcal.

(b) The relative uncertainty isσC

C=

0.176.57

= 0.026= 2.6%.

Alternatively, lnC = ln0.956+ ln(∆T)− lnm, so∂ lnC∂∆T

=1

∆T= 0.363636,

∂ lnC∂m

=−1m

= −2.5,

σC

C= σlnC =

√(∂ lnC∂∆T

)2

σ2∆T +

(∂ lnC∂m

)2

σ2m = 0.026= 2.6%

(c) σC =

√(∂C

∂∆T

)2

σ2∆T +

(∂C∂m

)2

σ2m,

∂C∂∆T

=0.956

m= 2.39

∂C∂m

=−0.956(∆T)

m2 = −16.4312

If σ∆T = 0.01 andσm = 0.01, thenσC = 0.17.

If σ∆T = 0.02 andσm = 0.005, thenσC = 0.10.

Reducing the uncertainty in the mass to 0.005 g provides the greater reduction.

8. (a) The estimate of the hardness is the average of the 22 measurements, which isX = 65.52. The uncertainty is


136 CHAPTER 3

σX = σ/√

n = 0.63/√

22= 0.13.

(b) The uncertainty in a single measurement isσ = 0.63.

9. (a) LetX be the measured northerly component of velocity and letY be the measured easterly component ofvelocity. The velocity of the earth’s crust is estimated with V =

√X2 +Y2.

Now X = 22.10, σX = 0.34 andY = 14.3, σY = 0.32, soV =√

22.102+14.32 = 26.32, and

∂V∂X

= X/√

X2 +Y2 = 0.83957

∂V∂Y

= Y/√

X2 +Y2 = 0.543251

σV =

√(∂V∂X

)2

σ2X +

(∂V∂Y

)2

σ2Y = 0.33

V = 26.32±0.33 mm/year

(b) Let T be the estimated number of years it will take for the earth’s crust to move 100 mm.

By part (a),V = 26.3230, σV = 0.334222. We are using extra precision inV andσV to get good precision forT andσT .

Now T = 100/V = 3.799,dTdV

= −100/V2 = −0.144321

σT =

∣∣∣∣dTdV

∣∣∣∣σV = 0.144321(0.334222)= 0.048

T = 3.799±0.048 years

10. (a) LetM1 represent the estimated molar mass of the unknown gas, and let M2 represent the molar mass of carbondioxide. ThenM2 = 44, andR= 1.66, σR = 0.03.

Now M1 = M2R2 = 44R2 = 121.2, anddM1

dR= 88R= 146.08, soσM =

∣∣∣∣dM1

dR

∣∣∣∣σR = 146.08(0.03) = 4.4.

M1 = 121.2±4.4 g/mol

(b) The relative uncertainty isσM1

M1= 4.4/121.2= 0.036= 3.6%.

Alternatively, lnM1 = ln44+2lnR, sod lnM1

dR= 2/R= 2/1.66= 1.204819.



σM1

M1= σlnM1 =

∣∣∣∣d lnM1

dR

∣∣∣∣σR = 1.204819(0.03)= 0.036= 3.6%

11. LetX1 andX2 represent the estimated thicknesses of the outer layers, and letY1,Y2,Y3,Y4 represent the estimatedthicknesses of the inner layers. ThenX1 = X2 = 1.25,Y1 = Y2 = Y3 = Y4 = 0.80, σXi = 0.10 for i = 1,2, andσYj = 0.05 for j = 1,2,3,4.

The estimated thickness of the item isT = X1 +X2+Y1+Y2+Y3 +Y4 = 5.70.

The uncertainty in the estimate isσT =√

σ2X1

+ σ2X2

+ σ2Y1

+ σ2Y2

+ σ2Y3

+ σ2Y4

= 0.17.

The thickness is 5.70±0.17 mm.

12. Let X be the estimated concentration in the first experiment and let Y be the estimated concentration in thesecond experiment. ThenX = 43.94, σX = 1.18, andY = 48.66, σY = 1.76.

The estimated difference isY−X = 48.66−43.94= 4.72.

The uncertainty in the estimate isσY−X =√

σ2Y + σ2

X =√

1.762+1.182 = 2.12.

The difference is 4.72±2.12 g/L.

13. (a) The relative uncertainty inλ is σλ/λ = σlnλ. The given relative uncertainties areσlnV = 0.0001,σln I = 0.0001,σlnA = 0.001,σln l = 0.01,σlna = 0.01. lnλ = lnV + ln I + lnA− lnπ− ln l − lna.

σlnλ =√

σ2lnV + σ2

ln I + σ2lnA + σ2

ln l + σ2lna =

√0.00012+0.00012+0.0012+0.012+0.012 = 0.014 = 1.4%

(b) If σlnV = 0.0001,σln I = 0.0001,σlnA = 0.001,σln l = 0.005, andσlna = 0.01, thenσlnV = 0.011.

If σlnV = 0, σln I = 0, σlnA = 0, σln l = 0.01, andσlna = 0.01, thenσlnV = 0.014.

Reducing the uncertainty inl to 0.5% provides the greater reduction.

14. Consider the cable to be made of four strands of each of three types of wire. LetX1 = X2 = X3 = X4 be theestimated strengths of the wires of the first type, soXi = 6000 andσXi = 20 for i = 1,2,3,4. LetY1 = Y2 =Y3 = Y4 be the estimated strengths of the wires of the second type, soYi = 5700 andσYi = 30 for i = 1,2,3,4.


138 CHAPTER 3

Let Z1 = Z2 = Z3 = Z4 be the estimated strengths of the wires of the third type, soZi = 6200 andσZi = 40 fori = 1,2,3,4.

(a) LetD denote the strength of the cable estimated by the ductile wire method.ThenD = X1 +X2+X3+X4+Y1 +Y2+Y3+Y4 +Z1+Z2 +Z3+Z4 = 71,600. The uncertainty is

σD =√

∑4i=1 σ2

Xi+ ∑4

i=1σ2Yi

+ ∑4i=1 σ2

Zi=√

4(202)+4(302)+4(402) = 108.

The strength of the cable is 71,600±108 pounds.

(b) The weakest wire has strengthYi = 5700± 30 lbs. The strength of the cable estimated by the brittle wiremethod, isB = 12Yi = 68,400. The uncertainty isσB = 12σY = 12(30) = 360. SoB = 68,400±360 pounds.

15. (a) Yes, since the strengths of the wires are all estimated to be the same, the sum of the strengths is the same as thestrength of one of them multiplied by the number of wires. Therefore the estimated strength is 80,000 poundsin both cases.

(b) No, for the ductile wire method the squares of the uncertainties of the 16 wires are added, to obtainσ =√16×202 = 80. For the brittle wire method, the uncertainty in the strength of the weakest wire is multiplied

by the number of wires, to obtainσ = 16×20= 320.

16. A = 80, σA = 5, B = 90, σB = 3. The relative increase isR= (B−A)/A= B/A−1.

∂R∂A

= −B/A2 = −0.0140625

∂R∂B

= 1/A= 0.0125

σR =

√(∂R∂A

)2

σ2A +

(∂R∂B

)2

σ2B = 0.080

The uncertainty is 0.080.

17. (a)r = 3.00, σr = 0.03,v = 4.0, σv = 0.2. Q = πr2v = 113.1.

∂Q∂r

= 2πrv = 75.3982



∂Q∂v

= πr2 = 28.2743

σQ =

√(∂Q∂r

)2

σ2r +

(∂Q∂v

)2

σ2v = 6.1

Q = 113.1±6.1 m3/s

(b) r = 4.00, σr = 0.04,v = 2.0, σv = 0.1. Q = πr2v = 100.5.

∂Q∂r

= 2πrv = 50.2655

∂Q∂v

= πr2 = 50.2655

σQ =

√(∂Q∂r

)2

σ2r +

(∂Q∂v

)2

σ2v = 5.4

Q = 100.5±5.4 m3/s

(c) The relative uncertainty isσQ

Q= σlnQ.

lnQ = lnπ +2lnr + lnv, σln r =σr

r= 0.01, σlnv =

σv

v= 0.05,

∂ lnQ∂r

=2r,

∂ lnQ∂v

=1v

σlnQ =

√(∂ lnQ

∂r

)2

σ2r +

(∂ lnQ

∂v

)2

σ2v =

√22(σr

r

)2+(σv

v

)2=√

22(0.012)+0.052 = 0.054

Yes, the relative uncertainty inQ can be determined from the relative uncertainties inr andv, and it is equal to5.4%.

18. (a)C0 = 0.2, t = 300,C = 0.174, σC = 0.005

k = (1/t)(lnC0− lnC) = −0.005365− (1/300) lnC = 0.000464

dkdC

=−1

300C= −0.019157

σk =

∣∣∣∣dkdC

∣∣∣∣σC = (0.019157)(0.005)= 0.000096

k = (46.4±9.6)×10−5 s−1

(b)σk

k=

9.6×10−5

46.4×10−5 = 0.21= 21%


140 CHAPTER 3

(c) From part (a),k = 46.4207×10−5 andσk = 9.5785×10−5. We are using extra precision ink andσk in orderto get good precision int1/2 andσt1/2

.

t1/2 =ln2k

, sodt1/2

dk=

− ln2k2 = −3.216639×106

σt1/2=

∣∣∣∣dt1/2

dk

∣∣∣∣σk = (3.216639×106)(9.5785×10−5) = 308

σt1/2= 308≈ 310 s

(d) ln(t1/2) = ln(ln2)− lnk. Thereforeσlnt1/2= σlnk =

σk

k= 0.21= 21% from part (b).

19. (a)C0 = 0.03,t = 40,C = 0.0023, σC = 0.0002.k = (1/t)(1/C−1/C0) = (1/40)(1/C)−5/6= 10.04

dkdC

=−1

40C2 = −4725.90,σk =

∣∣∣∣dkdC

∣∣∣∣σC = 4725.90(0.0002)= 0.95

k = 10.04±0.95 s−1

(b) C0 = 0.03,t = 50,C = 0.0018, σC = 0.0002.k = (1/t)(1/C−1/C0) = (1/50)(1/C)−2/3= 10.4

dkdC

=−1

50C2 = −6172.84,σk =

∣∣∣∣dkdC

∣∣∣∣σC = 6172.84(0.0002)= 1.2

k = 10.4±1.2 s−1

(c) Letk = (k1+ k2)/2= 0.5k1+0.5k2. From parts (a) and (b),σk1= 0.945180 andσk2

= 1.234568. We are usingextra precision inσk1

andσk2in order to get good precision inσk.

σk =√

0.52σ2k1

+0.52σ2k2

=√

0.52(0.9451802)+0.52(1.2345682) = 0.78

(d) The value ofc is cbest=σ2

k2

σ2k1

+ σ2k2

=1.2345682

0.9451802+1.2345682= 0.63.

20. (a)s= 1, v = 3.2, σv = 0.1, a1 = v2/2s= v2/2 = 5.12



da1

dv= v/s= 3.2, σa1 =

∣∣∣∣da1

dv

∣∣∣∣σv = (3.2)(0.1) = 0.32

a1 = 5.12±0.32 m/s2

(b) s= 1, t = 0.63, σt = 0.01,a2 = 2s/t2 = 2/t2 = 5.04

da2

dt= −4s/t3 = −15.997, σa2 =

∣∣∣∣da2

dt

∣∣∣∣σt = (15.997)(0.01) = 0.16

a2 = 5.04±0.16 m/s2

(c) The weighted average with the smallest uncertainty iscbesta1+(1−cbest)a2, wherecbest=σ2

a2

σ2a1

+ σ2a2

.

From part (a),σa1 = 0.32, and from part (b),σa2 = 0.16. Socbest=0.162

0.322+0.162 = 0.20.

The weighted average with the smallest uncertainty is 0.20a1+0.80a2.

The uncertainty is√

0.202σ2a1

+0.802σ2a2

= 0.14 m/s2.

21. (a) Letsbe the measured side of the square. Thens= 181.2, σs = 0.1.

The estimated area isS= s2 = 32,833.

dSds

= 2s= 362.4, σS =

∣∣∣∣dSds

∣∣∣∣σs = (362.4)(0.1) = 36

S= 32,833±36 m2

(b) The estimated area of a semicircle isC = (πs2)/8 = 12,894.

dCds

= πs/4 = 142.314, σC =

∣∣∣∣dCds

∣∣∣∣σs = (142.314)(0.1) = 14

C = 12,894±14 m2

(c) This is not correct. Lets denote the length of a side of the square. SinceSandC are both computed in termsof s, they are not independent. In order to computeσA correctly, we must expressA directly in terms ofs:

A = s2 +2πs2/8 = s2(1+ π/4). SoσA =

∣∣∣∣dAds

∣∣∣∣σs = 2s(1+ π/4)σs = 65 m2.


142 CHAPTER 3

22. (a)r = 3.0, σr = 0.1, A = πr2 = 28.27433,dAdr

= 2πr = 18.85

σA =

∣∣∣∣dAdr

∣∣∣∣σr = 18.85(0.1) = 1.9

A = 28.27433±1.9 cm2

(b)d2Adr2 = 2π. The bias corrected estimate isA−

(12

)d2Adr2 σ2

r = 28.27433−π(0.12) = 28.2429 cm2.

(c) No. The difference between the two estimates is much lessthan the uncertainty.

23. (a)P1 = 8.1, σP1 = 0.1, P2 = 15.4, σP2 = 0.2, P =√

P1P2 = 11.16871

∂P3

∂P1=

P2

2√

P1P2= 0.689426

∂P3

∂P2=

P1

2√

P1P2= 0.36262

σP3 =

√(∂P3

∂P1

)2

σ2P1

+

(∂P3

∂P2

)2

σ2P2

= 0.10

P3 = 11.16871±0.10 MPa

(b)∂2P3

∂P21

= −(0.25)P−3/21 P1/2

2 = −0.042557,∂2P3

∂P22

= −(0.25)P−3/22 P1/2

1 = −0.011773

The bias corrected estimate is

P3−(

12

)[(∂2P3

∂P21

)σ2

P1+

(∂2P3

∂P22

)σ2

P2

]= 11.16871−(0.5)[−0.042557(0.12)−0.011773(0.22)] = 11.16916.

(c) No. The difference between the two estimates is much lessthan the uncertainty.


SECTION 4.1 143

Chapter 4

Section 4.1

1. (a)X ∼ Bernoulli(p), wherep = 0.55. µX = p = 0.55,σ2X = p(1− p) = 0.2475.

(b) No. A Bernoulli random variable has possible values 0 and1. The possible values of Y are 0 and 2.

(c) Y = 2X, whereX ∼ Bernoulli(p).

ThereforeµY = 2µX = 2p = 1.10, andσ2Y = 22σ2

X = 4p(1− p) = 0.99.

2. (a) pX = 0.25

(b) pY = 0.35

(c) pZ = 0.60

(d) No. If the order is for a small drink, it cannot also be for amedium drink.

(e) Yes.pZ = 0.60= 0.25+0.35= pX + pY.

(f) Yes. If the drink is small, thenX = 1, Y = 0, andZ = 1, soZ = X +Y. If the drink is medium, thenX = 0,Y = 1, andZ = 1, soZ = X +Y. If the drink is large, thenX = 0,Y = 0, andZ = 0, soZ = X +Y.

3. (a) pX = 0.05

(b) pY = 0.20

(c) pZ = 0.23

(d) Yes, it is possible for there to be both discoloration anda crack.

(e) No. pZ = P(X = 1 orY = 1) 6= P(X = 1)+P(Y = 1) becauseP(X = 1 andY = 1) 6= 0.


144 CHAPTER 4

(f) No. If the surface has both discoloration and a crack, then X = 1,Y = 1, andZ = 1, butX +Y = 2.

4. (a) If X andY cannot both be equal to 1, the possible values forZ are 0 and 1, soZ is a Bernoulli random variable.

(b) pZ = P(Z = 1) = P(X = 1 orY = 1) = P(X = 1)+P(Y = 1)−P(X = 1 andY = 1) =P(X = 1)+P(Y = 1)−0= pX + pY

(c) P(Z = 2) = P(X = 1 andY = 1) 6= 0. SinceP(Z = 2) 6= 0, Z is not a Bernoulli random variable.

5. (a) pX = 1/2

(b) pY = 1/2

(c) pZ = 1/4

(d) Yes.P(X = x andY = y) = P(X = x)P(Y = y) for all values ofx andy.

(e) Yes.pZ = 1/4 = (1/2)(1/2) = pX pY.

(f) Yes. If both coins come up heads, thenX = 1,Y = 1, andZ = 1, soZ = XY. If not, thenZ = 0, and eitherX,Y, or both are equal to 0 as well, so againZ = XY.

6. (a) pX = 1/6

(b) pY = 5/36

(c) pZ = 1/36

(d) No. P(X = 1 andY = 1) = P(double 3) = 1/36, butP(X = 1)P(Y = 1) = (1/6)(5/36) 6= 1/36.

(e) No. pZ = P(Z = 1) = P(X = 1 andY = 1) 6= P(X = 1)P(Y = 1).

(f) Yes. If the dice come up double 3, thenX = 1,Y = 1, andZ = 1, soZ = XY. If not, thenZ = 0, and eitherX,Y, or both are equal to 0 as well, so againZ = XY.


SECTION 4.2 145

7. (a) Since the possible values ofX andY are 0 and 1, the possible values of the productZ = XY are also 0 and 1.ThereforeZ is a Bernoulli random variable.

(b) pZ = P(Z = 1) = P(XY = 1) = P(X = 1 andY = 1) = P(X = 1)P(Y = 1) = pX pY.

Section 4.2

1. (a)P(X = 2) =8!

2!(8−2)!(0.4)2(1−0.4)8−2 = 0.2090

(b) P(X = 4) =8!

4!(8−4)!(0.4)4(1−0.4)8−4 = 0.2322

(c) P(X < 2) = P(X = 0)+P(X = 1)

=8!

0!(8−0)!(0.4)0(1−0.4)8−0+

8!1!(8−1)!

(0.4)1(1−0.4)8−1

= 0.1064

(d) P(X > 6) = P(X = 7)+P(X = 8)

=8!

7!(8−7)!(0.4)7(1−0.4)8−7+

8!0!(8−0)!

(0.4)0(1−0.4)8−0

= 0.007864+0.000655

= 0.0085

(e) µX = (8)(0.4) = 3.2

(f) σ2X = (8)(0.4)(1−0.4) = 1.92

2. (a)P(X < 4) = P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)

=6!

0!(6−0)!(0.7)0(1−0.7)6−0+

6!1!(6−1)!

(0.7)1(1−0.7)6−1+6!

2!(6−2)!(0.7)2(1−0.7)6−2


146 CHAPTER 4

+6!

3!(6−3)!(0.7)3(1−0.7)6−3

= 0.2557

(b) P(X ≥ 2) = 1−P(X < 2)

= 1−P(X = 0)−P(X = 1)

= 1− 6!0!(6−0)!

(0.7)0(1−0.7)6−0− 6!1!(6−1)!

(0.7)1(1−0.7)6−1

= 0.9891

(c) P(1≤ X < 4) = P(X = 1)+P(X = 2)+P(X = 3)

=6!

1!(6−1)!(0.7)1(1−0.7)6−1+

6!2!(6−2)!

(0.7)2(1−0.7)6−2+6!

3!(6−3)!(0.7)3(1−0.7)6−3

= 0.2550

(d) P(1 < X ≤ 4) = P(X = 2)+P(X = 3)+P(X = 4)

=6!

2!(6−2)!(0.7)2(1−0.7)6−2+

6!3!(6−3)!

(0.7)3(1−0.7)6−3+6!

4!(6−4)!(0.7)4(1−0.7)6−4

= 0.5689

(e) P(X = 0) =6!

0!(6−0)!(0.7)0(1−0.7)6−0 = 0.000729

(f) P(X = 4) =6!

4!(6−4)!(0.7)4(1−0.7)6−4 = 0.3241

(g) µX = (6)(0.7) = 4.2

(h) σ2X = (6)(0.7)(0.3) = 1.26


SECTION 4.2 147

3. (a)P(X = 3) =5!

3!(5−3)!(0.2)3(1−0.2)5−3 = 0.0512

(b) P(X ≤ 2) = P(X = 0)+P(X = 1)+P(X = 2)

=10!

0!(10−0)!(0.6)0(1−0.6)10−0+

10!1!(10−1)!

(0.6)1(1−0.6)10−1+10!

2!(10−2)!(0.6)2(1−0.6)10−2

= 0.0123

(c) P(X ≥ 5) = P(X = 5)+P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9)

=9!

5!(9−5)!(0.5)5(1−0.5)9−5+

9!6!(9−6)!

(0.5)6(1−0.5)9−6+9!

7!(9−7)!(0.5)7(1−0.5)9−7

+9!

8!(9−8)!(0.5)8(1−0.5)9−8+

9!9!(9−9)!

(0.5)9(1−0.5)9−9

= 0.5000

(d) P(3≤ X ≤ 4) = P(X = 3)+P(X = 4)

=8!

3!(8−3)!(0.8)3(1−0.8)8−3+

8!4!(8−4)!

(0.8)4(1−0.8)8−4

= 0.0551

4. LetX be the number of defective items in the sample. ThenX ∼ Bin(5,0.1).

(a) P(X = 0) =5!

0!(5−0)!(0.1)0(1−0.1)5−0 = 0.5905

(b) P(X = 1) =5!

1!(5−1)!(0.1)1(1−0.1)5−1 = 0.3281

(c) P(X ≥ 1) = 1−P(X = 0) = 1− 5!0!(5−0)!

(0.1)0(1−0.1)5−0 = 1−0.5905= 0.4095

(d) P(X < 2) = P(X = 0)+P(X = 1)

=5!

0!(5−0)!(0.1)0(1−0.1)5−0+

5!1!(5−1)!

(0.1)1(1−0.1)5−1


148 CHAPTER 4

= 0.59049+0.32805

= 0.9185

5. LetX be the number of heads obtained. ThenX ∼ Bin(10,0.5).

(a) P(X = 3) =10!

3!(10−3)!(0.5)3(1−0.5)10−3 = 0.1172

(b) µX = 10(0.5) = 5

(c) σ2X = 10(0.5)(1−0.5) = 2.5

(d) σX =√

10(0.5)(1−0.5) = 1.58

6. Let X be the number of times the die comes up 6, and letY be the number of times the die comes up an oddnumber. ThenX ∼ Bin(8,1/6) andY ∼ Bin(8,1/2).

(a) P(X = 2) =8!

2!(8−2)!(1/6)2(1−1/6)8−2 = 0.2605

(b) P(Y = 5) =8!

5!(8−5)!(1/2)5(1−1/2)8−5 = 0.2188

(c) µX = 8(1/6) = 1.333

(d) µY = 8(1/2) = 4

(e) σX =√

(8)(1/6)(5/6) = 1.0541

(f) σY =√

(8)(1/2)(1/2) = 1.4142

7. LetX be the number of vehicles that require warranty repairs. Then X ∼ Bin(12,0.2).

(a) P(X = 4) =12!

4!(12−4)!(0.2)4(1−0.2)12−4 = 0.1329


SECTION 4.2 149

(b) P(X < 3) =12!

0!(12−0)!(0.2)0(1−0.2)12−0+

12!1!(12−1)!

(0.2)1(1−0.2)12−1

+12!

2!(12−2)!(0.2)2(1−0.2)12−2 = 0.5583

(c) P(X = 0) =12!

0!(12−0)!(0.2)0(1−0.2)12−0 = 0.0687

(d) µX = 12(0.2) = 2.4

(e) σX =√

(12)(0.2)(0.8) = 1.3856

8. LetX be the number of tires, out of the four chosen, that have a flaw.ThenX ∼ Bin(4,0.05).

(a) P(X = 0) =4!

0!(4−0)!(0.05)0(1−0.05)4−0 = 0.8145

(b) P(X = 1) =4!

1!(4−1)!(0.05)1(1−0.05)4−1 = 0.1715

(c) P(X ≥ 1) = 1−P(X = 0) = 1− 4!0!(4−0)!

(0.05)0(1−0.05)4−0 = 1−0.8145= 0.1855

9. LetX be the number of bits out of the eight that are equal to 1. ThenX ∼ Bin(8,0.5).

(a) P(X = 8) =8!

8!(8−8)!(0.5)8(1−0.5)8−8 = 0.0039

(b) P(X = 3) =8!

3!(8−3)!(0.5)3(1−0.5)8−3 = 0.2188

(c) P(X ≥ 6) = P(X = 6)+P(X = 7)+P(X = 8)

=8!

6!(8−6)!(0.5)6(1−0.5)8−6+

8!7!(8−7)!

(0.5)7(1−0.5)8−7+8!

8!(8−8)!(0.5)8(1−0.5)8−8

= 0.10938+0.03125+0.00391

= 0.1445

(d) P(X ≥ 2) = 1−P(X < 2)


150 CHAPTER 4

= 1−P(X = 0)−P(X = 1)

= 1− 8!0!(8−0)!

(0.5)0(1−0.5)8−0− 8!1!(8−1)!

(0.5)1(1−0.5)8−1

= 1−0.00391−0.03125

= 0.9648

10. LetX be the number of rods in the sample that meet specifications, and let p be the proportion of the day’sproduction that meets specifications. ThenX ∼ Bin(100, p). The observed value ofX is 92.

(a) p = X/100= 92/100= 0.92, σp =

√p(1− p)

n

Replacingp with p = 0.92 andn with 100,σp =

√0.92(1−0.92)

100= 0.027.

(b) We must findn so that

√p(1− p)

n= 0.01.

Substitutep = 0.92 for p to obtain

√0.92(1−0.92)

n= 0.01. Solving forn yieldsn = 736.

11. LetX be the number of defective parts in the sample from vendor A, and letY be the number of defective partsin the sample from vendor B. LetpA be the probability that a part from vendor A is defective, andlet pB be theprobability that a part from vendor B is defective. ThenX ∼ Bin(100, pA) andY ∼ Bin(200, pB). The observedvalue ofX is 12 and the observed value ofY is 10.

(a) pA = X/100= 12/100= 0.12, σpA =

√pA(1− pA)

100

ReplacingpA with pA = 0.12 andn with 100,σpA=

√0.12(1−0.12)

100= 0.032.

(b) pB = Y/200= 10/200= 0.05, σpB =

√pB(1− pB)

200

ReplacingpB with pB = 0.05 andn with 200,σpB =

√0.05(1−0.05)

200= 0.015.

(c) The difference is estimated withpA− pB = 0.12−0.05= 0.07.

The uncertainty is√

σ2pA

+ σ2pB

= 0.036.


SECTION 4.2 151

12. (a) LetD denote the event that the item is defective and letR denote the even that the item can be repaired. ThenP(D) = 0.2 andP(R|D) = 0.6. ThenP(D∩Rc) = P(Rc |D)P(D) = [1−P(R|D)]P(D) = (1−0.6)(0.2) = 0.08.

(b) LetX be the number of items out of 20 that are defective and cannot be repaired.ThenX ∼ Bin(20,0.08).

P(X = 2) =20!

2!(20−2)!(0.08)2(1−0.08)20−2 = 0.2711

13. (a)P(can be used) = P(meets spec)+P(long) = 0.90+0.06= 0.96.

(b) LetX be the number of bolts out of 10 that can be used. ThenX ∼ Bin(10,0.96).

P(X < 9) = 1−P(X ≥ 9)

= 1−P(X = 9)−P(X = 10)

= 1− 10!9!(10−9)!

(0.96)9(1−0.96)10−9− 10!10!(10−10)!

(0.96)10(1−0.96)10−10

= 1−0.27701−0.66483

= 0.0582

14. (a) LetX be the number of the ten gears that are scrap. ThenX ∼ Bin(10,0.05).

P(X ≥ 1) = 1−P(X = 0) = 1− 10!0!(10−0)!

(0.05)0(1−0.05)10−0 = 1−0.5987= 0.4013

(b) Eight or more are not scrap if and only if two or fewer are scrap.

P(X ≤ 2) =10!

0!(10−0)!(0.05)0(1−0.05)10−0+

10!1!(10−1)!

(0.05)1(1−0.05)10−1

+10!

2!(10−2)!(0.05)2(1−0.05)10−2 = 0.9885

(c) The probability that a gear is either degraded or scrap is0.15+0.05= 0.20. LetY be the number of gears thatare either degraded or scrap. ThenY ∼ Bin(10,0.2).

P(Y > 2) = 1−P(Y ≤ 2)

= 1−P(Y = 0)−P(Y = 1)−P(Y = 2)

= 1− 10!0!(10−0)!

(0.2)0(1−0.2)10−0− 10!1!(10−1)!

(0.2)1(1−0.2)10−1− 10!2!(10−2)!

(0.2)2(1−0.2)10−2

= 0.3222


152 CHAPTER 4

(d) The probability that a gear is either conforming or degraded is 0.80+ 0.15= 0.95. LetY be the number ofgears that are either conforming or degraded. ThenY ∼ Bin(10,0.95).

P(Y = 9) =10!

9!(10−9)!(0.95)9(1−0.95)10−9 = 0.3151

15. (a) LetY be the number of lights that are green. ThenY ∼ Bin(3,0.6).

P(Y = 3) =3!

3!(3−3)!(0.6)3(1−0.6)3−3 = 0.216

(b) X ∼ Bin(5,0.216)

(c) P(X = 3) =5!

3!(5−3)!(0.216)3(1−0.216)5−3 = 0.0619

16. (a) LetX be the number of components out of 10 that are defective. ThenX ∼ Bin(10,0.1).

P(return) = P(X > 1)

= 1−P(X ≤ 1)

= 1−P(X = 0)−P(X = 1)

= 1− 10!0!(10−0)!

(0.1)0(1−0.1)10−0− 10!1!(10−1)!

(0.1)1(1−0.1)10−1

= 1−0.34868−0.38742

= 0.2639

(b) LetX be the number of components out of 10 that are defective. ThenX ∼ Bin(10,0.2).


= 1−P(X ≤ 1)

= 1−P(X = 0)−P(X = 1)

= 1− 10!0!(10−0)!

(0.2)0(1−0.2)10−0− 10!1!(10−1)!

(0.2)1(1−0.2)10−1

= 1−0.10737−0.26844

= 0.6242

(c) LetX be the number of components out of 10 that are defective. ThenX ∼ Bin(10,0.02).



SECTION 4.2 153

= 1−P(X ≤ 1)

= 1−P(X = 0)−P(X = 1)

= 1− 10!0!(10−0)!

(0.02)0(1−0.02)10−0− 10!1!(10−1)!

(0.02)1(1−0.02)10−1

= 1−0.81707−0.16675

= 0.0162

(d) Letn be the required sample size. LetX be the number of sampled components that are defective.ThenX ∼ Bin(n,0.2).

P(accept) = P(X = 0) = (0.8)n ≤ 0.01. Sonln0.8≤ ln0.01,−0.2231n≤−4.605,n≥ 20.64.

The minimum sample size is 21.

17. (a) LetX be the number of components that function. ThenX ∼ Bin(5,0.9).

P(X ≥ 3) =5!

3!(5−3)!(0.9)3(1−0.9)5−3+

5!4!(5−4)!

(0.9)4(1−0.9)5−4

+5!

5!(5−5)!(0.9)5(1−0.9)5−5 = 0.9914

(b) We need to find the smallest value ofn so thatP(X ≤ 2) < 0.10 whenX ∼ Bin(n,0.9). Consulting Table A.1,we find that ifn = 3, P(X ≤ 2) = 0.271, and ifn = 4, P(X ≤ 2) = 0.052. The smallest value ofn is thereforen = 4.

18. (a) On a rainy day, the numberX of components that function is distributed Bin(6,0.7).

P(X ≥ 4) =6!

4!(6−4)!(0.7)4(1−0.7)6−4+

6!5!(6−5)!

(0.7)5(1−0.7)6−5

+6!

6!(6−6)!(0.7)6(1−0.7)6−6 = 0.7443

(b) On a nonrainy day, the numberX of components that function is distributed Bin(6,0.9).

P(X ≥ 4) =6!

4!(6−4)!(0.9)4(1−0.9)6−4+

6!5!(6−5)!

(0.9)5(1−0.9)6−5

+6!

6!(6−6)!(0.9)6(1−0.9)6−6 = 0.9842

(c) Let R denote the event of a rainy day, and letF denote the event that the system functions. From part (a) weknow thatP(F |R) = 0.7443, and from part (b) we know thatP(F |Rc) = 0.9842.

P(F) = P(F|R)P(R)+P(F|Rc)P(Rc) = (0.7443)(0.2)+ (0.9842)(0.8)= 0.9362


154 CHAPTER 4

19. (a)X ∼ Bin(10,0.15).

P(X ≥ 7) = P(X = 7)+P(X = 8)+P(X = 9)+P(X = 10)

=10!

7!(10−7)!(0.15)7(1−0.15)10−7+

10!8!(10−8)!

(0.15)8(1−0.15)10−8

+10!

9!(10−9)!(0.15)9(1−0.15)10−9+

10!10!(10−10)!

(0.15)10(1−0.15)10−10

= 1.2591×10−4+8.3326×10−6+3.2677×10−7+5.7665×10−9

= 1.346×10−4

(b) Yes, only about 13 or 14 out of every 100,000 samples of size 10 would have 7 or more defective items.

(c) Yes, because 7 defectives in a sample of size 10 is an unusually large number for a good shipment.

(d) P(X ≥ 2) = 1−P(X < 2)

= 1−P(X = 0)−P(X = 1)

= 1− 10!0!(10−0)!

(0.15)0(1−0.15)10−0− 10!1!(10−1)!

(0.15)1(1−0.15)10−1

= 1−0.19687−0.34743

= 0.4557

(e) No, in about 45% of the samples of size 10, 2 or more items would be defective.

(f) No, because 2 defectives in a sample of size 10 is not an unusually large number for a good shipment.

20. (a)X ∼ Bin(8,0.80).

P(X ≤ 1) = P(X = 0)+P(X = 1)

=8!

0!(8−0)!(0.8)0(1−0.8)8−0+

8!1!(8−1)!

(0.8)1(1−0.8)8−1

= 2.56×10−6+8.192×10−5

= 8.448×10−5


SECTION 4.2 155

(b) Yes, if the claim is true, then only about 8 or 9 out of every100,000 samples would have 1 or fewer policyholders with smoke detectors.

(c) Yes, because 1 out of 8 is an unusually small number of policy holders with smoke detectors if the claim istrue.

(d) P(X ≤ 6) = 1−P(X > 6)

= 1−P(X = 7)+P(X = 8)

= 1− 8!7!(8−7)!

(0.80)7(1−0.80)8−7− 8!8!(8−8)!

(0.80)8(1−0.80)8−8

= 1−0.33554−0.16777

= 0.4967

(e) No, for nearly half the samples of size 8, 6 or fewer policyholders would have smoke detectors.

(f) No, 6 out of 8 is not an unusually small number if the claim is true.

21. (a) LetX be the number of bits that are reversed. ThenX ∼ Bin(5,0.3). The correct value is assigned ifX ≤ 2.

P(X ≤ 2) = P(X = 0)+P(X = 1)+P(X = 2) =5!

0!(5−0)!(0.3)0(1−0.3)5−0+

5!1!(5−1)!

(0.3)1(1−0.3)5−1

+5!

2!(5−2)!(0.3)2(1−0.3)5−2 = 0.8369

(b) We need to find the smallest odd value ofn so thatP(X ≤ (n−1)/2)≥ 0.90 whenX ∼ Bin(n,0.3). ConsultingTable A.1, we find that ifn = 3, P(X ≤ 1) = 0.784, if n = 5, P(X ≤ 2) = 0.837, if n = 7, P(X ≤ 3) = 0.874,and if n = 9, P(X ≤ 4) = 0.901. The smallest value ofn is thereforen = 9.

22. P(Y = y) = P(n−X = y) = P(X = n−y) =n!

(n−y)!y!pn−y(1− p)y =

n!y!(n−y)!

(1− p)y[1− (1− p)]n−y

23. (a)Y = 10X +3(100−X) = 7X +300


156 CHAPTER 4

(b) X ∼ Bin(100,0.9), soµX = 100(0.9) = 90.

µY = 7µX +300= 7(90)+300= $930

(c) σX =√

100(0.9)(0.1) = 3. σY = 7σX = $21

24. LetX be the number of components of the two in the first design that work. LetY be the number of componentsof the four in the second design that work. ThenX ∼ Bin(2,0.9) andY ∼ Bin(4,0.9). The probability that thedesign with two components works is

P(X ≥ 1) = 1−P(X = 0) = 1− 2!0!(2−0)!

(0.9)0(1−0.9)2−0 = 1−0.01= 0.99.

The probability that the design with four components works is

P(Y ≥ 2) = 1−P(Y < 2)

= 1−P(Y = 0)−P(Y = 1)

= 1− 4!0!(4−0)!

(0.9)0(1−0.9)4−0− 4!1!(4−1)!

(0.9)1(1−0.9)4−1

= 1−0.0001−0.0036

= 0.9963

The design with four components has the greater probabilityof functioning.

25. Let p be the probability that a tire has no flaw. The results of Example 4.14 show that the sample proportionis p = 93/100= 0.93 andσp = 0.0255. LetX be the number of tires out of four that have no flaw. ThenX ∼ Bin(4, p). Let q be the probability that exactly one of four tires has a flaw.

Thenq = P(X = 3) =4!

3!(4−3)!(p)3(1− p)4−3 = 4p3(1− p) = 4p3−4p4.

The estimate ofq is q = 4p3−4p4 = 4(0.93)3−4(0.93)4 = 0.225.

dqdp

= 12p2−16p3 = 12(0.93)2−16(0.93)3 = −2.4909

σq =

∣∣∣∣dqdp

∣∣∣∣σp = (2.4909)(0.0255)= 0.064

The probability is 0.225±0.064.

26. (a) The sample size isn = 612 andp = 88/612= 0.143791.


SECTION 4.3 157

σp =√

p(1− p)/n =√

(0.143791)(0.856209)/612= 0.014.

p = 0.144±0.014

(b) LetO =p

1− pbe the odds. ThenO is estimated with

p1− p

=88/612

1−88/612=

88524

= 0.167939.

dOdp

= (1− p)−2 = (1−0.143791)−2 = 1.364081.

σO =

∣∣∣∣dOdp

∣∣∣∣σp = (1.364081)(0.014)= 0.019.

O = 0.168±0.019

Section 4.3

1. (a)P(X = 1) = e−4 41

1!= 0.0733

(b) P(X = 0) = e−4 40

0!= 0.0183

(c) P(X < 2) = P(X = 0)+P(X = 1) = e−440

0!+e−441

1!= 0.0183+0.0733= 0.0916

(d) P(X > 1) = 1−P(X ≤ 1) = 1−P(X = 0)−P(X = 1) = 1−0.0183−0.0733= 0.9084

(e) SinceX ∼ Poisson(4), µX = 4.

(f) SinceX ∼ Poisson(4), σX =√

4 = 2.

2. The mean number of particles to be withdrawn is (2 per mL)(3mL) = 6. ThereforeX ∼ Poisson(6).

(a) P(X = 5) = e−6 65

5!= 0.1606

(b) P(X ≤ 2) = P(X = 0)+P(X = 1)+P(X = 2)


158 CHAPTER 4

= e−6 60

0!+e−661

1!+e−662

2!= 0.0024788+0.014873+0.044618

= 0.0620

(c) P(X > 1) = 1−P(X ≤ 1)

= 1−P(X = 0)−P(X = 1)

= 1−e−660

0!−e−661

1!= 1−0.0024788−0.014873

= 0.9826

(d) SinceX ∼ Poisson(6), µX = 6.

(e) SinceX ∼ Poisson(6), σX =√

6 = 2.45.

3. X is the number of successes inn = 10,000 independent Bernoulli trials, each of which has successprobabilityp = 0.0003. The mean ofX is np= (10,000)(0.0003) = 3. Sincen is large andp is small,X ∼ Poisson(3) toa very close approximation.

(a) P(X = 3) = e−3 33

3!= 0.2240

(b) P(X ≤ 2) = P(X = 0)+P(X = 1)+P(X = 2)

= e−3 30

0!+e−331

1!+e−332

2!= 0.049787+0.14936+0.22404

= 0.4232

(c) P(1≤ X < 4) = P(X = 1)+P(X = 2)+P(X = 3)

= e−3 31

1!+e−332

2!+e−333

3!= 0.14936+0.22404+0.22404

= 0.5974


SECTION 4.3 159



3 = 1.73.

4. (a)X ∼ Poisson(6), soP(X = 7) = e−667

7!= 0.1377

(b) P(X ≥ 3) = 1−P(X < 3)

= 1−P(X = 0)−P(X = 1)−P(X = 2)

= 1−e−660

0!−e−661

1!−e−662

2!= 1−0.00248−0.01487−0.04462

= 0.9380

(c) P(2 < X < 7) = P(X = 3)+P(X = 4)+P(X = 5)+P(X = 6)

= e−6 63

3!+e−664

4!+e−665

5!+e−666

6!= 0.08924+0.13385+0.16062+0.16062

= 0.5443



6 = 2.4495.

5. LetX be the number of messages that fail to reach the base station.ThenX is the number of successes inn = 1000 Bernoulli trials, each of which has success probability p = 0.005.The mean ofX is np= (1000)(0.005) = 5.Sincen is large andp is small,X ∼ Poisson(5) to a very close approximation.

(a) P(X = 3) = e−5 53

3!= 0.14037

(b) The event that fewer than 994 messages reach the base station is the same as the event that more than 6 messagesfail to reach the base station, or equivalently, thatX > 6.

P(X > 6) = 1−P(X ≤ 6) = 1−e−550

0!−e−551

1!−e−552

2!−e−553

3!−e−554

4!−e−555

5!−e−556

6!= 1−0.00674−0.03369−0.08422−0.14037−0.17547−0.17547−0.14622= 0.2378


160 CHAPTER 4

(c) µX = 5

(d) σX =√

5 = 2.2361

6. Let X be the number of individuals in the sample that carry the gene. ThenX is the number of successes inn = 1000 Bernoulli trials, each of which has success probability p = 0.0002. The mean ofX isnp = (1000)(0.0002) = 0.2. Sincen is large andp is small,X ∼ Poisson(0.2) to a very close approxima-tion.

(a) P(X = 1) = e−0.2 0.21

1!= 0.1637

(b) P(X = 0) = e−0.2 0.20

0!= 0.8187

(c) P(X > 2) = 1−P(X ≤ 2)

= 1−P(X = 0)−P(X = 1)−P(X = 2)

= 1−e−0.20.20

0!−e−0.20.21

1!−e−0.20.22

2!= 1−0.81873−0.16375−0.016375

= 0.0011

(d) SinceX ∼ Poisson(0.2), µX = 0.2.

(e) SinceX ∼ Poisson(0.2), σX =√

0.2 = 0.45.

7. (a) LetX be the number of messages received in one hour.

Since the mean rate is 8 messages per hour,X ∼ Poisson(8).

P(X = 5) = e−8 85

5!= 0.0916

(b) LetX be the number of messages received in 1.5 hours.


P(X = 10) = e−121210

10!= 0.1048


SECTION 4.3 161

(c) LetX be the number of messages received in one half hour.


P(X < 3) = P(X = 0)+P(X = 1)+P(X = 2)

= e−440

0!+e−441

1!+e−442

2!= 0.018316+0.073263+0.14653

= 0.2381

8. LetX be the number of diodes that fail. ThenX is the number of successes inn = 300 Bernoulli trials, each ofwhich has success probabilityp = 0.002. The mean ofX is np= (300)(0.002) = 0.6. Sincen is large andp issmall,X ∼ Poisson(0.6) to a very close approximation.

(a) P(X = 2) = e−0.6 0.62

2!= 0.0988

(b) SinceX ∼ Poisson(0.6), µX = 0.6.

(c) SinceX ∼ Poisson(0.6), σX =√

0.6 = 0.7746.

(d) P(X = 0) = e−0.6 0.60

0!= 0.548812

(e) From part (d), the probability that a given board works is0.548812.

Let Y be the number of boards out of five that work. ThenY ∼ Bin(5,0.548812).

P(Y ≥ 4) =5!

4!(5−4)!(0.548812)4(1−0.548812)5−4+

5!5!(5−5)!

(0.548812)5(1−0.548812)5−5 = 0.2544.

Note that the Poisson distribution cannot be used forY becausen = 5 is not large.

9. (ii). Let X ∼ Bin(n, p) whereµX = np= 3. Thenσ2X = np(1− p), which is less than 3 because 1− p< 1. Now

let Y have a Poisson distribution with mean 3. The variance ofY is also equal to 3, because the variance of aPoisson random variable is always equal to its mean. ThereforeY has a larger variance thanX.


162 CHAPTER 4

10. LetX represent the number of particles observed in 3 mL. Letλ represent the true concentration in particlesper mL. ThenX ∼ Poisson(3λ). The observed value ofX is 48. The estimated concentration isλ = 48/3= 16.The uncertainty isσλ =

√16/3 = 2.3. λ = 16.0±2.3.

11. LetX represent the number of bacteria observed in 0.5 mL. Letλ represent the true concentration in bacteria permL. ThenX ∼ Poisson(0.5λ). The observed value ofX is 39. The estimated concentration isλ = 39/0.5= 78.The uncertainty isσλ =

√78/0.5= 12.49. λ = 78±12.

12. (a) LetX be the number of plants in a two-acre region. ThenX ∼ Poisson(20).

P(X = 18) = e−202018

18!= 0.08439

(b) The area of the circle is 10,000π = 31,415.9 ft2. In units of acres, the area is 31,415.9/43,560= 0.72121acres. LetX be the number of plants in the circle. ThenX ∼ Poisson(7.2121).

P(X = 12) = e−7.21217.212112

12!= 0.0305

(c) LetY represent the number of plants observed in 0.1 acres. Letλ represent the true concentration in plants peracre. ThenY ∼ Poisson(0.1λ). The observed value ofY is 5. The estimated concentration isλ = 5/0.1 = 50.The uncertainty isσλ =

√50/0.1= 22.36. λ = 50±22.

13. (a) LetN be the number of defective components produced. ThenN ∼ Poisson(20).

P(N = 15) = e−202015

15!= 0.0516

(b) LetX be the number that are repairable. ThenX ∼ Bin(15,0.6).

P(X = 10) =15!

10!(15−10)!(0.6)10(1−0.6)15−10 = 0.1859

(c) GivenN, X ∼ Bin(N,0.6).

(d) LetN be the number of defective components, and letX be the number that are repairable.

P(N = 15∩X = 10) = P(N = 15)P(X = 10|N = 15)

=

(e−202015

15!

)(15!

10!(15−10)!(0.6)10(1−0.6)15−10

)

= 0.00960


SECTION 4.3 163

14. Letλ be the mean number of particles emitted in one minute. LetX be the number of particles actually emittedduring a given minute.

ThenX ∼ Poisson(λ), soP(X = 0) = e−λ λ0

0!= e−λ.

Now e−λ = 0.1353, soλ = − ln0.1353= 2.000.

15. LetV be the required volume, in mL. LetX be the number of particles in a volumeV.

P(X ≥ 1) = 1−P(X = 0) = 0.99, soP(X = 0) = 0.01.

ThenX ∼ Poisson(0.5V), soP(X = 0) = e−(0.5V) (0.5V)0

0!= e−0.5V .

Thereforee−0.5V = 0.01, soV = −2ln0.01= 9.210 mL.

16. (a) LetX be the number of raisins in a randomly chosen slice.

Since the mean number of raisins per slice is 100/60= 5/3, X ∼ Poisson(5/3).

P(X = 0) = e−(5/3) (5/3)0

0!= 0.1889

(b) LetX be the number of raisins in a randomly chosen slice.

Since the mean number of raisins per slice is 200/60= 10/3,X ∼ Poisson(10/3).

P(X = 5) = e−(10/3) (10/3)5

5!= 0.1223

(c) Letn be the required number of raisins. LetX be the number of raisins in a randomly chosen slice.

ThenX ∼ Poisson(n/60).

P(X = 0) = e−(n/60) (n/60)0

0!= e−(n/60) = 0.01

−n/60= ln0.01= −4.605, son = 276.3

The smallest value ofn for whichP(X = 0) ≤ 0.01 isn = 277.

17. Let λ1 be the mean number of chips per cookie in one of Mom’s cookies,and letλ2 be the mean numberof chips per cookie in one of Grandma’s cookies. LetX1 andX2 be the numbers of chips in two of Mom’s


164 CHAPTER 4

cookies, and letY1 andY2 be the numbers of chips in two of Grandma’s cookies. ThenXi ∼ Poisson(λ1) andYi ∼ Poisson(λ2). The observed values areX1 = 14,X2 = 11,Y1 = 6, andY2 = 8.

(a) The estimate isλ1 = X = (14+11)/2= 12.5.

(b) The estimate isλ2 = Y = (6+8)/2= 7.0.

(c) σX1 = σX2 =√

λ1. The uncertainty isσX =√

λ1/2. Replacingλ1 with λ1, σX =√

12.5/2= 2.5.

(d) σY1 = σY2 =√

λ2. The uncertainty isσY =√

λ2/2. Replacingλ2 with λ2, σY =√

7.0/2 = 1.9.

(e) The estimate isλ1− λ2 = 12.5−7.0= 5.5.

The uncertainty isσλ1−λ2=√

σ2λ1

+ σ2λ2

=√

6.25+3.5= 3.1.

λ1−λ2 = 5.5±3.1

18. If the mean decay rate is exactly 1 per second, then the mean number of events in 10 seconds is 10, soX ∼ Poisson(10).

(a) P(X ≤ 1) = P(X = 0)+P(X = 1) = e−10100

0!+e−10101

1!= 4.994×10−4

(b) Yes. If the mean rate is 1 event per second, then there willbe 1 or fewer events in 10 seconds only about 5times in 10,000.

(c) Yes, because 1 event in 10 seconds is an unusually small number if the mean rate is 1 event per second.

(d) P(X ≤ 8) = ∑8x=0P(X = x) = ∑8

x=0e−1010x

x!= 0.3328

(e) No. If the mean rate is 1 event per second, then there will be 8 or fewer events in 10 seconds about one-third ofthe time.

(f) No, because 8 events in 10 seconds is not an unusually small number if the mean rate is 1 event per second.

19. If the mean number of particles is exactly 7 per mL, thenX ∼ Poisson(7).

(a) P(X ≤ 1) = P(X = 0)+P(X = 1) = e−770

0!+e−771

1!= 7.295×10−3


SECTION 4.3 165

(b) Yes. If the mean concentration is 7 particles per mL, thenonly about 7 in every thousand 1 mL samples willcontain 1 or fewer particles.

(c) Yes, because 1 particle in a 1 mL sample is an unusually small number if the mean concentration is 7 particlesper mL.

(d) P(X ≤ 6) = ∑6x=0P(X = x) = ∑6

x=0e−77x

x!= 0.4497

(e) No. If the mean concentration is 7 particles per mL, then about 45% of all 1 mL samples will contain 6 orfewer particles.

(f) No, because 6 particles in a 1 mL sample is not an unusuallysmall number if the mean concentration is 7particles per mL.

20. LetλB represent the background rate, letλS represent the source rate, and letλT = λB + λS represent the totalrate, all in particles per second.

(a) LetX be the number of background events counted in 100 seconds. ThenX ∼ Poisson(100λB).

The observed value ofX is 36. The estimated background rate isλB = X/100= 36/100= 0.36.

The uncertainty isσλB=√

λB/100. ReplacingλB with λB yieldsσλB=√

0.36/100= 0.06.

λB = 0.36±0.06

(b) LetY be the number of events, both background and source, countedin 100 seconds.

ThenY ∼ Poisson(100λT).

The observed value ofY is 324. The estimated total rate isλT = Y/100= 324/100= 3.24.

The uncertainty isσλT=√

λT/100. ReplacingλT with λT yieldsσλT=√

3.24/100= 0.18.

λT = 3.24±0.18

(c) λS = λT −λB, soλS = λT − λB = 3.24−0.36= 2.88.

Now λT andλB are independent, because they are based on independent measurements (X andY).

ThereforeσλS=√

σ2λT

+ σ2λB

=√

(0.06)2+(0.18)2 =√

0.0036+0.0324= 0.19.

λS = 2.88±0.19

(d) If background and source plus background are each counted for 150 seconds,

thenσ2λB

= λB/150 andσ2λT

= λT/150.


166 CHAPTER 4

ApproximatingλB with λB = 0.36,σ2λB

= 0.36/150= 0.0024.

ApproximatingλT with λT = 3.24,σ2λT

= 3.24/150= 0.0216.

σλS=√

σ2λT

+ σ2λB

=√

0.0024+0.0216= 0.15

If background is counted for 100 seconds and source plus background is counted for 200 seconds, then

σ2λB

= λB/100 andσ2λT

= λT/200.

ApproximatingλB with λB = 0.36,σ2λB

= 0.36/100= 0.0036.

ApproximatingλT with λT = 3.24,σ2λT

= 3.24/200= 0.0162.

σλS=√

σ2λT

+ σ2λB

=√

0.0036+0.0162= 0.14. Counting background for 100 seconds and source plus back-

ground for 200 seconds produces the smaller uncertainty.

(e) It is not possible. Assume that the background emissionsare counted for 100 seconds, and the source plus back-ground is counted forN seconds. Then the uncertainty in the estimate of the source rate is

√0.36/100+3.24/N.

No matter how largeN is, this quantity is always greater than√

0.36/100= 0.06, and thus greater than 0.03.

21. LetX be the number of flaws in a one-square-meter sheet of aluminum. Let λ be the mean number of flaws persquare meter. ThenX ∼ Poisson(λ).

Let p be the probability that a one-square-meter sheet of aluminum has exactly one flaw.

Thenp = P(X = 1) = e−λ λ1

1!= λe−λ. From Example 4.27,λ = 2, andσλ

2 = 0.02.

Thereforep is estimated withp = λe−λ = 2e−2 = 0.271.

dp

dλ= e−λ − λe−λ = e−2−2e−2 = −0.135335

σp =

∣∣∣∣dp

dλ

∣∣∣∣σλ = 0.135335√

0.02= 0.019

p = 0.271±0.019

Section 4.4

1. LetX be the number of cars with serious problems among the five chosen. ThenX ∼ H(15,5,6).


SECTION 4.4 167

P(X = 2) =

(52

)(15−56−2

)

(156

) = 0.4196

2. LetX be the number of defective items among the five chosen.

(a) If there are five defective items in the shipment, thenX ∼ H(40,5,5).

P(X < 2) = P(X = 0)+P(X = 1) =

(50

)(40−55−0

)

(405

) +

(51

)(40−55−1

)

(405

) = 0.8912

(b) If there are ten defective items in the shipment, thenX ∼ H(40,10,5).

P(X ≥ 2) = 1−P(X < 2) = 1−P(X = 0)−P(X = 1) = 1−

(100

)(40−105−0

)

(405

) −

(101

)(40−105−1

)

(405

) = 0.3669

3. LetX be the number of the day on which the computer crashes. ThenX ∼ Geom(0.1).

P(X = 12) = (0.1)(0.9)11 = 0.0314

4. Let X be the number days that pass up to and including the first time the car encounters a red light. ThenX ∼ Geom(0.4).

(a) P(X = 3) = (0.4)(0.6)2 = 0.144

(b) P(X ≤ 3) = ∑3x=0P(X = x) = ∑3

x=0(0.4)(0.6)x = 0.784

Alternatively,P(X ≤ 3) = 1−P(X > 3) = 1−P(no red light for first 3 days) = 1− (0.6)3 = 0.784.


168 CHAPTER 4

(c) µX = 1/0.4= 2.5

(d) σ2X = (1−0.4)/(0.42) = 3.75

5. (a)Y ∼ NB(3,0.4). P(Y = 7) =

(7−13−1

)(0.4)3(1−0.4)7−3 = 0.1244

(b) µY = 3/0.4 = 7.5

(c) σ2Y = 3(1−0.4)/(0.42) = 11.25

6. Let X1 be the number of green lights,X2 be the number of yellow lights, andX3 be the number of red lightsencountered. Then(X1,X2,X3) ∼ MN(10,0.5,0.1,0.4).

P(X1 = 4,X2 = 1,X3 = 5) =10!

4!1!5!(0.5)4(0.1)1(0.4)5 = 0.0806

7. (iv). P(X = x) = p(1− p)x−1 for x≥ 1. Since 1− p < 1, this quantity is maximized whenx = 1.

8. LetX be then number of packages that will be filled before the process is stopped.ThenX ∼ Geom(0.01).

(a) µX = 1/0.01= 100

(b) σ2X = (1−0.01)/(0.012) = 9900

(c) LetY be the number of packages up to and including the fourth one whose weight falls outside the specification.ThenY ∼ NB(4,0.01).

µY = 4/0.01= 400, σ2 = 4(1−0.01)/(0.012) = 39,600


SECTION 4.4 169

9. LetX be the number of hours that have elapsed when the fault is detected.ThenX ∼ Geom(0.8).

(a) P(X ≤ 3) = P(X = 1)+P(X = 2)+P(X = 3) = (0.8)(1−0.8)0+(0.8)(1−0.8)1+(0.8)(1−0.8)2 = 0.992

(b) P(X = 3|X > 2) =P(X = 3∩X > 2)

P(X > 2)=

P(X = 3)

P(X > 2)

Now P(X = 3) = 0.8(1−0.8)2 = 0.032, andP(X > 2) = 1−P(X ≤ 2) = 1−P(X = 1)−P(X = 2) = 1−0.8−0.8(1−0.8)= 0.04.

ThereforeP(X = 3|X > 2) =0.0320.04

= 0.8.

(c) µX = 1/0.8= 1.25

10. LetX be the number of runs up to and including the fifth failure.ThenX ∼ NB(5,0.001).

(a) µX = 5/0.001= 5000

(b) σX =

√5(1−0.001)

0.0012 = 2234.95

11. (a)X ∼ H(10,3,4). P(X = 2) =

(32

)(10−34−2

)

(104

) = 0.3

(b) µX = 4(3/10) = 1.2

(c) σX =

√

4

(310

)(710

)(10−410−1

)= 0.7483

12. (a)X ∼ H(500,100,20), soP(X = 5) =

(1005

)(500−100

20−5

)

(50020

) =

100!5!95!

400!15!385!

500!20!480!

.


170 CHAPTER 4

(b) The binomial approximation isX ∼ Bin(20,0.2). SoP(X = 5) ≈ 20!5!(20−5)!

(0.2)5(1−0.2)20−5 = 0.1746.

13. (a)X ∼ H(60,5,10), so

P(X > 1) = 1−P(X = 0)−P(X = 1) = 1−

(50

)(60−510−0

)

(6010

) −

(51

)(60−510−1

)

(6010

) = 0.1904

(b) X ∼ H(60,10,10), so

P(X > 1) = 1−P(X = 0)−P(X = 1) = 1−

(100

)(60−1010−0

)

(6010

) −

(101

)(60−1010−1

)

(6010

) = 0.5314

(c) X ∼ H(60,20,10), so

P(X > 1) = 1−P(X = 0)−P(X = 1) = 1−

(200

)(60−2010−0

)

(6010

) −

(201

)(60−2010−1

)

(6010

) = 0.9162

14. (a)(X1,X2,X3,X4) ∼ MN(15,0.20,0.30,0.15,0.35)

P(X1 = 3,X2 = 4,X3 = 2,X4 = 6) =15!

3!4!2!6!(0.20)3(0.30)4(0.15)2(0.35)6 = 0.0169

(b) X1 ∼ Bin(15,0.2). P(X1 = 3) =15!

3!(15−3)!(0.2)3(1−0.2)15−3 = 0.2501

15. Let X1 denote the number of orders for a 2.8 liter engine, letX2 denote the number of orders for a 3.0 literengine, letX3 denote the number of orders for a 3.3 liter engine, and letX4 denote the number of orders for a3.8 liter engine.

(a) (X1,X2,X3,X4) ∼ MN(20,0.1,0.4,0.3,0.2)

P(X1 = 3,X2 = 7,X3 = 6,X4 = 4) =20!

3!7!6!4!(0.1)3(0.4)7(0.3)6(0.2)4 = 0.0089


SECTION 4.5 171

(b) X3 ∼ Bin(20,0.4). P(X3 > 10) = ∑20x=11

20!x!(20−x)!

(0.4)x(1−0.4)20−x = 0.1275

Alternatively, use Table A.1 to findP(X3 ≤ 10) = 0.872. ThenP(X3 > 10) = 1−P(X ≤ 10) = 0.128

16. (a) LetX1 be the number of readings that are within 0.1C of the true temperature, letX2 be the number of readingsthat are more than 0.1C above the true temperature, letX3 be the number of readings that are more than 0.1Cbelow the true temperature. Then(X1,X2,X3) ∼ MN(10,0.7,0.1,0.2).

P(X1 = 5,X2 = 2,X3 = 3) =10!

5!2!3!(0.7)5(0.1)2(0.2)3 = 0.0339

(b) X1 ∼ Bin(10,0.7)

P(X1 > 8) = P(X = 9)+P(X = 10)

=10!

9!(10−9)!(0.7)9(1−0.7)10−9+

10!10!(10−10)!

(0.7)10(1−0.7)10−10

= 0.1493

17. P(X = n) = p(1− p)n−1. P(Y = 1) =

(n1

)p1(1− p)n−1 = np(1− p)n−1. SoP(X = n) = (1/n)P(Y = 1).

18. LetY ∼ Bin(10,0.3). From Table A.1,P(Y = 1) = 0.121, soP(X = 10) = (1/10)(0.121) = 0.0121.

Section 4.5

1. (a) Using Table A.2: 1−0.1977= 0.8023

(b) Using Table A.2: 0.9032−0.6554= 0.2478

(c) Using Table A.2: 0.8159−0.3821= 0.4338


172 CHAPTER 4

(d) Using Table A.2: 0.0668+(1−0.3264)= 0.7404

2. (a) Using Table A.2: 0.7123

(b) Using Table A.2: 0.0197−0.0017= 0.0180

(c) Using Table A.2: 0.7580−0.1401= 0.6179

(d) Using Table A.2: 0.8315+(1−0.9474)= 0.8841

3. (a) Using Table A.2:c = 1

(b) Using Table A.2:c = −2.00

(c) Using Table A.2:c = 1.50

(d) Using Table A.2:c = 0.83

(e) Using Table A.2:c = 1.45

4. (a)z= (2−2)/√

9 = 0. The area to the right ofz= 0 is 0.5000.

(b) For 1,z= (1−2)/√

9 = −0.33. For 7,z= (7−2)/√

9 = 1.67.

The area betweenz= −0.33 andz= 1.67 is 0.9525−0.3707= 0.5818.

(c) For−2.5, z= (−2.5−2)/√

9 = −1.50. For−1, z= (−1−2)/√

9 = −1.00.

The area betweenz= −1.50 andz= −1.00 is 0.1587−0.0668= 0.0919.

(d) P(−3≤ X−2 < 3) = P(−1≤ X < 5).

For−1, z= (−1−2)/√

9 = −1.00. For 5,z= (5−2)/√

9 = 1.00.



SECTION 4.5 173

5. (a)z= (19−16)/2= 1.50. The area to the right ofz= 1.50 is 0.0668.

(b) Thez-score of the 10th percentile is≈−1.28.

The 10th percentile is therefore≈ 16−1.28(2) = 13.44.

(c) z= (14.5−16)/2= −0.75. The area to the right ofz= −0.75 is 0.2266.

Therefore a lifetime of 14.5 is on the 23rd percentile, approximately.

(d) For 14.5,z= (14.5−16)/2= −0.75. For 17,z= (17−16)/2= 0.5.


6. (a)z= (100−99.80)/0.1= 2.00. The area to the right ofz= 2.00 is 0.0228.

(b) For 99.95,z= (99.95−99.80)/0.1= 1.50. For 100.05,z= (100.05−99.80)/0.1= 2.50.

The area betweenz= 1.50 andz= 2.50 is 0.9938−0.9332= 0.0606.

7. (a)z= (700−480)/90= 2.44. The area to the right ofz= 2.44 is 0.0073.


The 25th percentile is therefore≈ 480−0.67(90)= 419.7.

(c) z= (600−480)/90= 1.33. The area to the left ofz= 1.33 is 0.9082.

Therefore a score of 600 is on the 91st percentile, approximately.

(d) For 420,z= (420−480)/90= −0.67. For 520,z= (520−480)/90= 0.44.


8. (a) For 60,z= (60−64.3)/2.6= −1.65. For 66,z= (66−64.3)/2.6= 0.65.



174 CHAPTER 4

(b) The area to the right ofz= 0.5 is 1−0.6915= 0.3085.

The proportion of women who are taller is 0.3085.

(c) Thez-score of the 90th percentile is≈ 1.28.

The 90th percentile is therefore≈ 64.3+1.28(2.6)= 67.63 inches.

(d) z= (67−64.3)/2.6= 1.04. The area to the right ofz= 1.04 is 1−0.8508= 0.1492.

(e) From part (d), the probability that a randomly chosen woman is more than 67 inches tall is 0.1492.

Let X be the number of women out of the five chosen who are more than 67inches tall.

ThenX ∼ Bin(5,0.1492) soP(X = 1) =5!

1!(5−1)!(0.1492)1(1−0.1492)5−1 = 0.3909.

9. (a)z= (12−10)/1.4= 1.43. The area to the right ofz= 1.43 is 1−0.9236= 0.0764.


The 25th percentile is therefore≈ 10−0.67(1.4)= 9.062 GPa.

(c) Thez-score of the 95th percentile is≈ 1.645.

The 25th percentile is therefore≈ 10+1.645(1.4)= 12.303 GPa.

10. No. The maximum possible score of 800 is 1.5 standard deviations above the mean of 650. If the populationwere normal, about 7% of the scores would therefore be above 800. In fact, no scores are above 800. Thereforethe population cannot be normal.

11. (a)z= (6−4.9)/0.6= 1.83. The area to the right ofz= 1.83 is 1−0.9664= 0.0336.

The process will be shut down on 3.36% of days.


SECTION 4.5 175

(b) z= (6−5.2)/0.4= 2.00. The area to the right ofz= 2.00 is 1−0.9772= 0.0228.

Since a process with this broth will be shut down on 2.28% of days, this broth will result in fewer days ofproduction lost.

12. (a)z= (0.30−0.55)/0.15= −1.67. The area to the left ofz= −1.67 is 0.0475.

Since 0.0475 is less than 0.0500 = 5%, this process meets the requirement.

(b) Letσ be the required standard deviation. Since the recovery willbe greater than 0.50 95% of the time, the value0.50 is the 5th percentile of the distribution. Thez-score of the 5th percentile is−1.645. Since the mean is 0.60,the standard deviation satisfies the equation−1.645= (0.50−0.60)/σ. Solving forσ yieldsσ = 0.06079.

13. Let X be the diameter of a hole, and letY be the diameter of a piston. ThenX ∼ N(15, 0.0252), andY ∼ N(14.88, 0.0152). The clearance is given byC = 0.5X−0.5Y.

(a) µC = µ0.5X−0.5Y = 0.5µX −0.5µY = 0.5(15)−0.5(14.88)= 0.06 cm

(b) σC =√

0.52σ2X +(−0.5)2σ2

Y =√

0.52(0.0252)+0.52(0.0152) = 0.01458 cm

(c) SinceC is a linear combination of normal random variables, it is normally distributed, withµC andσC as givenin parts (a) and (b).

Thez-score of 0.05 is(0.05−0.06)/0.01458= −0.69. The area to the left ofz= −0.69 is 0.2451.

ThereforeP(C < 0.05) = 0.2451.

(d) Thez-score of the 25th percentile is≈−0.67.

The 25th percentile is therefore≈ 0.06−0.67(0.01458)= 0.0502 cm.

(e) Thez-score of 0.05 isz= (0.05−0.06)/0.01458=−0.69. Thez-score of 0.09 isz = (0.09−0.06)/0.01458= 2.06.


ThereforeP(0.05< C < 0.09) = 0.7352.

(f) The probability is maximized whenµC = 0.07 cm, the midpoint between 0.05 and 0.09 cm. NowµY = 14.88,and µX must be adjusted so thatµC = 0.5µX − 0.5µY = 0.07. SubstitutingµC = 0.07 andµY = 14.88, andsolving forµX yieldsµX = 15.02 cm.


176 CHAPTER 4

To find the probability that the clearance will be between 0.05 and 0.09:

Thez-score of 0.05 isz= (0.05−0.07)/0.01458= −1.37.

Thez-score of 0.09 isz= (0.09−0.07)/0.01458= 1.37.


ThereforeP(0.05< C < 0.09) = 0.8294.

14. (a) The specification interval is 0.645 to 0.655 cm.

Thez-score of 0.645 is(0.645−0.652)/0.003= −2.33.

Thez-score of 0.655 is(0.655−0.652)/0.003= 1.00.


Therefore 83.14% of shafts will meet the specification.

(b) Thez-score of 0.645 is(0.645−0.650)/0.003= −1.67.

Thez-score of 0.655 is(0.655−0.650)/0.003= 1.67.


Therefore 90.50% of shafts will meet the specification.

(c) Let σ be the required standard deviation. Since the mean of 0.650 is in the center of the specification interval,the proportion of diameters that are too thin will be equal tothe proportion of diameters that are too thick.Thereforeσ must be chosen so that the proportion of diameters that are greater than 0.655 is 0.005.

Thez-score for which 0.005 of the area is to the right is approximately z= 2.58. Therefore we findσ so thatthez-score of 0.655 is 2.58 by solving 2.58= (0.655−0.650)/σ, soσ = 0.00194 cm.

Note that ifz= 2.57 is used instead ofz= 2.58, thenσ = 0.00195 cm.

15. (a) Thez-score of 12 is(12−12.05)/0.03= −1.67. The area to the left ofz= −1.67 is 0.0475. The proportion is0.0475.

(b) Letµ be the required value of the mean. This value must be chosen sothat the 1st percentile of the distributionis 12. Thez-score of the 1st percentile is approximatelyz= −2.33. Therefore−2.33= (12−µ)/0.03. Solvingfor µ yieldsµ= 12.07 ounces.

(c) Let σ be the required standard deviation. The value ofσ must be chosen so that the 1st percentile of thedistribution is 12. Thez-score of the 1st percentile is approximatelyz = −2.33. Therefore−2.33 =(12−12.05)/σ. Solving forσ yieldsσ = 0.0215 ounces.


SECTION 4.5 177

16. (a) Thez-score of 90 is(90−110)/10= −2.00. The area to the left ofz= −2.00 is 0.0228. The proportion is0.0228.

(b) Letµ be the required value of the mean. This value must be chosen sothat the 1st percentile of the distributionis 90. Thez-score of the 1st percentile is approximatelyz= −2.33. Therefore−2.33= (90−µ)/10. Solvingfor µ yieldsµ= 113.3 microns.

(c) Let σ be the required standard deviation. The value ofσ must be chosen so that the 1st percentile of thedistribution is 90. Thez-score of the 1st percentile is approximatelyz = −2.33. Therefore−2.33= (90−110)/σ. Solving forσ yieldsσ = 8.584 microns.

17. (a) The proportion of strengths that are less than 65 is 0.10. Therefore 65 is the 10th percentile of strength. Thez-score of the 10th percentile is approximatelyz= −1.28. Letσ be the required standard deviation. The valueof σ must be chosen so that thez-score for 65 is−1.28. Therefore−1.28= (65−75)/σ. Solving forσ yieldsσ = 7.8125 N/m2.

(b) Let σ be the required standard deviation. The value ofσ must be chosen so that the 1st percentile of thedistribution is 65. Thez-score of the 1st percentile is approximatelyz = −2.33. Therefore−2.33 =(65−75)/σ. Solving forσ yieldsσ = 4.292 N/m2.

(c) Letµ be the required value of the mean. This value must be chosen sothat the 1st percentile of the distributionis 65. Thez-score of the 1st percentile is approximatelyz= −2.33. Therefore−2.33= (65−µ)/5. Solvingfor µ yieldsµ= 76.65 N/m2.

18. LetRbe the length of a rod.

(a) SinceX = R+3,µX = µR+3 = µR+3 = 10+3= 13 cm

(b) SinceX = R+3, σ2X = σ2

R+3 = σ2R = 0.12 = 0.01 cm2

(c) SinceR∼ N(10,0.12) andX = R+3,X ∼ N(13,0.12)

19. Leta = 1/σ and letb= −µ/σ. ThenZ = aX+b. Now aµ+b= (1/σ)µ−µ/σ = 0, anda2σ2 = (1/σ)2σ2 = 1.Equation (4.25) now shows thatZ ∼ N(0, 1).


178 CHAPTER 4

20. (a) Thez-score of 160 is(160−200)/10= −4.00. The area to the left ofz= −4.00 is negligible.

ThereforeP(X ≤ 160) ≈ 0.

(b) Yes. If the procedure is functioning correctly, values of 160 N or less would almost never occur.

(c) Yes, because 160 N is unusually small if the process is functioning correctly.

(d) Thez-score of 203 is(203−200)/10= 0.30. The area to the right ofz= 0.30 is 1−0.6179= 0.3821.

ThereforeP(X ≥ 203) = 0.3821.

(e) No. If the procedure is functioning correctly, values of203 N or more would occur about 38% of the time.

(f) No, because 203 N is not unusually large if the process is functioning correctly.

(g) Thez-score of 195 is(195−200)/10= −0.50. The area to the left ofz= −0.50 is 0.3085.

ThereforeP(X ≤ 195) = 0.3085.

(h) No. If the procedure is functioning correctly, values of195 N or less would occur about 31% of the time.

(i) No, because 195 N is not unusually small if the process is functioning correctly.

21. LetX be the lifetime of bulb A and letY be the lifetime of bulb B.

ThenX ∼ N(800, 1002) andY ∼ N(900, 1502).

(a) LetD = Y−X. The event that bulb B lasts longer than bulb A is the event that D > 0.

µD = µY −µX = 900−800= 100.

SinceX andY are independent,σD =√

σ2X + σ2

Y =√

1002+1502 = 180.278.

SinceD is a linear combination of independent normal random variables,D is normally distributed.

Thez-score of 0 is(0−100)/180.278= −0.55. The area to the right ofz= −0.55 is 1−0.2912= 0.7088.

ThereforeP(D > 0) = 0.7088.

(b) LetD = Y−X. The event that bulb B lasts longer than bulb A is the event that D > 200.

µD = µY −µX = 900−800= 100.


SECTION 4.5 179

SinceX andY are independent,σD =√

σ2X + σ2

Y =√

1002+1502 = 180.278.

SinceD is a linear combination of independent normal random variables,D is normally distributed.

Thez-score of 200 is(200−100)/180.278= 0.55.

The area to the right ofz= 0.55 is 1−0.7088= 0.2912.

ThereforeP(D > 200) = 0.2912.

(c) LetS= X +Y be the total lifetime of the two bulbs. It is necessary to findP(S> 2000).

µS = µX +µY = 800+900= 1700.

SinceX andY are independent,σS =√

σ2X + σ2

Y =√

1002+1502 = 180.278.

SinceS is a linear combination of independent normal random variables,S is normally distributed.

Thez-score of 2000 is(2000−1700)/180.278= 1.66.


ThereforeP(S> 2000) = 0.0485.

22. (a) SinceM is a linear combination of independent normal random variables,M is normally distributed.

µM = 0.5µX +µY = 0.5(0.450)+0.250= 0.475.

σ2M = 0.52σ2

X + σ2Y = 0.52(0.0502)+0.0252 = 0.00125.

ThereforeM ∼ N(0.475, 0.00125).

(b) Thez-score of 0.5 is(0.5−0.475)/√

0.00125= 0.71.


ThereforeP(M > 0.5) = 0.2389.

23. (a) Ifm= 0, thenX = E, soX ∼ N(0,0.25).

P(error) = P(X > 0.5). Thez-score of 0.5 is(0.5−0)/√

0.25= 1.00.


ThereforeP(error) = 0.1587.

(b) If m= 0, thenX = E, soX ∼ N(0,σ2).

P(error) = P(X > 0.5) = 0.01. Thez-score of 0.5 is(0.5−0)/σ.

SinceP(X > 0.5) = 0.01, thez-score of 0.5 is 2.33.

Therefore(0.5−0)/σ = 2.33. Solving forσ yieldsσ = 0.2146, andσ2 = 0.04605.


180 CHAPTER 4

24. (a) Ifm= 0, thenX = −1.5+E, soX ∼ N(−1.5,0.64).

P(error) = P(X > 0.5). Thez-score of 0.5 is[0.5− (−1.5)])/√

0.64= 2.50.



(b) If m= 1, thenX = 1.5+E, soX ∼ N(1.5,0.64).

P(error) = P(X ≤ 0.5). Thez-score of 0.5 is(0.5−1.5)/√

0.64= −1.25.

The area to the left ofz= −1.25 is 0.1056.


(c) LetSbe the value from the sent string (0 or 1). LetC be the event that the value is received correctly.

From part (a) we know thatP(C|S= 0) = 1−0.0062= 0.9938.From part (b) we know thatP(C|S= 1) = 1−0.1056= 0.8944.Now P(C) = P(C|S= 0)P(S= 0)+P(C|S= 1)P(S= 1) = (0.9938)(0.4)+ (0.8944)(0.6)= 0.9342.

(d) LetSbe the value from the sent string (0 or 1), and letR be the value received. We must findP(S= 0|R= 1).

P(S= 0|R= 1) =P(R= 1|S= 0)P(S= 0)

P(R= 1|S= 0)P(S= 0)+P(R= 1|S= 1)P(S= 1)

From part (a) we know thatP(R= 1|S= 0) = 0.0062.From part (b) we know thatP(R= 1|S= 1) = 1−0.1056= 0.8944.

ThereforeP(S= 0|R= 1) =(0.0062)(0.4)

(0.0062)(0.4)+ (0.8944)(0.6)= 0.00460.

(e) LetSbe the value from the sent string (0 or 1), and letR be the value received. We must findP(S= 1|R= 0).

P(S= 1|R= 0) =P(R= 0|S= 1)P(S= 1)

P(R= 0|S= 1)P(S= 1)+P(R= 0|S= 0)P(S= 0)

From part (a) we know thatP(R= 0|S= 0) = 1−0.0062= 0.9938.From part (b) we know thatP(R= 0|S= 1) = 0.1056.

ThereforeP(S= 1|R= 0) =(0.1056)(0.6)

(0.1056)(0.6)+ (0.9938)(0.4)= 0.1375.

25. (a) The sample mean is 114.8 J and the sample standard deviation is 5.006 J.

(b) Thez-score of 100 is(100−114.8)/5.006= −2.96. The area to the left ofz= −2.96 is 0.0015. Thereforeonly 0.15% of bolts would have breaking torques less than 100J, so the shipment would be accepted.

(c) The sample mean is 117.08 J; the sample standard deviation is 8.295 J. Thez-score of 100 is(100−117.08)/8.295=−2.06. The area to the left ofz=−2.06 is 0.0197. Therefore about 2% of the bolts would have breaking torquesless than 100 J, so the shipment would not be accepted.


SECTION 4.6 181

(d) The bolts in part (c) are stronger. In fact, the weakest bolt in part (c) is stronger than the weakest bolt in part (a),the second-weakest bolt in part (c) is stronger than the second-weakest bolt in part (a), and so on.

(e) The method is certainly not valid for the bolts in part (c). This sample contains an outlier (140), so the normaldistribution should not be used.

26.

k P(|X−µX| ≥ kσX) 1/k2

1 0.3174 12 0.0456 0.253 0.0026 0.111

The actual probabilities are much smallerthan the Chebyshev bounds.

Section 4.6

1. LetY be the lifetime of the component.

(a) E(Y) = eµ+σ2/2 = e1.2+(0.4)2/2 = 3.5966

(b) P(3 < Y < 6) = P(ln3 < lnY < ln6) = P(1.0986< lnY < 1.7918). lnY ∼ N(1.2,0.42).

Thez-score of 1.0986 is(1.0986−1.2)/0.4= −0.25.

Thez-score of 1.7918 is(1.7918−1.2)/0.4= 1.48.


ThereforeP(3 < Y < 6) = 0.5293.

(c) Letm be the median ofY. ThenP(Y ≤ m) = P(lnY ≤ lnm) = 0.5.

Since lnY ∼ N(1.2,0.42), P(lnY < 1.2) = 0.5. Therefore lnm= 1.2, som= e1.2 = 3.3201.

(d) Lety90 be the 90th percentile ofY. ThenP(Y ≤ y90) = P(lnY ≤ lny90) = 0.90.

Thez-score of the 90th percentile is approximatelyz= 1.28.

Therefore thez-score of lny90 must be 1.28, so lny90 satisfies the equation 1.28= (lny90−1.2)/0.4.

lny90 = 1.712, soy90 = e1.712= 5.540.


182 CHAPTER 4

2. LetY be the percent absorbed.

(a) E(Y) = eµ+σ2/2 = e1.5+(0.5)2/2 = 5.0784

(b) Letm be the median ofY. ThenP(Y ≤ m) = P(lnY ≤ lnm) = 0.5.

Since lnY ∼ N(1.5,0.52), P(lnY < 1.5) = 0.5. Therefore lnm= 1.5, som= e1.5 = 4.4817.

(c) P(Y > 10) = P(lnY > ln10) = P(lnY > 2.3026).

Thez-score of 2.3026 is(2.3026−1.5)/0.5= 1.61.

The area to the right ofz= 1.61 is 1−0.9463= 0.0537. ThereforeP(Y > 10) = 0.0537.

(d) P(Y < 5) = P(lnY < ln5) = P(lnY < 1.6094).

Thez-score of 1.6094 is(1.6094−1.5)/0.5= 0.22.

The area to the left ofz= 0.22 is 0.5871. ThereforeP(Y < 5) = 0.5871.

(e) Lety75 be the 75th percentile ofY. ThenP(Y ≤ y75) = P(lnY ≤ lny75) = 0.75.



lny75 = 1.835, soy75 = e1.835= 6.265.

(f) V(Y) = e2µ+2σ2 −e2µ+σ2= e2(1.5)+2(0.5)2 −e2(1.5)+(0.5)2

= 7.32511.

The standard deviation is√

V(Y) =√

e2(1.5)+2(0.5)2 −e2(1.5)+(0.5)2=√

7.32511= 2.7065.

3. LetY represent the BMI for a randomly chosen man aged 25–34.

(a) E(Y) = eµ+σ2/2 = e3.215+(0.157)2/2 = 25.212

(b) V(Y) = e2µ+2σ2 −e2µ+σ2= e2(3.215)+2(0.157)2−e2(3.215)+(0.157)2

= 15.86285.


V(Y) =√

e2(3.215)+2(0.157)2−e2(3.215)+(0.157)2=√

15.86285= 3.9828.

(c) Letm be the median ofY. ThenP(Y ≤ m) = P(lnY ≤ lnm) = 0.5.

Since lnY ∼ N(3.215,0.1572), P(lnY < 3.215) = 0.5.

Therefore lnm= 3.215, som= e3.215= 24.903.


SECTION 4.6 183

(d) P(Y < 22) = P(lnY < ln22) = P(lnY < 3.0910).

Thez-score of 3.0910 is(3.0910−3.215)/0.157= −0.79.


ThereforeP(Y < 22) = 0.2148.




lny75 = 3.3202, soy75 = e3.3202= 27.666.

4. LetY be the increase in the risk.

(a) E(Y) = eµ+σ2/2 = e−15.65+(0.79)2/2 = 2.1818×10−7

(b) Letm be the median ofY. ThenP(Y ≤ m) = P(lnY ≤ lnm) = 0.5.

Since lnY ∼ N(−15.65,0.792), P(lnY < −15.65) = 0.5.

Therefore lnm= −15.65, som= e−15.65 = 1.5970×10−7.

(c) V(Y) = e2µ+2σ2 −e2µ+σ2= e2(−15.65)+2(0.79)2−e2(−15.65)+(0.79)2

= 4.1250×10−14.


V(Y)=√

e2(−15.65)+2(0.79)2−e2(−15.65)+(0.79)2=√

4.1250×10−14= 2.0310× 10−7.

(d) Lety5 be the 5th percentile ofY. ThenP(Y ≤ y5) = P(lnY ≤ lny5) = 0.05.

Thez-score of the 5th percentile is approximatelyz= −1.645.

Therefore thez-score of lny5 must be−1.645,

so lny5 satisfies the equation−1.645= [ lny5− (−15.65)]/0.79.

lny5 = −16.9496, soy5 = e−16.9496= 4.354×10−8.



Therefore thez-score of lny95 must be 1.645,

so lny95 satisfies the equation 1.645= [ lny95− (−15.65)]/0.79.

lny95 = −14.35045, soy95 = e−14.35045= 5.857×10−7.


184 CHAPTER 4

5. (a) lnI ∼ N(1, 0.2), lnR∼ N(4, 0.1), andI andRare independent. Therefore lnV ∼ N(5,0.3).Since lnV is normal,V is lognormal, withµV = 5 andσ2

V = 0.3.

(b) P(V < 200) = P(lnV < ln200) = P(lnV < 5.298317).

Now lnV ∼ N(5,0.3), so thez-score of 5.298317 is(5.298317−5)/√

0.3 = 0.54.

The area to the left ofz= 0.54 is 0.7054.

ThereforeP(V < 200) = 0.7054.

(c) P(150< V < 300) = P(ln150< lnV < ln300) = P(5.010635< lnV < 5.703782).

Now lnV ∼ N(5,0.3), so thez-score of 5.010635 is(5.010635−5)/√

0.3 = 0.02, and thez-score of 5.703782is (5.703782−5)/

√0.3 = 1.28.

The area betweenz= 0.02 andz= 1.28 is 0.8997−0.5080= 0.3917.

ThereforeP(150< V < 300) = 0.3917.

(d) The mean ofV is E(V) = eµ+σ2/2 = e5+0.3/2 = 172.43.

(e) Since lnV ∼ N(5,0.3), the median of lnV is 5.

Therefore the median ofV is e5 = 148.41.

(f) The standard deviation ofV is√

e2µ+2σ2 −e2µ+σ2=√

e2(5)+2(0.3)−e2(5)−0.3 = 101.99.

(g) Letv10 be the 10th percentile ofV. ThenP(V ≤ v10) = P(lnV ≤ lnv10) = 0.10.


Therefore thez-score of lnv10 must be−1.28, so lnv10 satisfies the equation−1.28= ( lnv10−5)/√

0.3.

lnv10 = 4.2989, sov10 = e4.2989= 73.619.

(h) Letv90 be the 90th percentile ofV. ThenP(V ≤ v90) = P(lnV ≤ lnv90) = 0.90.


Therefore thez-score of lnv90 must be 1.28, so lnv90 satisfies the equation 1.28= ( lnv90−5)/√

0.3.

lnv90 = 5.7011, sov90 = e5.7011= 299.19.

6. The probability that a given circuit has a voltage less than 200 volts isP(V < 200) = 0.7054.

Let X be the number of circuits whose voltage is less than 200 volts. ThenX ∼ Bin(10,0.7054).

P(X ≥ 8) = P(X = 8)+P(X = 9)+P(X = 10)

=10!

8!(10−8)!(0.7054)8(1−0.7054)10−8+

10!9!(10−9)!

(0.7054)9(1−0.7054)10−9

+10!

10!(10−10)!(0.7054)10(1−0.7054)10−10

= 0.3973


SECTION 4.6 185

7. LetX represent the withdrawal strength for a randomly chosen annularly threaded nail, and letY represent thewithdrawal strength for a randomly chosen helically threaded nail.

(a) E(X) = e3.82+(0.219)2/2 = 46.711 N/mm

(b) E(Y) = e3.47+(0.272)2/2 = 33.348 N/mm

(c) First find the probability for annularly threaded nails.

P(X > 50) = P(lnX > ln50) = P(lnX > 3.9120).

Thez-score of 3.9120 is(3.9120−3.82)/0.219= 0.42.


Therefore the probability for annularly threaded nails is 0.3372.

Now find the probability for helically threaded nails.

P(Y > 50) = P(lnY > ln50) = P(lnY > 3.9120).

Thez-score of 3.9120 is(3.9120−3.47)/0.272= 1.63.


Therefore the probability for helically threaded nails is 0.0516.

Annularly threaded nails have the greater probability. Theprobability is 0.3372 vs. 0.0516 for helically threadednails.

(d) First find the median strength for annularly threaded nails.

Let m be the median ofX. ThenP(X ≤ m) = P(lnX ≤ lnm) = 0.5.

Since lnX ∼ N(3.82,0.2192), P(lnY < 3.82) = 0.5.

Therefore lnm= 3.82, som= e3.82.

Now P(Y > e3.82) = P(lnY > lne3.82) = P(lnY > 3.82).

Thez-score of 3.82 is(3.82−3.47)/0.272= 1.29.

The area to the right ofz= 1.29 is 1−0.9015= 0.0985. Therefore the probability is 0.0985.

(e) The log of the withdrawal strength for this nail is ln20= 2.996.

For annularly threaded nails, thez-score of 2.996 is(2.996−3.82)/0.219= −3.76.

The area to the left ofz= −3.76 is less than 0.0001.

Therefore a strength of 20 is extremely small for an annularly threaded nail; less than one in ten thousand suchnails have strengths this low.

For helically threaded nails, thez-score of 2.996 is(2.996−3.47)/0.272= −1.74.


Therefore about 4.09% of helically threaded nails have strengths of 20 or below.


186 CHAPTER 4

We can be pretty sure that it was a helically threaded nail. Only about 0.01% of annularly threaded nails havestrengths as small as 20, while about 4.09% of helically threaded nails do.

8. i. Since the lognormal population is skewed to the right, the mean will always be larger than the median. Moreprecisely, the mean iseµ+σ2/2, while the median iseµ. Sinceσ2/2 > 0, the mean is greater than the median.

9. LetX represent the price of a share of company A one year from now. LetY represent the price of a share ofcompany B one year from now.

(a) E(X) = e0.05+(0.1)2/2 = $1.0565

(b) P(X > 1.20) = P(lnX > ln1.20) = P(lnX > 0.1823).

Thez-score of 0.1823 is(0.1823−0.05)/0.1= 1.32.


ThereforeP(X > 1.20) = 0.0934.

(c) E(Y) = e0.02+(0.2)2/2 = $1.0408

(d) P(Y > 1.20) = P(lnY > ln1.20) = P(lnY > 0.1823).

Thez-score of 0.1823 is(0.1823−0.02)/0.2= 0.81.


ThereforeP(Y > 1.20) = 0.2090.

10. (a)P(X < 20) = P(lnX < ln20) = P(lnX < 2.996).

Thez-score of 2.996 is(2.996−5)/0.5= −4.01.

The area to the left ofz= −4.01 is negligible.

ThereforeP(X < 20) ≈ 0.

(b) Yes. Almost none of the specimens will have tensile strengths less than 20 MPa if the claim is true.

(c) Yes, because 20 MPa is unusually small if the claim is true.


SECTION 4.7 187

(d) P(X < 130) = P(lnX < ln130) = P(lnX < 4.868).

Thez-score of 4.868 is(4.868−5)/0.5= −0.26.


ThereforeP(X < 130) = 0.3974.

(e) No. Nearly 40% of the specimens will have tensile strengths less than 130 MPa if the claim is true.

(f) No, because 130 MPa is not unusually small if the claim is true.

11. lnX1, ..., lnXn are independent normal random variables, so lnP = a1 lnX1 + · · ·+ an lnXn is a normal randomvariable. It follows thatP is lognormal.

Section 4.7

1. (a)µT = 1/0.45= 2.2222

(b) σ2T = 1/(0.452) = 4.9383

(c) P(T > 3) = 1−P(T ≤ 3) = 1− (1−e−0.45(3)) = 0.2592

(d) Letm be the median. ThenP(T ≤ m) = 0.5.

P(T ≤ m) = 1−e−0.45m = 0.5, soe−0.45m = 0.5.

Solving form yieldsm= 1.5403.

2. (a)λ = 1/µ= 1/2 = 0.5 year−1

(b) LetT be the lifetime of a fuse, and letm be the median ofT. ThenP(T ≤ m) = 0.5.

P(T ≤ m) = 1−e−0.5m = 0.5, soe−0.5m = 0.5.

Solving form yieldsm= 1.3863 years.


188 CHAPTER 4

(c) σ = 1/λ = 1/0.5= 2 years

(d) LetT be the lifetime of a fuse, and lett60 be the 60th percentile ofT. ThenP(T ≤ m) = 0.60.

P(T ≤ m) = 1−e−0.5m = 0.6, soe−0.5m = 0.4.

Solving form yieldsm= 1.8326 years.

(e) P(T > 5) = 1−P(T ≤ 5) = 1− (1−e−0.5(5)) = 0.0821

(f) The probability isP(T > 3|T > 1). By the lack of memory property,P(T > 3|T > 1) = P(T > 2).

P(T > 2) = 1−P(T ≤ 2) = 1− (1−e−0.5(2)) = 0.3679

3. LetX be the diameter in microns.

(a) µX = 1/λ = 1/0.25= 4 microns

(b) σX = 1/λ = 1/0.25= 4 microns

(c) P(X < 3) = 1−e−0.25(3) = 0.5276

(d) P(X > 11) = 1− (1−e−0.25(11)) = 0.0639

(e) Letm be the median. ThenP(T ≤ m) = 0.5.

P(T ≤ m) = 1−e−0.25m = 0.5, soe−0.25m = 0.5.

Solving form yieldsm= 2.7726 microns.

(f) Let x75 be the 75th percentile, which is also the third quartile. Then P(T ≤ x75) = 0.75.

P(T ≤ x75) = 1−e−0.25x75 = 0.75, soe−0.25x75 = 0.25.

Solving forx75 yieldsx75 = 5.5452 microns.

(g) Letx99 be the 99th percentile. ThenP(T ≤ x99) = 0.75.

P(T ≤ x99) = 1−e−0.25x99 = 0.99, soe−0.25x99 = 0.01.

Solving forx99 yieldsx99 = 18.4207 microns.


SECTION 4.7 189

4. LetT be the lifetime of a transistor, in months.

(a) P(T > 8) = 1−P(T ≤ 8) = 1− (1−e−0.2(8)) = 0.2019

(b) P(4 < T < 10) = P(T < 10)−P(T ≤ 4)

= P(T ≤ 10)−P(T ≤ 4)

= (1−e−0.2(10))− (1−e−0.2(4))

= 0.3140

(c) µT = 1/λ = 1/0.2 = 5 months.

(d) The mediantm solves the equationFT(tm) = 0.5.

Therefore 1−e−0.2tm = 0.5. Solving fortm yieldstm = 3.4657 months.

(e) σT = 1/λ = 5 months

(f) The 25th percentilet25 solves the equationFT(t25) = 0.25.

Therefore 1−e−0.2t25 = 0.25. Solving fortm yieldstm = 1.4383 months.

5. (a) LetX be the number of transistor that last longer than 8 months. The probability that a transistor lasts longerthan 8 months is 0.2019.

ThereforeX ∼ Bin(10,0.2019).

P(X ≥ 3) = 1−P(X = 0)−P(X = 1)−P(X = 2)

= 1− 10!0!(10−0)!

(0.2019)0(1−0.2019)10−0− 10!1!(10−1)!

(0.2019)1(1−0.2019)10−1

= − 10!2!(10−2)!

(0.2019)2(1−0.2019)10−2

= 0.3279

(b) Let T be the lifetime of a transistor, in months. The probability that a transistor fails before 10 months isP(T < 10) = P(T ≤ 10) = 1−e−0.2(10) = 0.86466. ThereforeX ∼ Bin(10,0.86466).

P(X > 8) = P(X = 9)+P(X = 10)

=10!

9!(10−9)!(0.86466)9(1−0.86466)10−9+

10!10!(10−10)!

(0.86466)10(1−0.86466)10−10

= 0.5992


190 CHAPTER 4

6. (a)P(X ≥ 5) = 1−P(X < 5) = 1−P(X ≤ 5) = 1− (1−e−1(5)) = 0.0067

(b) Yes, it would happen less than 1% of the time.

(c) No, because 5 minutes is unusually long if the claim is true.

7. No. If the lifetimes were exponentially distributed, theproportion of used components lasting longer than 5years would be the same as the proportion of new components lasting longer than 5 years, because of the lackof memory property.

8. (a)T ∼ Exp(2), soµT = 1/2 = 0.5 seconds.

(b) Letm be the median. ThenP(T ≤ m) = 0.5.

P(T ≤ m) = 1−e−2m = 0.5, soe−2m = 0.5.

Solving form yieldsm= 0.3466 seconds.

(c) P(T > 2) = 1−P(T ≤ 2) = 1− (1−e−2(2)) = 0.0183

(d) P(T < 0.1) = P(T ≤ 0.1) = 1−e−2(0.1) = 0.1813

(e) P(0.3 < T < 1.5) = P(T < 1.5)−P(T ≤ 0.3)

= P(T ≤ 1.5)−P(T ≤ 0.3)

= (1−e−2(1.5))− (1−e−2(0.3))

= 0.4990

(f) The probability isP(T ≤ 4|T > 3) = 1−P(T > 4|T > 3)

By the lack of memory property,P(T > 4|T > 3) = P(T > 1),

soP(T ≤ 4|T > 3) = 1−P(T > 1) = P(T ≤ 1) = 1−e−2(1) = 0.8647.


SECTION 4.7 191

9. LetT be the waiting time between accidents. ThenT ∼ Exp(3).

(a) µT = 1/3 year

(b) σT = 1/3 year

(c) P(T > 1) = 1−P(T ≤ 1) = 1− (1−e−3(1)) = 0.0498

(d) P(T < 1/12) = 1−e−3(1/12) = 0.2212

(e) The probability isP(T ≤ 1.5|T > 0.5) = 1−P(T > 1.5|T > 0.5)

By the lack of memory property,P(T > 1.5|T > 0.5) = P(T > 1),

soP(T ≤ 1.5|T > 0.5) = 1−P(T > 1) = P(T ≤ 1) = 1−e−3(1) = 0.9502.

10. (a)λ = 1/µX = 1/3

(b) LetY be the number of flaws in a five-meter length of aluminum.

ThenY ∼ Poisson(5λ), soY ∼ Poisson(5/3).

P(Y = 2) = e−(5/3) (5/3)2

2!= 0.2623

11. X1, ...,X5 are each exponentially distributed withλ = 1/200= 0.005.

(a) P(X1 > 100) = 1−P(X1 ≤ 100) = 1− (1−e−0.005(100)) = e−0.5 = 0.6065

(b) P(X1 > 100 andX2 > 100 andX3 > 100 andX4 > 100 andX5 > 100) =5

∏i=1

P(Xi > 100)

= (e−0.5)5

= e−2.5

= 0.0821


192 CHAPTER 4

(c) The time of the first replacement will be greater than 100 hours if and only if each of the bulbs lasts longer than100 hours.

(d) Using parts (b) and (c),P(T ≤ 100) = 1−P(T > 100) = 1−0.0821= 0.9179.

(e) P(T ≤ t) = 1−P(T > t) = 1−5

∏i=1

P(Xi > t) = 1− (e−0.005t)5 = 1−e−0.025t

(f) Yes,T ∼ Exp(0.025)

(g) µT = 1/0.025= 40 hours

(h) T ∼ Exp(nλ)

Section 4.8

1. LetT be the waiting time.

(a) µT = (0+15)/2= 7.5 minutes.

(b) σT =

√(15−0)2

12= 4.3301 minutes

(c) P(5 < T < 11) =11−515−0

= 0.4

(d) The probability that the waiting time is less than 5 minutes on any given morning is(5−0)/(15−0)= 1/3.Let X be the number of mornings on which the waiting time is less than 5 minutes. ThenX ∼ Bin(10,1/3).

P(X = 4) =10!

4!(10−4)!(1/3)4(1−1/3)10−4 = 0.2276.

2. LetX be the diameter of a rod.

(a) µX = (15+17)/2= 16 mm


SECTION 4.8 193

(b) σX =

√(17−15)2

12= 0.5774 mm

(c) P(15< X < 15.5) =15.5−1517−15

= 0.25

(d) The probability that the diameter of any given rod is greater than 16 mm is(16−15)/(17−15)= 0.5.Let Y be the number of rods whose diameters are greater than 16 mm. ThenY ∼ Bin(5,0.5).

P(Y ≥ 4) = P(X = 4)+P(X = 5) =5!

4!(5−4)!(0.5)4(1−0.5)5−4+

5!5!(5−5)!

(0.5)5(1−0.5)5−5 = 0.1875.

3. (a)µT = 4/0.5 = 8

(b) σT =√

4/0.52 = 4

(c) P(T ≤ 1) = 1−4−1

∑j=0

e−(0.5)(1) [(0.5)(1)] j

j!

= 1−e−(0.5)(1) [(0.5)(1)]0

0!−e−(0.5)(1) [(0.5)(1)]1

1!−e−(0.5)(1) [(0.5)(1)]2

2!−e−(0.5)(1) [(0.5)(1)]3

3!= 1−0.60653−0.30327−0.075816−0.012636

= 0.00175

(d) P(T ≥ 4) = 1−P(T ≤ 3)

= 1−(

1−4−1

∑j=0

e−(0.5)(3) [(0.5)(3)] j

j!

)

= e−(0.5)(3) [(0.5)(3)]0

0!+e−(0.5)(3) [(0.5)(3)]1

1!+e−(0.5)(3) [(0.5)(3)]2

2!+e−(0.5)(3) [(0.5)(3)]3

3!= 0.22313+0.33470+0.25102+0.12551

= 0.9344

4. LetT be the time at which the sixth motor fails. ThenT ∼ Γ(6,3.6).

The probability that fewer than 6 motors fail within one yearis P(T > 1).

P(T > 1) = 1−P(T ≤ 1)


194 CHAPTER 4

= 1−(

1−6−1

∑j=0

e−(3.6)(1) [(3.6)(1)] j

j!

)

= e−(3.6)(1) [(3.6)(1)]0

0!+e−(3.6)(1) [(3.6)(1)]1

1!+e−(3.6)(1) [(3.6)(1)]2

2!+e−(3.6)(1) [(3.6)(1)]3

3!

+ e−(3.6)(1) [(3.6)(1)]4

4!+e−(3.6)(1) [(3.6)(1)]5

5!= 0.027324+0.098365+0.17706+0.21247+0.19122+0.13768

= 0.8441

5. (a)α = 0.5, β = 3, so 1/α = 2 which is an integer.

µT = (1/3)2! = 2/3 = 0.6667.

(b) σT =√

(1/32)[4!− (2!)2] =√

(1/9)(24−4) = 1.4907

(c) P(T < 1) = P(T ≤ 1) = 1−e−[(3)(1)]0.5= 1−e−30.5

= 0.8231

(d) P(T > 5) = 1−P(T ≤ 5) = 1−(

1−e−[(3)(5)]0.5)

= e−150.5= 0.0208

(e) P(2 < T < 4) = P(T < 4)−P(T < 2)

= P(T ≤ 4)−P(T ≤ 2)

=(

1−e−[(3)(4)]0.5)−(

1−e−[(3)(2)]0.5)

= e−[(3)(2)]0.5 −e−[(3)(4)]0.5

= e−60.5 −e−120.5

= 0.086338−0.031301

= 0.0550

6. (a) f (t) = αβαtα−1e−(βt)α, andF(t) = 1−e−(βt)α

. Soh(t) =αβαtα−1e−(βt)α

e−(βt)α = αβαtα−1.

(b) h′(t) = (α−1)αβαtα−2. Now α > 0, β > 0, andt > 0, soh

′(t) > 0 if α−1 > 0 andh

′(t) < 0 if α−1 < 0.

Thereforeh(t) is increasing int for α > 1 and decreasing int for α < 1.


SECTION 4.8 195

(c) If T has an exponential distribution, thenT has a Weibull distribution withα = 1. Whenα = 1, h′(t) = 0 for

all t, so the hazard function is constant.

7. LetT be the lifetime in hours of a bearing.

(a) P(T > 1000) = 1−P(T ≤ 1000) = 1− (1−e−[(0.0004474)(1000)]2.25) = e−[(0.0004474)(1000)]2.25

= 0.8490

(b) P(T < 2000) = P(T ≤ 2000) = 1−e−[(0.0004474)(2000)]2.25= 0.5410

(c) Letm be the median.

ThenP(T ≤ m) = 0.5, so 1−e−[(0.0004474)(m)]2.25= 0.5, ande−[(0.0004474)(m)]2.25

= 0.5.

(0.0004474m)2.25 = − ln0.5 = 0.693147

0.0004474m= (0.693147)1/2.25 = 0.849681

m= 0.849681/0.0004474= 1899.2 hours

(d) h(t) = αβαtα−1 = 2.25(0.00044742.25)(20002.25−1) = 8.761×10−4

8. LetT be the lifetime of a battery.

(a) P(T > 10) = 1−P(T ≤ 10) = 1− (1−e−[(0.1)(10)]2) = e−1 = 0.3679

(b) P(T < 5) = P(T ≤ 5) = 1−e−[(0.1)(5)]2 = 0.2212

(c) P(T > 20) = 1−P(T ≤ 20) = 1− (1−e−[(0.1)(20)]2) = e−4 = 0.0183

(d) h(10) = αβα10α−1 = 2(0.12)(102−1) = 0.2000

9. LetT be the lifetime of a fan.

(a) P(T > 10,000) = 1− (1−e−[(0.0001)(10,000)]1.5) = e−[(0.0001)(10,000)]1.5

= 0.3679


196 CHAPTER 4

(b) P(T < 5000) = P(T ≤ 5000) = 1−e−[(0.0001)(5000)]1.5= 0.2978

(c) P(3000< T < 9000) = P(T ≤ 9000)−P(T ≤ 3000)

= (1−e−[(0.0001)(9000)]1.5)− (1−e−[(0.0001)(3000)]1.5

)

= 0.4227

10. (a)P(T ≤ 1) = 1−e−[(0.01)(1)]3 = 1.000×10−6

(b) Yes. Only one component in 106 will have a lifetime of 1 day or less if the model is correct.

(c) No, because a lifetime of 1 day is unusually short if the model is correct.

(d) P(T ≤ 90) = 1−e−[(0.01)(90)]3 = 0.5176

(e) No. About half the components will have lifetimes of 90 days or less if the model is correct.

(f) Yes, since 90 days is neither an unusually short nor an unusually long lifetime.

11. (a)P(X1 > 5) = 1−P(X1 ≤ 5) = 1− (1−e−[(0.2)(5)]2) = e−1 = 0.3679

(b) SinceX2 has the same distribution asX1, P(X2 > 5) = P(X1 > 5) = e−1. SinceX1 andX2 are independent,P(X1 > 5 andX2 > 5) = P(X1 > 5)P(X2 > 5) = (e−1)2 = e−2 = 0.1353.

(c) The lifetime of the system will be greater than 5 hours if and only if the lifetimes of both components aregreater than 5 hours.

(d) P(T ≤ 5) = 1−P(T > 5) = 1−e−2 = 0.8647

(e) P(T ≤ t) = 1−P(T > t) = 1−P(X1 > t)P(X2 > t) = 1− (e−[(0.2)(t)]2)2 = 1−e−0.08t2 = 1−e−(√

0.08t)2


SECTION 4.8 197

(f) Yes,T ∼ Weibull(2,√

0.08) = Weibull(2,0.2828).

12. µX =

Z b

ax

1b−a

dx=a+b

2

13. µX2 =Z b

ax2 1

b−adx=

a2 +ab+b2

3

Now µX =a+b

2, thereforeσ2

X = µX2 −µ2X =

a2 +ab+b2

3−(

a+b2

)2

=(b−a)2

12.

14. (a) SinceP(X ≤ a) = 0, P(X ≤ x) = 0 if x≤ a.

(b) SinceP(X ≤ b) = 1, P(X ≤ x) = 1 if x > b.

(c) If a < x≤ b, P(X ≤ x) = P(X ≤ a)+P(a < X ≤ x) = 0+

Z x

a

1b−a

dx=x−ab−a

.

Combining the results of (a), (b), and (c) shows that the cumulative distribution function ofX is

F(x) =

0 x≤ ax−ab−a

a < x≤ b

1 x > b

15. (a) The cumulative distribution function ofU is

FU(x) =

0 x≤ 0x 0 < x≤ 11 x > 1


198 CHAPTER 4

(b) FX(x) = P(X ≤ x) = P((b−a)U +a≤ x) = P

(U ≤ x−a

b−a

)= FU

(x−ab−a

)=

0 x≤ ax−ab−a

a < x≤ b

1 x > b

(c) The cdf ofX is that of a random variable distributedU(a,b).

Section 4.9

1. (a) We denote the mean ofµ1 by E(µ1) and the variance ofµ1 byV(µ1).

E(µ1) =µX1 +µX2

2=

µ+µ2

= µ.

The bias ofµ1 is E(µ1)−µ= µ−µ= 0.

The variance ofµ1 is V(µ1) =σ2 + σ2

4=

σ2

2=

12

.

The mean squared error ofµ1 is the sum of the variance and the square of the bias, so

MSE(µ1) =12

+02 =12

.

(b) We denote the mean ofµ2 by E(µ2) and the variance ofµ2 byV(µ2).

E(µ2) =µX1 +2µX2

3=

µ+2µ3

= µ.

The bias ofµ2 is E(µ2)−µ= µ−µ= 0.

The variance ofµ2 is V(µ2) =σ2 +4σ2

9=

5σ2

9=

59

.


MSE(µ2) =59

+02 =59

.

(c) We denote the mean ofµ3 by E(µ3) and the variance ofµ3 byV(µ3).

E(µ3) =µX1 +µX2

4=

µ+µ4

=µ2

.

The bias ofµ3 is E(µ3)−µ=µ2−µ= −µ

2.

The variance ofµ3 is V(µ3) =σ2 + σ2

16=

σ2

8.


MSE(µ3) =σ2

8+(−µ

2

)2=

2µ2+18

.


SECTION 4.9 199

(d) µ3 has smaller mean squared error thanµ1 whenever2µ2+1

8<

12

.

Solving forµ yields−1.2247< µ< 1.2247.

(e) µ3 has smaller mean squared error thanµ2 whenever2µ2+1

8<

59

.

Solving forµ yields−1.3123< µ< 1.3123.

2. (a) We denote the mean ofσ2k by E(σ2

k).

E(σ2k) = E

((n−1)s2

k

)=

(n−1)µs2

k=

(n−1)σ2

k

The bias ofσ2k is E(σ2

k)−σ2 =(n−1)σ2

k−σ2 =

(n−1−k)σ2

k.

(b) We denote the variance ofσ2k byV(σ2

k).

V(σ2k) = V

((n−1)s2

k

)=

(n−1)2

k2 σ2s2 =

(n−1)2

k2

2σ4

n−1=

2(n−1)σ4

k2

(c) The mean squared error ofσ2k is the sum of the variance and the square of the bias, so

MSE(σ2k) =

2(n−1)σ4

k2 +(n−1−k)2σ4

k2 .

(d) MSE(σ2k) =

2(n−1)σ4

k2 +(n−1−k)2σ4

k2 =2(n−1)+ (n−1)2

k2 σ4− 2(n−1)

kσ4.

We differentiateMSE(σ2k) with respect tok and set the derivative equal to 0.

ddk

MSE(σ2k) =

ddk

(2(n−1)+ (n−1)2

k2 σ4− 2(n−1)

kσ4)

=−4(n−1)−2(n−1)2

k3 σ4 +2(n−1)

k2 σ4 = 0

It follows that−4−2(n−1)+2k= 0. Solving fork yieldsk = n+1.

3. The probability mass function ofX is f (x; p) = p(1− p)x−1.

The MLE is the value ofp that maximizesf (x; p), or equivalently, lnf (x; p).

ddp

ln f (x; p) =d

dp[ln p+(x−1) ln(1− p)] =

1p− x−1

1− p= 0.

Solving forp yields p =1x

. The MLE is p =1X

.


200 CHAPTER 4

4. The joint probability mass function ofX1, ...,Xn is

f (x1, ...,xn; λ) =n

∏i=1

e−λ λxi

xi != e−nλ λ∑n

i=1 xi

∏ni=1xi !

.

The MLE is the value ofλ that maximizesf (x1, ...,xn; λ), or equivalently, lnf (x1, ...,xn; λ).

ddλ

ln f (x1, ...,xn; λ) =ddλ

(−nλ +n

∑i=1

xi lnλ−n

∑i=1

lnxi !) = −n+∑n

i=1xi

λ= 0.

Now we solve forλ:

−n+∑n

i=1xi

λ= 0.

λ =∑n

i=1xi

n= x.

The MLE isλ = X.

5. (a) The probability mass function ofX is f (x; p) =n!

x!(n−x)!(p)x(1− p)n−x.

The MLE of p is the value ofp that maximizesf (x; p), or equivalently, lnf (x; p).

ddp

ln f (x; p) =d

dp[lnn! − lnx! − ln(n−x)! +xln p+(n−x) ln(1− p)] =

xp− n−x

1− p= 0.

Solving forp yields p =xn

. The MLE of p is p =Xn

.

The MLE ofp

1− pis therefore

p1− p

=X

n−X.

(b) The MLE of p is p =1X

. The MLE ofp

1− pis therefore

p1− p

=1

X−1.

(c) The MLE ofλ is λ = X. The MLE ofe−λ is thereforee−λ = e−X.

6. The joint probability density function ofX1, ...,Xn is

f (x1, ...,xn; µ) =n

∏i=1

1√2π

e−(xi−µ)2/2 = (2π)−n/2e−∑ni=1(xi−µ)2/2.

The MLE is the value ofµ that maximizesf (x1, ...,xn; µ), or equivalently, lnf (x1, ...,xn; µ).

ddµ

ln f (x1, ...,xn; µ) =ddµ

[−(n/2) ln2π−

n

∑i=1

(xi −µ)2

2

]=

n

∑i=1

(xi −µ) = 0.


SECTION 4.9 201

Now we solve forµ:

∑ni=1(xi −µ) = ∑n

i=1xi −nµ= 0, soµ=∑n

i=1xi

n= x.

The MLE ofµ is µ= X.


f (x1, ...,xn; σ) =n

∏i=1

1

σ√

2πe−x2

i /2σ2= (2π)−n/2σ−ne−∑n

i=1x2i /2σ2

.

The MLE is the value ofσ that maximizesf (x1, ...,xn; σ), or equivalently, lnf (x1, ...,xn; σ).

ddσ

ln f (x1, ...,xn; σ) =d

dσ

[−(n/2) ln2π−nlnσ−

n

∑i=1

x2i

2σ2

]= − n

σ+

∑ni=1x2

i

σ3 = 0.

Now we solve forσ:

− nσ

+∑n

i=1x2i

σ3 = 0

−nσ2 + ∑ni=1x2

i = 0

σ2 =∑n

i=1x2i

n

σ =

√∑n

i=1x2i

n

The MLE ofσ is σ =

√∑n

i=1X2i

n.


f (x1, ...,xn; µ,σ) =n

∏i=1

1

σ√

2πe−(xi−µ)2/2σ2

= (2π)−n/2σ−ne−∑ni=1(xi−µ)2/2σ2

.

The MLEs are the values ofµ andσ that maximizesf (x1, ...,xn; µ,σ), or equivalently, lnf (x1, ...,xn; µ,σ).

∂∂µ

ln f (x1, ...,xn; µ,σ) =∂∂µ


n

∑i=1

(xi −µ)2

2σ2

]=

∑ni=1(xi −µ)

σ2 = 0.

Now we solve forµ:

∑ni=1(xi −µ)

σ2 = 0.

∑ni=1(xi −µ) = 0.

∑ni=1xi −nµ= 0

µ=∑n

i=1xi

n= x.


202 CHAPTER 4

The MLE ofµ is µ= X.

Now we substituteµ for µ in ln f (x,...,xn; µ,σ) and maximize overσ:

∂∂σ

ln f (x1, ...,xn; µ,σ) =∂

∂σ


n

∑i=1

(xi − µ)2

2σ2

]= − n

σ+

∑ni=1(xi − µ)2

σ3 = 0.

Now we solve forσ:

− nσ

+∑n

i=1(xi − µ)2

σ3 = 0

−nσ2 + ∑ni=1(xi − µ)2 = 0

σ2 =∑n

i=1(xi − µ)2

n

σ =

√∑n

i=1(xi − µ)2

n

Now µ= X, so the MLE ofσ is σ =

√∑n

i=1(Xi −X)2

n.

Section 4.10

1. (a) No

(b) No

(c) Yes

2.

12 14 16 18 20

0.001

0.01

0.050.1

0.25

0.5

0.75

0.90.95

0.99

0.999

These data appear to come from an approxi-mately normal distribution.

3.

2 2.5 3 3.5 4 4.5

0.001

0.01

0.050.1

0.25

0.5

0.75

0.90.95

0.99

0.999

These data do not appear to come from an ap-proximately normal distribution.


SECTION 4.11 203

4.

50 60 70 80 90

0.001

0.01

0.050.1

0.25

0.5

0.75

0.90.95

0.99

0.999

These data do not appear to come from an ap-proximately normal distribution.

5.

0 5 10 15 20 25

0.001

0.01

0.050.1

0.25

0.5

0.75

0.90.95

0.99

0.999

The PM data do not appear to come from anapproximately normal distribution.

6.

0 0.5 1 1.5 2 2.5 3

0.001

0.01

0.050.1

0.25

0.5

0.75

0.90.95

0.99

0.999

The logs of the PM data do appear to comefrom an approximately normal distribution.

7. Yes. If the logs of the PM data come from a normal population, then the PM data come from a lognormalpopulation, and vice versa.

Section 4.11

1. (a) LetX1, ...,X50 be the weights of the 50 bags.

ThenX is approximately normally distributed with meanµX = 100 andσX = 0.5/√

50= 0.0707.

Thez-score of 99.9 is(99.9−100)/0.0707= −1.41.


204 CHAPTER 4


P(X < 99.9) = 0.0793.

(b) X is approximately normally distributed with meanµX = 100.15, andσX = 0.5/√

50= 0.0707.

Thez-score of 100 is(100−100.15)/0.0707= −2.12.


P(X < 100) = 0.0170.

2. (a) LetX1, ...,X250 denote the thicknesses of the 250 sheets of paper.

Let S= X1 + · · ·+X250 be the total thickness of the 250 pages.

ThenSis approximately normally distributed with meanµS= 250(0.08)= 20 andσX = 0.01√

250= 0.158114.

Thez-score of 20.2 is(20.2−20)/0.158114= 1.26.


P(S> 20.2) = 0.1038.

(b) Lets10 denote the 10th percentile


Therefores10 satisfies the equation−1.28= (s10−20)/0.158114.

s10 = 19.798 mm.

(c) No, because we don’t know the distribution of page thicknesses. In particular, we do not know whether thepage thicknesses are normally distributed.

3. LetX1, ...,X100 be the heights of the 100 men.

ThenX is approximately normally distributed with meanµX = 70 andσX = 2.5/√

100= 0.25.

Thez-score of 69.5 is(69.5−70)/0.25= −2.00.

The area to the right ofz= −2.00 is 1−0.0228= 0.9772.

P(X > 69.5) = 0.9772.

4. (a) LetX1, ...,X625 be the amounts of tax on the 625 forms.

ThenX is approximately normally distributed with meanµX = 2000 andσX = 500/√

625= 20.

Thez-score of 1980 is(1980−2000)/20= −1.00.



SECTION 4.11 205

P(X > 1980) = 0.8413.

(b) LetY be the number of forms whose tax is greater than $3000.

ThenY ∼ Bin(625,0.10), soY is approximately normal with mean 625(0.10) = 62.5 and standard deviation√625(0.1)(0.9) = 7.5.

To findP(Y > 60), use the continuity correction and find thez-score of 60.5.

Thez-score of 60.5 is(60.5−62.5)/7.5= −0.27.


P(Y > 60) = 0.6064.

5. (a) LetX1, ...,X80 be the breaking strengths of the 80 fabric pieces.

ThenX is approximately normally distributed with meanµX = 1.86 andσX = 0.27/√

80= 0.030187.

Thez-score of 1.8 is(1.8−1.86)/0.030187= −1.99.


P(X < 1.8) = 0.0233.

(b) Letx80 denote the 80th percentile


Thereforex80 satisfies the equation 0.84= (x80−1.86)/0.030187.

x80 = 1.8854 mm.

(c) Let n be the necessary sample size. ThenX is approximately normally distributed with meanµX = 1.86 andσX = 0.27/

√n.

SinceP(X < 1.8) = 0.01, 1.8 is the 1st percentile of the distribution ofX.

Thez-score of the 1st percentile is approximatelyz= −2.33.

Therefore 1.8 = 1.86−2.33(0.27/√

n). Solving forn yieldsn≈ 110.

6. (a) LetX1, ...,X200 be the amounts of warpage in the 200 wafers.

ThenX is approximately normally distributed with meanµX = 1.3 andσX = 0.1/√

200= 0.00707107.

Thez-score of 1.8 is(1.305−1.3)/0.00707107= 0.71.


P(X > 1.305) = 0.2389.

(b) Letx25 denote the 25th percentile.


Thereforex25 satisfies the equation−0.67= (x25−1.3)/0.00707107.

x25 = 1.2953 mm.


206 CHAPTER 4

(c) Let n be the necessary sample size. ThenX is approximately normally distributed with meanµX = 1.3 andσX = 0.1/

√n.

SinceP(X > 1.305) = 0.05, 1.305 is the 95th percentile of the distribution ofX.


Therefore 1.305= 1.3+1.645(0.1/√

n). Solving forn yieldsn≈ 1083.

7. From the results of Example 4.70, the probability that a randomly chosen wire has no flaws is 0.48.

Let X be the number of wires in a sample of 225 that have no flaws.

ThenX ∼ Bin(225,0.48), soµX = 225(0.48) = 108, andσ2X = 225(0.48)(0.52) = 56.16.

To findP(X < 110), use the continuity correction and find thez-score of 109.5.

Thez-score of 109.5 is(109.5−108)/√

56.16= 0.20.


P(X < 110) = 0.5793.

8. (a) LetX1, ...,X50 be the amounts of solution in 50 drums. ThenµXi = 30.01 andσXi = 0.1.

Let S= X1 + · · ·+X50 be the total amount of solution in 50 drums.

ThenS is approximately normally distributed withµS = 50(30.01) = 1500.5, σS = 0.1√

50= 0.7071.

Thez-score of 1500 is(1500−1500.5)/0.7071= −0.71.


P(S> 1500) = 0.7611.

(b) LetX1, ...,X80 be the amounts of solution in 80 drums. ThenµXi = 30.01 andσXi = 0.1.

Let T = X1 + · · ·+X80 be the total amount of solution in 80 drums.

ThenT is approximately normally distributed withµT = 80(30.01) = 2400.8, σT = 0.1√

80= 0.8944.

The probability that 80 drums will be filled without running out isP(T < 2401).

Thez-score of 2401 is(2401−2400.8)/0.8944= 0.22.


P(T < 2401) = 0.5871.

(c) Let s be the required amount of solution. ThenP(T < s) = 0.9, sos is the 90th percentile of the distributionof T.


From part (b),T is approximately normally distributed withµT = 80(30.01) = 2400.8 andσT = 0.1√

80=0.8944.

Therefores satisfies the equation 1.28= (s−2400.8)/0.8944.

s= 2401.9 L.


SECTION 4.11 207

9. Letn be the required number of measurements. LetX be the average of then measurements.

Then the true value isµX, and the standard deviation isσX = 1/√

n.

Now P(µX −0.25< X < µX + 0.25) = 0.95. In any normal population, 95% of the population is within1.96standard deviations of the mean.

Therefore 1.96σX = 0.25. SinceσX = 1/√

n, n = 61.47. The smallest value ofn is thereforen = 62.

10. LetX be the number of cans that fail the test.

ThenX ∼ Bin(100,0.2), soX is approximately normal with meanµX = 100(0.20)= 20 and standard deviationσX =

√100(0.2)(0.8) = 4.

To findP(X ≥ 12), use the continuity correction and find thez-score of 11.5.

Thez-score of 11.5 is(11.5−20)/4= −2.125; we will usez= −2.13.


P(X ≥ 12) = 0.9834.

11. (a) LetX represent the number of defective parts in a shipment.

ThenX ∼Bin(400,0.20), soX is approximately normal with meanµX = 400(0.20)= 80 and standard deviationσX =

√400(0.2)(0.8) = 8.

To findP(X > 90), use the continuity correction and find thez-score of 90.5.

Thez-score of 90.5 is(90.5−80)/8= 1.31.


P(X > 90) = 0.0951.

(b) LetY represent the number of shipments out of 500 that are returned.

From part (a) the probability that a shipment is returned is 0.0951, soY ∼ Bin(500,0.0951).

It follows that Y is approximately normal with meanµY = 500(0.0951) = 47.55 and standard deviationσY =

√500(0.0951)(0.9049)= 6.5596.

To findP(Y ≥ 60), use the continuity correction and find thez-score of 59.5.

Thez-score of 59.5 is(59.5−47.55)/6.5596= 1.82.


P(X ≥ 60) = 0.0344.

(c) Let p be the required proportion of defective parts, and letX represent the number of defective parts in ashipment.

Then X ∼ Bin(400, p), so X is approximately normal with meanµX = 400p and standard deviationσX =

√400p(1− p).


208 CHAPTER 4

The probability that a shipment is returned isP(X > 90) = 0.01.

Using the continuity correction, 90.5 is the 1st percentileof the distribution ofX.


Thez-score can be expressed in terms ofp by−2.33= (90.5−400p)/√

400p(1− p).

This equation can be rewritten as 162,171.56p2−74,571.56p+8190.25= 0.

Solving forp yields p = 0.181. (0.278 is a spurious root.)

12. (a) LetX be the number of O-rings that meet the specification.

Then X ∼ Bin(1000,0.9), so X is approximately normal with meanµX = 1000(0.90) = 900 and standarddeviationσX =

√1000(0.9)(0.1) = 9.486833.


Thez-score of 889.5 is(889.5−900)/9.486833= −1.11.


P(X < 890) = 0.1335.

(b) Letx60 denote the 60th percentile


Thereforex60 satisfies the equation 0.25= (x60−900)/9.486833.

x60 = 902 or 903.

(c) From part (a), the probability that fewer than 890 O-rings meet the specification on any given day is 0.1335.LetY be the number of days on which fewer than 890 O-rings meet the specification. ThenY ∼ Bin(5,0.1335).

P(Y ≥ 3) = P(Y = 3)+P(Y = 4)+P(Y = 5)

=5!

3!(5−3)!(0.1335)3(1−0.1335)5−3+

5!4!(5−4)!

(0.1335)4(1−0.1335)5−4

= +5!

5!(5−5)!(0.1335)5(1−0.1335)5−5

= 0.01928

13. (a) LetX be the number of particles emitted by mass A and letY be the number of particles emitted by mass B ina five-minute time period. ThenX ∼ Poisson(100) andY ∼ Poisson(125).

Now X is approximately normal with mean 100 and variance 100, andY is approximately normal with mean125 and variance 125.

The number of particles emitted by both masses together isS= X +Y.

SinceX andY are independent, and both approximately normal,S is approximately normal as well, with meanµS = 100+125= 225 and standard deviationσS =

√100+125= 15.


SECTION 4.11 209

Thez-score of 200 is(200−225)/15= −1.67. The area to the left ofz= −1.67 is 0.0475.

P(S< 200) = 0.0475.

(b) Let X be the number of particles emitted by mass A and letY be the number of particles emitted by mass B ina two-minute time period. ThenX ∼ Poisson(40) andY ∼ Poisson(50).

Now X is approximately normal with mean 40 and variance 40, andY is approximately normal with mean 50and variance 50.

The difference between the numbers of particles emitted by the two masses isD = Y−X.

Mass B emits more particles than mass A ifD > 0.

SinceX andY are independent, and both approximately normal,D is approximately normal as well, with meanµD = 50−40= 10 and standard deviationσD =

√50+40= 9.486833.

Thez-score of 0 is(0−10)/9.486833= −1.05. The area to the right ofz= −1.05 is 1−0.1469= 0.8531.

P(D > 0) = 0.8531.

14. (a) LetX be the number of particles contained in a 2 ml sample. ThenX ∼ Poisson(60),

soX is approximately normal with meanµX = 60 and standard deviationσX =√

60= 7.745967.

Thez-score of 50 is(50−60)/7.7459674= −1.29.


P(X > 50) = 0.9015.

(b) LetY be the number of samples that contain more than 50 particles.

From part (a), the probability that a sample contains more than 50 particles is 0.9015.

ThereforeY ∼ Bin(10,0.9015).

P(Y ≥ 9) = P(Y = 9)+P(Y = 10)

=10!

9!(10−9)!(0.9015)9(1−0.9015)10−9+

10!10!(10−10)!

(0.9015)10(1−0.9015)10−10

= 0.7419

(c) LetW be he number of samples that contain more than 50 particles.

From part (a), the probability that a sample contains more than 50 particles is 0.9015.

ThereforeW ∼ Bin(100,0.9015), soW is approximately normal with meanµW = 100(0.9015) = 90.15 andstandard deviation

√100(0.9015)(0.0985)= 2.979895.

To findP(W ≥ 90), use the continuity correction and find thez-score of 89.5.

Thez-score of 89.5 is(89.5−90.15)/2.979895= −0.22.


P(W ≥ 90) = 0.5871.


210 CHAPTER 4

15. (a) LetX be the number of particles withdrawn in a 5 mL volume.

Then the mean ofX is 50(5) = 250, soX ∼ Poisson(250), andX is approximately normal with meanµX = 250and standard deviationσX =

√250= 15.8114.

Thez-score of 235 is(235−250)/15.8114= −0.95, and thez-score of 265 is(265−250)/15.8114= 0.95.


The probability is 0.6578.

(b) Since the withdrawn sample contains 5 mL, the average number of particles per mL will be between 48 and 52if the total number of particles is between 5(48) = 240 and 5(52) = 260.

From part (a),X is approximately normal with meanµX = 250 and standard deviationσX =√

250= 15.8114.




(c) LetX be the number of particles withdrawn in a 10 mL volume.

Then the mean ofX is 50(10)= 500, soX ∼Poisson(500), andX is approximately normal with meanµX = 500and standard deviationσX =

√500= 22.3607.

The average number of particles per mL will be between 48 and 52 if the total number of particles is between10(48) = 480 and 10(52) = 520.




(d) Letv be the required volume. LetX be the number of particles withdrawn in a volume ofv mL.

ThenX ∼Poisson(50v), soX is approximately normal with meanµX = 50v and standard deviationσX =√

50v.

The average number of particles per mL will be between 48 and 52 if the total number of particlesX is between48v and 52v.

SinceµX = 50v, P(48v < X < 52v) = 0.95 if thez-score of 48v is−1.96 and thez-score of 52v is 1.96.

Thez-score of 1.96 therefore satisfies the equation 1.96= (52v−50v)/√

50v, or equivalently,√

v= 1.96√

50/2.

Solving forv yieldsv = 48.02 ml.

16. (a) If the claim is true, thenX is approximately normal with meanµX = 40 andσX = 5/√

100= 0.5.

Thez-score of 36.7 is(36.7−40)/0.5= −6.6.


P(X ≤ 36.7) ≈ 0.


SECTION 4.11 211

(b) Yes. Almost none of the samples will have mean lifetimes 36.7 hours or less if the claim is true.

(c) No, because a sample mean lifetime of 36.7 hours is very unusual if the claim is true.

(d) If the claim is true, thenX is approximately normal with meanµX = 40 andσX = 5/√

100= 0.5.

Thez-score of 39.8 is(39.8−40)/0.5= −0.4.


P(X ≤ 39.8) = 0.3446.

(e) No. More than 1/3 of the samples will have mean lifetimes 39.8 hours or less if the claim is true.

(f) Yes, because a sample mean lifetime of 39.8 hours or less is not unusual if the claim is true.

17. (a) If the claim is true, thenX ∼ Bin(1000,0.05), soX is approximately normal with meanµX = 1000(0.05) = 50andσX =

√1000(0.05)(0.95)= 6.89202.


Thez-score of 74.5 is(74.5−50)/6.89202= 3.55.


P(X ≥ 75) = 0.0002.

(b) Yes. Only about 2 in 10,000 samples of size 1000 will have 75 or more nonconforming tiles if the goal hasbeen reached.

(c) No, because 75 nonconforming tiles in a sample of 1000 is an unusually large number if the goal has beenreached.

(d) If the claim is true, thenX ∼ Bin(1000,0.05), soX is approximately normal with meanµX = 1000(0.05) = 50andσX =

√1000(0.05)(0.95)= 6.89202.


Thez-score of 52.5 is(52.5−50)/6.89202= 0.36.


P(X ≥ 53) = 0.3594.

(e) No. More than 1/3 of the samples of size 1000 will have 53 ormore nonconforming tiles if the goal has beenreached.


212 CHAPTER 4

(f) Yes, because 53 nonconforming tiles in a sample of 1000 isnot an unusually large number if the goal has beenreached.

18. LetX1, ...,X100be the times taken on machine 1 by the 100 parts. LetY1, ...,Y100be the times taken on machine 2by the 100 parts.

Let SX = X1 + · · ·+ X100 be the total time on machine 1, and letSY = Y1 + · · ·+Y100 be the total time onmachine 2.

ThenSX is approximately normal with meanµSX = 100(0.5) = 50 and standard deviationσSX = 0.4√

100= 4,andSY is approximately normal with meanµSY = 100(0.6) = 60 and standard deviationσSY = 0.5

√100= 5.

(a) Thez-score of 55 is(55−50)/4= 1.25.


P(SX > 55) = 0.1056.

(b) Thez-score of 55 is(55−60)/5= −1.00.


P(SY < 55) = 0.1587.

(c) LetT = SX +SY be the total time used by both machines.

SinceSX andSY are independent and approximately normal,T is approximately normal with mean

µT = µSX +µSY = 50+60= 110, and standard deviationσT =√

σ2SX

+ σ2SY

=√

42 +52 = 6.403124.

Thez-score of 115 is(115−110)/6.403124= 0.78.

The area to the right ofz= 0.78 is 1−0.7823= 0.2177.P(T > 115) = 0.2177.

(d) LetD = SX −SY be the difference between the time on machine 1 and the time onmachine 2.

SinceSX andSY are independent and approximately normal,D is approximately normal with mean

µD = µSX −µSY = 50−60= −10, and standard deviationσT =√

σ2SX

+ σ2SY

=√

42+52 = 6.403124.

The time on machine 1 is greater than the time on machine 2 ifD > 0. Thez-score of 0 is[0−(−10)]/6.403124=1.56.

The area to the right ofz= 1.56 is 1−0.9406= 0.0594.P(D > 0) = 0.0594.

19. LetX be the number of rivets from vendor A that meet the specification, and letY be the number of rivets fromvendor B that meet the specification.

ThenX ∼ Bin(500,0.7), andY ∼ Bin(500,0.8).

It follows thatX is approximately normal with meanµX = 500(0.7) = 350 and varianceσ2X = 500(0.7)(0.3)=

105, andY is approximately normal with meanµY = 500(0.8) = 400 and varianceσ2X = 500(0.8)(0.2) = 80.

Let T = X +Y be the total number of rivets that meet the specification.


SECTION 4.11 213

ThenT is approximately normal with meanµT = µX + µY = 350+ 400= 750, and standard deviationσT =√σ2

X + σ2Y =

√105+80= 13.601471.

To findP(T > 775), use the continuity correction and find thez-score of 775.5.

Thez-score of 775.5 is(775.5−750)/13.601471= 1.87.


P(T > 775) = 0.0307.

20. (a)λ solves the equation 11,000= (ln15.3− lnλ)/0.0001210.

Thereforeλ = 4.0425 emissions per minute.

(b) From part (a),λ = 4.0425 emissions per minute, andX ∼ Poisson(25λ).

The mean ofX is µX = 25λ = 101.06, and the standard deviation isσX =√

25λ = 10.053.

(c) µλ = µX/25= 4.0425,σλ = σX/25= 0.4021

(d) The required value ofλ solves the equation 10,000= (ln15.3− lnλ)/0.0001210.


(e) The required value ofλ solves the equation 12,000= (ln15.3− lnλ)/0.0001210.


(f) The age estimate is correct to within±1000 years if it is between 10,000 and 12,000, or equivalently, if3.5818< λ < 4.5624.

Sinceλ = X/25,P(3.5818< λ < 4.5624) = P(3.5818< X/25< 4.5624) = P(89.545< X < 114.06).

Thez-score of 89.545 is(89.545−101.06)/10.053= −1.15.

Thez-score of 114.06 is(114.06−101.06)/10.053= 1.29.


Note that one may computeP(90≤ X ≤ 114) sinceX must be an integer. In this case the answer obtained is0.7658.


214 CHAPTER 4

Section 4.12

1. (a)X ∼ Bin(100,0.03), Y ∼ Bin(100,0.05)

(b) Answers will vary.

(c) ≈ 0.72

(d) ≈ 0.18

(e) The distribution deviates somewhat from the normal.

2. (a)≈ 0.84

(b) ≈ 0.77

3. (a)µA = 6 exactly (simulation results will be approximate),σ2A ≈ 0.25.

(b) ≈ 0.16

(c) The distribution is approximately normal.

4. (a)≈ 36

(b) ≈ 36

(c) ≈ 2.8

(d) Yes, the cable is acceptable. The probability that the cable breaks under a load of 28 kN is approximately 0.005.

5. (a)≈ 0.25

(b) ≈ 0.25


SECTION 4.12 215

(c) ≈ 0.61

6. (a–d) Answers will vary.

7. (a–c) Answers will vary.

(d) ≈ 0.025

8. (a,b) Answers will vary.

(c) Replacing B results in a longer mean system lifetime (≈ 3.25 compared with≈ 3.0).

(d) Replacing A results in a smaller probability that the system fails within a month (≈0.15 compared with≈ 0.18).

(e) Replacing A results in a greater 10th percentile (≈ 0.76 compared with≈ 0.68).

9. (a) Answers will vary.

(b) ≈ 2.7

(c) ≈ 0.34

(d) ≈ 1.6

(e) System lifetime is not approximately normally distributed.

(f) Skewed to the right.


(b) ≈ 4.75

(c) ≈ 3.75


216 CHAPTER 4

(d) ≈ 0.28

(e)≈ 9.7

(f) ≈ 0.46


(b) ≈ 10,090

(c) ≈ 1250

(d) ≈ 0.58

(e)≈ 0.095

(f) The distribution differs somewhat from normal.

12. (a,b) Answers will vary.

(c) µmedian= 0 exactly (simulation results will be approximate),σmedian≈ 0.536.

(d) Answers will vary.

(e) µX = 0 exactly, σX = 1/√

5 exactly (simulation results will be approximate).

(f) Bias is exactly 0 in both cases (simulation results will be approximate). σmedian≈ 0.536, σX = 1/√

5 exactly(simulation results will be approximate).

(g) Bias estimate should be close to 0 becauseµX = µ so the true bias is 0. The uncertainty should be close to1/

√5 becauseσX = σ/

√n = 1/

√5.

13. (a)λ = 0.25616

(b–d) Answers will vary.



(e) Bias≈ 0.037, σλ ≈ 0.12


1. LetX be the number of people out of 105 who appear for the flight.

ThenX ∼Bin(105,0.9), soX is approximately normal with meanµX = 105(0.9)= 94.5 and standard deviationσX =

√105(0.9)(0.1) = 3.0741.

To findP(X ≤ 100), use the continuity correction and find thez-score for 100.5.

Thez-score of 100.5 is(100.5−94.5)/3.0741= 1.95.


P(X ≤ 100) = 0.9744.

2. (a) LetX be the number of cracks in a 500 meter length of pavement.

Then the mean ofX is 5, soX ∼ Poisson(5).

P(X = 8) = e−5 58

8!= 0.06528.

(b) LetY be the number of cracks in a 100 meter length of pavement.

Then the mean ofY is 1, soY ∼ Poisson(1).

P(X = 0) = e−1 10

0!= 0.3679.

(c) SinceT is the distance between two consecutive events in a Poisson process,T has an exponential distribution.Since the rate at which events occur is 0.01 per meter,T ∼ Exp(0.01).

The probability density function isfT(t) = 0.01e−0.01t for t > 0, fT(t) = 0 for t ≤ 0.

(d) P(T > 50) = 1−P(T ≤ 50) = 1− (1−e−0.01(50)) = e−0.5 = 0.6065

3. (a) LetX be the number of plants out of 10 that have green seeds. ThenX ∼ Bin(10,0.25).

P(X = 3) =10!

3!(10−3)!(0.25)3(1−0.25)10−3 = 0.2503.

(b) P(X > 2) = 1−P(X ≤ 2)


218 CHAPTER 4

= 1−P(X = 0)−P(X = 1)−P(X = 2)

= 1− 10!0!(10−0)!

(0.25)0(1−0.25)10−0− 10!1!(10−1)!

(0.25)1(1−0.25)10−1

− 10!2!(10−2)!

(0.25)2(1−0.25)10−2

= 0.4744

(c) LetY be the number of plants out of 100 that have green seeds.

ThenY ∼ Bin(100,0.25) soY is approximately normal with meanµY = 100(0.25)= 25 and standard deviationσY =

√100(0.25)(0.75) = 4.3301.

To findP(Y > 30), use the continuity correction and find thez-score for 30.5.

Thez-score of 30.5 is(30.5−25)/4.3301= 1.27.


P(Y > 30) = 0.1020.

(d) To findP(30≤Y ≤ 35), use the continuity correction and find thez-scores for 29.5 and 35.5.

Thez-score of 29.5 is(29.5−25)/4.3301= 1.04.

Thez-score of 35.5 is(35.5−25)/4.3301= 2.42.

The area to betweenz= 1.04 andz= 2.42 is 0.9922−0.8508= 0.1414.

P(30≤Y ≤ 35) = 0.1414.

(e) Fewer than 80 have yellow seeds if more than 20 have green seeds.

To findP(Y > 20), use the continuity correction and find thez-score for 20.5.

Thez-score of 20.5 is(20.5−25)/4.3301= −1.04.


P(Y > 20) = 0.8508.

4. (a) True. If lnX1, ..., lnXn come from an approximately normal population, thenX1, ...,Xn come from an approxi-mately lognormal population.

(b) False.X1, ...,Xn come from an approximately lognormal population.

(c) False. lnX1, ..., lnXn come from an approximately normal population.

(d) True. lnX1, ..., lnXn come from an approximately normal population.

5. LetX denote the number of devices that fail. ThenX ∼ Bin(10,0.01).



(a) P(X = 0) =10!

0!(10−0)!(0.01)0(1−0.01)10−0 = 0.9910 = 0.9044

(b) P(X ≥ 2) = 1−P(X ≤ 1)

= 1−P(X = 0)−P(X = 1)

= 1− 10!0!(10−0)!

(0.01)0(1−0.01)10−0− 10!1!(10−1)!

(0.01)1(1−0.01)10−1

= 0.00427

(c) Let p be the required probability. ThenX ∼ Bin(10, p).

P(X = 0) =10!

0!(10−0)!p0(1− p)10−0 = (1− p)10 = 0.95.

Solving forp yields p = 0.00512.

6. The concentrations are not approximately normally distributed. There are no concentrations lower than 0, and0 is only 0.71/1.09 = 0.65 standard deviations below the mean. In a normal distribution, approximately 26%of the values would be farther below the mean than this. In addition, the mean is much larger than the median,which is uncharacteristic of a sample from a normal distribution.

7. (a) The probability that a normal random variable is within one standard deviation of its mean is the area under thenormal curve betweenz= −1 andz= 1. This area is 0.8413−0.1587= 0.6826.

(b) The quantityµ+zσ is the 90th percentile of the distribution ofX. The 90th percentile of a normal distributionis 1.28 standard deviations above the mean. Thereforez= 1.28.

(c) Thez-score of 15 is(15−10)/√

2.6 = 3.10.


P(X > 15) = 0.0010.

8. (a) Thez-score of 9.3 is (9.3−10)/1= −0.70.

Thez-score of 10.7 is (10.7−10)/1= 0.70.


The proportion of resistors with resistances between 9.3 and 10.7Ω is 0.5160.


220 CHAPTER 4

(b) LetX be the number of resistors in the sample whose resistances are between 9.3 and 10.7Ω.

From part (a), the probability that a resistor has resistance between 9.3 and 10.7Ω is 0.5160.

ThenX ∼ Bin(100,0.5160), soX is approximately normal with meanµX = 100(0.5160)= 51.60 and standarddeviationσX =

√100(0.5160)(0.4840)= 4.99744.

To findP(X ≥ 50), use the continuity correction and find thez-score for 49.5.

Thez-score of 49.5 is(49.5−51.6)/4.99744= −0.42.


P(X ≥ 50) = 0.6628.

(c) Letn be the required number. LetX be the number of resistors out ofn whose resistances are between 9.3 and10.7Ω.

X ∼ Bin(n,0.5160), so X is approximately normal with meanµX = 0.5160n and standard deviationσX =

√n(0.5160)(0.4840)= 0.499744

√n.



Thez-score can be expressed in terms ofn by−2.33= (49.5−0.5160n)/(0.499744√

n).

This equation can be rewritten as 0.516n−1.1644√

n−49.5= 0.

Solving for√

n yields√

n = 10.987 (−8.731 is a spurious root).

n = 120.7, or 121 to the nearest integer.

9. (a) Thez-score of 215 is(215−200)/10= 1.5.


The probability that the clearance is greater than 215µm is 0.0668.

(b) Thez-score of 180 is(180−200)/10= −2.00.

Thez-score of 205 is(205−200)/10= 0.50.


The probability that the clearance is between 180 and 205µm is 0.6687.

(c) LetX be the number of valves whose clearances are greater than 215µm.

From part (a), the probability that a valve has a clearance greater than 215µm is 0.0668, soX ∼ Bin(6,0.0668).

P(X = 2) =6!

2!(6−2)!(0.0668)2(1−0.0668)6−2 = 0.0508.

10. (a) No, we do not know the distribution of the population of beam stiffnesses. In particular, we do not know



whether the population is normal.

(b) Yes, the mean of a large sample is normally distributed, so the Central Limit Theorem can be used.

Let X be the sample mean stiffness. TheX is approximately normally distributed with meanµX = 30 andstandard deviationσX = 2/

√100= 0.2.

Thez-score of 30.2 is(30.2−30)/0.2= 1.00.


P(X > 30.2) = 0.1587.

11. (a) LetX be the number of assemblies in a sample of 300 that are oversize.

ThenX ∼Bin(300,0.05), soX is approximately normal with meanµX = 300(0.05)= 15 and standard deviationσX =

√300(0.05)(0.95) = 3.7749.


Thez-score of 19.5 is(19.5−15)/3.7749= 1.19.


P(X < 20) = 0.8830.

(b) LetY be the number of assemblies in a sample of 10 that are oversize. ThenX ∼ Bin(10,0.05).

P(X ≥ 1) = 1−P(X = 0) = 1− 10!0!(10−0)!

(0.05)0(1−0.05)10−0 = 0.4013.

(c) Let p be the required probability, and letX represent the number of assemblies in a sample of 300 that areoversize.

Then X ∼ Bin(300, p), so X is approximately normal with meanµX = 300p and standard deviationσX =

√300p(1− p).

P(X ≥ 20) = 0.01.



Thez-score can be expressed in terms ofp by−2.33= (19.5−300p)/√

300p(1− p).

This equation can be rewritten as 91,628.67p2−13,328.67p+380.25= 0.

Solving forp yields p = 0.0390. (0.1065 is a spurious root.)

12. (a) The mean concentration of bubbles is 0.3 per m2, so the mean number of bubbles in a 15 m2 area is 15(0.3) = 4.5.

Let X be the number of bubbles in a 15 m2 area. ThenX ∼ Poisson(4.5).

P(X > 2) = 1−P(X ≤ 2)


222 CHAPTER 4

= 1−P(X = 0)−P(X = 1)−P(X = 2)

= 1−e−4.54.50

0!−e−4.54.51

1!−e−4.54.52

2!= 1−0.01111−0.04999−0.11248

= 0.8264

(b) The mean number of bubbles in a 24 m2 area is 24(0.3) = 7.2.

Let Y be the number of bubbles in a 24 m2 area. ThenY ∼ Poisson(7.2).

P(Y = 0) = e−7.2 7.20

0!= e−7.2 = 0.000747.

(c) The total area of the 50 pieces of glass is 50(18) = 900 m2.

The mean number of bubbles in a 900 m2 area is 900(0.3) = 270.

Let W be the number of bubbles in a 900 m2 area. ThenW ∼ Poisson(270), soW is approximately normalwith meanµX = 270 and standard deviationσX =

√270= 16.4317.

Thez-score of 300 is(300−270)/16.4317= 1.83.


P(W > 300) = 0.0336.

13. (a) Letλ be the true concentration. Thenλ = 56/2 = 28.

The uncertainty isσλ =√

λ/2. Substitutingλ for λ, σλ =√

28/2= 3.7.

λ = 28.0±3.7.

(b) Letv be the required volume, in mL. Thenσλ =√

λ/v= 1.

Substitutingλ = 28 forλ and solving forv yieldsv = 28 mL.

14. (a)X ∼ NB(10,0.9).

P(X = 12) =

(12−110−1

)(0.9)10(1−0.9)12−10 = 0.1918

(b) µX = 10/0.9= 11.11

(c) σX =√

10(1−0.9)/0.92 = 1.1111



15. (a) LetX1,X2,X3 be the three thicknesses. LetS= X1 +X2 +X3 be the thickness of the stack. ThenS is normallydistributed with meanµS = 3(1.5) = 4.5 and standard deviation 0.2

√3 = 0.34641.

Thez-score of 5 is(5−4.5)/0.34641= 1.44.


P(S> 5) = 0.0749.

(b) Letx80 denote the 80th percentile. Thez-score of the 80th percentile is approximatelyz= 0.84.

Thereforex80 satisfies the equation 0.84= (x80−4.5)/0.34641.

Solving forx80 yieldsx80 = 4.7910 cm.

(c) Letn be the required number. ThenµS = 1.5n andσS = 0.2√

n.

P(S> 5) ≥ 0.99, so 5 is less than or equal to the 1st percentile of the distribution ofS.

Thez-score of the 1st percentile isz= −2.33.

Thereforen satisfies the inequality−2.33≥ (5−1.5n)/(0.2√

n).

The smallest value of n satisfying this inequality is the solution to the equation−2.33= (5−1.5n)/(0.2

√n).

This equation can be rewritten 1.5n−0.466√

n−5 = 0.

Solving for√

n with the quadratic formula yields√

n = 1.9877, son = 3.95. The smallest integer value ofn isthereforen = 4.

16. LetT be the lifetime of a microprocessor.

(a) T ∼ Exp(1/3000), soP(T ≤ 300) = 1−e−300/3000= 0.0952

(b) P(T > 6000) = 1−P(T ≤ 6000) = 1− (1−e−6000/3000) = 0.1353

(c) Let S represent the time to failure of the new microprocessor, andlet T represent the additional time, past thefirst 1000 hours, to failure of the old microprocessor. By thelack of memory property, bothSandT have thesame exponential distribution. SinceSandT are independent,P(S< T) = 1/2.

17. (a) LetT represent the lifetime of a bearing.

P(T > 1) = 1−P(T ≤ 1) = 1− (1−e−[(0.8)(1)]1.5) = 0.4889

(b) P(T ≤ 2) = 1−e−[(0.8)(2)]1.5= 0.8679


224 CHAPTER 4

18. LetX1, ...,X16 be the times needed to complete 16 oil changes.

Let T = X1 + · · ·+X16 be the total time needed to complete all 16 oil changes.

ThenT is normally distributed with meanµT = 16(29.5) = 472 and standard deviationσT = 3√

16 = 12.

The mechanic completes 16 oil changes in an eight-hour day ifT < 480.

Thez-score of 480 is(480−472)/12= 0.67.


P(T < 480) = 0.7486.

19. (a)S is approximately normal with meanµS = 75(12.2) = 915 andσS = 0.1√

75= 0.86603.

Thez-score of 914.8 is(914.8−915)/0.86603= −0.23.


P(S< 914.8) = 0.4090.

(b) No. More than 40% of the samples will have a total weight of914.8 ounces or less if the claim is true.

(c) No, because a total weight of 914.8 ounces is not unusually small if the claim is true.

(d) S is approximately normal with meanµS = 75(12.2) = 915 andσS = 0.1√

75= 0.86603.

Thez-score of 910.3 is(910.3−915)/0.86603= −5.43.


P(S< 910.3)≈ 0.

(e) Yes. Almost none of the samples will have a total weight of910.3 ounces or less if the claim is true.

(f) Yes, because a total weight of 910.3 ounces is unusually small if the claim is true.

20. (a) The mean number of hits in five hours is 100, soX ∼ Poisson(100). ThereforeX is approximately normal withmeanµX = 100 andσX =

√100= 10.

Thez-score of 95 is(95−100)/10= −0.50.


P(X ≤ 95) = 0.3085.

(b) No. More than 30% of five-hour periods will have 95 or fewerhits if the claim is true.

(c) No, because a total of 95 hits is not unusually small if theclaim is true.

(d) The mean number of hits in five hours is 100, soX ∼ Poisson(100). ThereforeX is approximately normal withmeanµX = 100 andσX =

√100= 10.



Thez-score of 65 is(65−100)/10= −3.50.


P(X ≤ 65) = 0.0002.

(e) Yes. Only about 2 in 10,000 five-hour periods will have 65 or fewer hits if the claim is true.

(f) Yes, because a total of 65 hits is unusually small if the claim is true.

21. (a)P(X ≤ 0) = F(0) = e−e−0= e−1

(b) P(X > ln2) = 1−P(X ≤ ln2) = 1−F(ln2) = 1−e−e− ln2= 1−e−1/2

(c) Letxm be the median ofX. ThenF(xm) = e−e−xm= 0.5.

Solving forxm yieldse−xm = − ln0.5 = ln2, soxm = − ln(ln2) = 0.3665.

22. (a) The cumulative distribution function ofX is F(x) =Z x

−∞f (t)dt.

If x < θ, F(x) =

Z x

−∞0dt = 0.

If x≥ θ, F(x) =

Z x

θ

rθr

tr+1 dt = rθrZ x

θt−r−1dt = rθr

(t−r

−r

x

θ

)= rθr

(−x−r + θ−r

r

)= 1−

(θx

)r

SoF(x) =

0 x < θ

1−(

θx

)r

x≥ θ

(b) µX =Z ∞

θx

rθr

xr+1 dx= rθrZ ∞

θx−r dx= rθr x1−r

1− r

∞

θ

= rθr θ1−r

r −1=

rθr −1

(c) σ2X =

Z ∞

θx2 rθr

xr+1 dx−µ2X

= rθrZ ∞

θx−r+1dx−

(rθ

r −1

)2

= rθr x2−r

2− r

∞

θ

−(

rθr −1

)2

= rθr θ2−r

r −2−(

rθr −1

)2


226 CHAPTER 4

=rθ2

(r −1)2(r −2)

(d) If r ≤ 1, then the integralZ ∞

θx

rθr

xr+1 dx= rθrZ ∞

θx−r dx= rθr x1−r

1− r

∞

θ

diverges, because limx→∞x1−r

1− r= ∞.

(e) If r ≤ 2, then the integralZ ∞

θx2 rθr

xr+1 dx= rθrZ ∞

θx−r+1dx= rθr x2−r

2− r

∞

θ

diverges, because limx→∞x2−r

2− r= ∞.

23. (a) fX(x) = F ′(x) =e−(x−α)/β

β[1+e−(x−α)/β]2

(b) fX(α−x) = fX(α+x) =ex/β

β[1+ex/β]2

(c) SincefX(x) is symmetric aroundα, its center of mass is atx = α.

24. (a)µX =

Z ∞

0x[pλ1e−λ1x +(1− p)λ2e

−λ2x]dx

= −pxe−λ1x∞

0

+ pZ ∞

0e−λ1x dx− (1− p)xe−λ2x

∞

0

+(1− p)

Z ∞

0e−λ2x dx

= −p(0)+ p1λ1

− (1− p)(0)+ (1− p)1λ2

=p

λ1+

1− pλ2

(b) The cumulative distribution function ofX is F(x) =

Z x

−∞f (t)dt.

If x < 0, thenF(x) =Z x

−∞0dt = 0.

If x≥ 0, then

F(x) =

Z x

0pλ1e−λ1t +(1− p)λ2e

−λ2t dt = p(1−e−λ1x)+ (1− p)(1−e−λ2x) = 1− pe−λ1x− (1− p)e−λ2x

(c) P(X ≤ 2) = F(2) = 1−0.5e−2(2)− (1−0.5)e−1(2) = 1−0.5e−4−0.5e−2 = 0.9232



(d) LetM1 be the event that mass 1 was selected.

P(M1|X ≤ 2) =P(X ≤ 2∩M1)

P(X ≤ 2).

NowP(X ≤2∩M1)= P(X ≤2|M1)P(M1)= (1−e−2(2))(0.5). From part (c),P(X ≤2)= 1−0.5e−4−0.5e−2.

ThereforeP(M1|X ≤ 2) =1−0.5e−4−0.5e−2

(1−e−2(2))(0.5)= 0.5317.

25. (a)P(X > s) = P(First s trials are failures) = (1− p)s

(b) P(X > s+ t |X > s) = P(X > s+ t andX > s)/P(X > s)

= P(X > s+ t)/P(X > s)

= (1− p)s+t/(1− p)s

= (1− p)t

= P(X > t)

Note that ifX > s+ t, it must be the case thatX > s, which is the reason thatP(X > s+ t andX > s) = P(X > s+ t).

(c) LetX be the number of tosses of the penny needed to obtain the first head.

ThenP(X > 5|X > 3) = P(X > 2) = 1/4.

The probability that the nickel comes up tails twice is also 1/4.

26. LetU be the length of the part of the stick used as the length of the rectangle. Then the width is 1−U , so thearea isU(1−U).

SinceU ∼U(0,1), the probability density function ofU is f (u) = 1 for 0< u < 1 and f (u) = 0 otherwise.

The mean area is thereforeZ 1

0u(1−u)(1)du=

16

.

27. (a)FY(y) = P(Y ≤ y) = P(7X ≤ y) = P(X ≤ y/7).

SinceX ∼ Exp(λ), P(X ≤ y/7) = 1−e−λy/7.

(b) fY(y) = F ′Y(y) = (λ/7)e−λy/7


228 CHAPTER 4

28. (a)P(X = x)

P(X = x−1)=

n!x!(n−x)!

px(1− p)n−x

n!(x−1)!(n−x+1)!

px−1(1− p)n−x+1

=

[n!(x−1)!(n−x+1)!

n!x!(n−x)!

](px

px−1

)[(1− p)n−x

(1− p)n−x+1

]

=

(n−x+1

x

)(p

1− p

)

(b) Letx be an integer such thatP(X = x) ≥ P(X = x−1). Then

(n−x+1

x

)(p

1− p

)≥ 1

p(n−x+1) ≥ x(1− p)

np−xp+ p ≥ x−xp

np+ p ≥ x

Reversing the above steps shows that ifx≤ np+ p thenP(X = x) ≥ P(X = x−1).

29. (a)P(X = x)

P(X = x−1)=

e−λλx/x!

e−λλx−1/(x−1)!=

e−λλx(x−1)!

e−λλx−1x!=

λx

(b) P(X = x) ≥ P(X = x−1) if and only ifλx≥ 1 if and only ifx≤ λ.

30. (a)FX(x) = P(X ≤ x) = P(σZ+µ≤ x) = P(Z ≤ x−µσ ) = Φ( x−µ

σ )

(b) F ′X(x) = (1/σ)Φ′( x−µ

σ ) = (1/σ)φ( x−µσ ) = 1

σ√

2π e− (x−µ)2

2σ2

(c) FX(x) = P(X ≤ x) = P(−σZ+µ≤ x) = P(Z ≥ x−µ−σ ) = 1−Φ( x−µ

−σ ),

soF ′X(x) = (1/σ)Φ′( x−µ

−σ ) = (1/σ)φ( x−µ−σ ) = 1

σ√

2π e− (x−µ)2

2σ2 .


SECTION 5.1 229

Chapter 5

Section 5.1

1. (a) 1.96

(b) 2.33

(c) 2.57 or 2.58

(d) 1.28

2. (a) 95%

(b) 97%

(c) 80%

(d) 99.9%

3. The level is the proportion of samples for which the confidence interval will cover the true value. Therefore asthe level goes up, the reliability goes up. This increase in reliability is obtained by increasing the width of theconfidence interval. Therefore as the level goes up the precision goes down.

4. (a)X = 3.8, s= 4.8, n = 74,z.025 = 1.96.

The confidence interval is 3.8±1.96(4.8/√

74), or (2.706, 4.894).

(b) X = 3.8, s= 4.8, n = 74,z.01 = 2.33.


74), or (2.500, 5.100).

(c) X = 3.8, s= 4.8, n = 74, so the upper confidence bound 4.4 satisfies 4.4 = 3.8+zα/2(4.8/√

74).

Solving forzα/2 yieldszα/2 = 1.08.

The area to the right ofz= 1.08 is 1−0.8599= 0.1401, soα/2 = 0.1401.

The level is 1−α = 1−2(0.1401)= 0.7198, or 71.98%.


230 CHAPTER 5

(d) z.025 = 1.96. 1.96(4.8/√

n) = 0.7, son = 181.

(e) z.01 = 2.33. 2.33(4.8/√

n) = 0.7, son = 256.

5. (a)X = 6230,s= 221,n = 100,z.025 = 1.96.

The confidence interval is 6230±1.96(221/√

100), or (6186.7, 6273.3).

(b) X = 6230,s= 221,n = 100,z.005 = 2.58.


100), or (6173.0, 6287.0).

(c) X = 6230,s= 221,n = 100, so the upper confidence bound 6255 satisfies 6255= 6230+zα/2(221/√

100).



The level is 1−α = 1−2(0.1292)= 0.7416, or 74.16%.

(d) z.025 = 1.96. 1.96(221/√

n) = 25, son = 301.

(e) z.005 = 2.58. 2.58(221/√

n) = 25, son = 521.

6. (a)X = 1.57,s= 0.23,n = 150,z.025 = 1.96.


150), or (1.5332, 1.6068).

(b) X = 1.57,s= 0.23,n = 150,z.0025= 2.81.


150), or (1.5172, 1.6228).

(c) X = 1.57,s= 0.23,n = 150, so the upper confidence bound 1.60 satisfies 1.60= 1.57+zα/2(0.23/√

150).



The level is 1−α = 1−2(0.0548)= 0.8904, or 89.04%.

(d) z.025 = 1.96. 1.96(0.23/√

n) = 0.02, son = 509.

(e) z.0025= 2.81. 2.81(0.23/√

n) = 0.02, son = 1045.


SECTION 5.1 231

7. (a)X = 150,s= 25,n = 100,z.025 = 1.96.


100), or (145.10, 154.90).

(b) X = 150,s= 25,n = 100,z.005 = 2.58.


100), or (143.55, 156.45).


100).



The level is 1−α = 1−2(0.1151)= 0.7698, or 76.98%.

(d) z.025 = 1.96. 1.96(25/√

n) = 2, son = 601.

(e) z.005 = 2.58. 2.58(25/√

n) = 2, son = 1041.

8. (a)X = 21.6, s= 3.2, n = 53,z.05 = 1.645.


53), or (20.88, 22.32).

(b) X = 21.6, s= 3.2, n = 53,z.025 = 1.96.


53), or (20.74, 22.46).

(c) X = 21.6, s= 3.2, n = 53, so the upper confidence bound 22.2 satisfies 22.2 = 21.6+zα/2(3.2/√

53).



The level is 1−α = 1−2(0.0853)= 0.8294, or 82.94%.

(d) z.05 = 1.645. 1.645(3.2/√

n) = 0.3, son = 308.

(e) z.025 = 1.96. 1.96(3.2/√

n) = 0.3, son = 438.

9. (a)X = 51.72,s= 2.52,n = 80,z.025 = 1.96.


80), or (51.168, 52.272).


232 CHAPTER 5

(b) X = 51.72,s= 2.52,n = 80,z.01 = 2.33.


80), or (51.064, 52.376).

(c) X = 51.72,s= 2.52,n = 80, so the upper confidence bound 52.14 satisfies 52.14= 51.72+zα/2(2.52/√

80).



The level is 1−α = 1−2(0.0681)= 0.8638, or 86.38%.

(d) z.025 = 1.96. 1.96(2.52/√

n) = 0.25, son = 391.

(e) z.01 = 2.33. 2.33(2.52/√

n) = 0.25, son = 552.

10. (a)X = 72,s= 10,n = 150,z.025 = 1.96.


150), or (70.40, 73.60).

(b) X = 72,s= 10,n = 150,z.0025= 2.81.


150), or (69.71, 74.29).


150).



The level is 1−α = 1−2(0.1112)= 0.7776, or 77.76%.

(d) z.025 = 1.96. 1.96(10/√

n) = 1.5, son = 171.

(e) z.0025= 2.81. 2.81(10/√

n) = 1.5, son = 351.

11. (a)X = 29,s= 9, n = 81,z.025 = 1.96.


81), or (27.04, 30.96).

(b) X = 29,s= 9, n = 81,z.005 = 2.58.


81), or (26.42, 31.58).


SECTION 5.1 233

(c) X = 29,s= 9, n = 81, so the upper confidence bound 30.5 satisfies 30.5 = 29+zα/2(9/√

81).



The level is 1−α = 1−2(0.0668)= 0.8664, or 86.64%.

(d) z.025 = 1.96. 1.96(9/√

n) = 1, son = 312.

(e) z.005 = 2.58. 2.58(9/√

n) = 1, son = 540.

12. (a)X = 6230,s= 221,n = 100,z.05 = 1.645.

The lower confidence bound is 6230−1.645(221/√

100) = 6913.6.

(b) The lower confidence bound 6220 satisfies 6220= 6230−zα(221/√

100).

Solving forzα yieldszα = 0.45.

The area to the left ofz= 0.45 is 1−α = 0.6736.

The level is 0.6736, or 67.36%.

13. (a)X = 1.57,s= 0.23,n = 150,z.02 = 2.05.

The lower confidence bound is 1.57−2.05(0.23/√

150) = 1.5315.

(b) The upper confidence bound 1.58 satisfies 1.58= 1.57+zα(0.23/√

150).



The level is 0.7019, or 70.19%.

14. (a)X = 150,s= 25,n = 100,z.05 = 1.645.


100) = 145.89.


100).



The level is 0.7881, or 78.81%.


234 CHAPTER 5

15. (a)X = 21.6, s= 3.2, n = 53,z.01 = 2.33.

The upper confidence bound is 21.6+2.33(3.2/√

53) = 22.62.

(b) The upper confidence bound 22.7 satisfies 22.7 = 21.6+zα(3.2/√

53).



The level is 0.9938, or 99.38%.

16. (a)X = 51.72,s= 2.52,n = 80,z.10 = 1.28.

The upper confidence bound is 51.72+1.28(2.52/√

80) = 52.081.

(b) The upper confidence bound 52 satisfies 52= 51.72+zα(2.52/√

80).



The level is 0.8389, or 83.89%.

17. (a)X = 72,s= 10,n = 150,z.02 = 2.05.


150) = 70.33.


150).



The level is 0.9929, or 99.29%.

18. (a)X = 29,s= 9, n = 81,z.05 = 1.645.

The upper confidence bound is 29+1.645(9/√

81) = 30.645.


81).



The level is 0.8413, or 84.13%.


SECTION 5.1 235

19. With a sample size of 70, the standard deviation ofX is σ/√

70. To make the interval half as wide, the standarddeviation ofX will have to beσ/(2

√70) = σ/

√280. The sample size needs to be 280.

20. (ii) one-half as wide. The ratio of the widths is equal to the ratio of the standard deviations of the sample mean,

which isσ/

√400

σ/√

100=√

100/400= 1/2.

21. The sample meanX is the midpoint of the interval, soX = 0.227. The upper confidence bound 0.241 satisfies0.241= 0.227+1.96(s/

√n).

Therefores/√

n= 0.00714286. A 90% confidence interval is 0.227±1.645(s/√

n)= 0.227±1.645(0.00714286),or (0.21525, 0.23875).

22. (a) True. This results from the expressionX±1.96(s/√

n), which is a 95% confidence interval.

(b) False. This is a specific interval, so it does not make sense to talk about the probability that it covers the truevalue.

(c) False. A confidence interval says something about where the true value is likely to be. It does not say wherethe next measurement is likely to be.

23. (a) False. The confidence interval is for the population mean, not the sample mean. The sample mean is known,so there is no need to construct a confidence interval for it.

(b) True. This results from the expressionX±1.96(s/√

n), which is a 95% confidence interval for the populationmean.

(c) False. The standard deviation of the mean involves the square root of the sample size, not of the populationsize.

24. LetX be the number of 90% confidence intervals that fail to cover the true mean.

Then X ∼ Bin(200,0.10), so X is approximately normal with meanµX = 200(0.10) = 20 and standarddeviation

√200(0.10)(0.90) = 4.242641.



236 CHAPTER 5

Thez-score of 15.5 is(15.5−20)/4.242641= −1.06.


P(X > 15) = 0.8554.

25. The supervisor is underestimating the confidence. The statement that the mean cost is less than $160 is aone-sided upper confidence bound with confidence level 97.5%.

26. It is not a valid confidence interval because the data uponwhich it is based are not a simple random sample.

Section 5.2

1. (a) p = 37/50= 0.74

(b) X = 37,n = 50, p = (37+2)/(50+4)= 0.72222,z.025 = 1.96.

The confidence interval is 0.72222±1.96√

0.72222(1−0.72222)/(50+4), or (0.603, 0.842).

(c) Letn be the required sample size.

Thenn satisfies the equation 0.10= 1.96√

p(1− p)/(n+4).

Replacing ˜p with 0.72222 and solving forn yieldsn = 74.

(d) X = 37,n = 50, p = (37+2)/(50+4)= 0.72222,z.005 = 2.58.


0.72222(1−0.72222)/(50+4), or (0.565, 0.879).

(e) Letn be the required sample size.


p(1− p)/(n+4).


2. (a)X = 17,n = 75, p = (17+2)/(75+4)= 0.24051,z.025 = 1.96.


0.24051(1−0.24051)/(75+4), or (0.146, 0.335).

(b) X = 17,n = 75, p = (17+2)/(75+4)= 0.24051,z.01 = 2.33.


0.24051(1−0.24051)/(75+4), or (0.128, 0.353).


SECTION 5.2 237



p(1− p)/(n+4).


(d) Letn be the required sample size.


p(1− p)/(n+4).


(e) LetX be the number of 95% confidence intervals that cover the true proportion.


√200(0.95)(0.05) = 3.082207.


Thez-score of 192.5 is(192.5−190)/3.082207= 0.81.


P(X > 192) = 0.2090.

3. (a)X = 52,n = 70, p = (52+2)/(70+4)= 0.72973,z.025 = 1.96.


0.72973(1−0.72973)/(70+4), or (0.629, 0.831).

(b) X = 52,n = 70, p = (52+2)/(70+4)= 0.72973,z.05 = 1.645.


0.72973(1−0.72973)/(70+4), or (0.645, 0.815).



p(1− p)/(n+4).




p(1− p)/(n+4).


(e) LetX be the number of 90% confidence intervals that cover the true proportion.


√300(0.90)(0.10) = 5.196152.


Thez-score of 280.5 is(280.5−270)/5.196152= 2.02.


238 CHAPTER 5


P(X > 192) = 0.0217.

4. (a)X = 85,n = 100, p = (85+2)/(100+4)= 0.83654,z.025 = 1.96.


0.83654(1−0.83654)/(100+4), or (0.765, 0.908).

(b) X = 85,n = 100, p = (85+2)/(100+4)= 0.83654,z.01 = 2.33.


0.83654(1−0.83654)/(100+4), or (0.752, 0.921).



p(1− p)/(n+4).




p(1− p)/(n+4).


5. (a)X = 859,n = 10501,p = (859+2)/(10501+4)= 0.081961,z.025 = 1.96.


0.081961(1−0.081961)/(10501+4), or (0.0767, 0.0872).

(b) X = 859,n = 10501,p = (859+2)/(10501+4)= 0.081961,z.005 = 2.58.


0.081961(1−0.081961)/(10501+4), or (0.0751, 0.0889).

(c) The upper confidence bound 0.085 satisfies the equation 0.085= 0.081961+zα√

0.081961(1−0.081961)/(10501+4)

Solving forzα yieldszα = 1.14. The area to the left ofz= 1.14 is 1−α = 0.8729.

The level is 0.8729, or 87.29%.

6. X = 37,n = 50, p = (37+2)/(50+4)= 0.72222,z.05 = 1.645.

The lower confidence bound is 0.72222−1.645√

0.72222(1−0.72222)/(50+4), or 0.622.

7. X = 17,n = 75, p = (17+2)/(75+4)= 0.24051,z.02 = 2.05.

The upper confidence bound is 0.24051+2.05√

0.24051(1−0.24051)/(75+4), or 0.339.


SECTION 5.2 239

8. X = 52,n = 70, p = (52+2)/(70+4)= 0.72973,z.01 = 2.33.

The lower confidence bound is 0.72973−2.33√

0.72973(1−0.72973)/(70+4), or 0.609.

9. (a)X = 30,n = 400, p = (30+2)/(400+4)= 0.079208,z.025 = 1.96.


0.079208(1−0.079208)/(400+4), or (0.0529, 0.1055).

(b) Letn be the required sample size.


p(1− p)/(n+4).


(c) LetX be the number of defective components in a lot of 200.

Let p be the population proportion of components that are defective. ThenX ∼ Bin(200, p), soX is approxi-mately normally distributed with meanµX = 200p andσX =

√200p(1− p).

Let r represent the proportion of lots that are returned.

Using the continuity correction,r = P(X > 20.5).

To find a 95% confidence interval forr, express thez-score ofP(X > 20.5) as a function ofp and substitute theupper and lower confidence limits forp.

Thez-score of 20.5 is(20.5−200p)/√

200p(1− p). Now find a 95% confidence interval forzby substitutingthe upper and lower confidence limits forp.

From part (a), the 95% confidence interval forp is (0.052873, 0.10554). Extra precision is used for this confi-dence interval to get good precision in the final answer.

Substituting 0.052873 forp yieldsz= 3.14. Substituting 0.10554 forp yieldsz= −0.14.

Since we are 95% confident that 0.052873< p < 0.10554, we are 95% confident that−0.14< z< 3.14.

The area to the right ofz= −0.14 is 1−0.4443= 0.5557. The area to the right ofz= 3.14 is 1−0.9992=0.0008.

Therefore we are 95% confident that 0.0008< r < 0.5557.

The confidence interval is (0.0008, 0.5557).

10. q = (1− p)2. From part (a), we are 95% confident that 0.0529< p < 0.1055.

We are therefore 95% confident that(1−0.1055)2 < q< (1−0.0529)2, or equivalently, that 0.800< q< 0.897.


11. (a)X = 26,n = 42, p = (26+2)/(42+4)= 0.60870,z.05 = 1.645.


0.60870(1−0.60870)/(42+4), or (0.490, 0.727).


240 CHAPTER 5

(b) X = 41,n = 42, p = (41+2)/(42+4)= 0.93478,z.025 = 1.96.

The expression for the confidence interval yields 0.93478± 1.96√

0.93478(1−0.93478)/(42+4),or (0.863, 1.006).

Since the upper limit is greater than 1, replace it with 1.

The confidence interval is (0.863, 1).

(c) X = 32,n = 42, p = (32+2)/(42+4)= 0.73913,z.005 = 2.58.


0.73913(1−0.73913)/(42+4), or (0.572, 0.906).

12. (a)X = 63,n = 150, p = (63+2)/(150+4)= 0.42208,z.025 = 1.96.


0.42208(1−0.42208)/(150+4), or (0.344, 0.500).

(b) X = 63,n = 150, p = (63+2)/(150+4)= 0.42208,z.005 = 2.58.


0.42208(1−0.42208)/(150+4), or (0.319, 0.525).



p(1− p)/(n+4).




p(1− p)/(n+4).


13. (a) Letn be the required sample size.


p(1− p)/(n+4).

Since there is no preliminary estimate of ˜p available, replace ˜p with 0.5.

Solving forn yieldsn = 381.

(b) X = 20,n = 100, p = (20+2)/(100+4)= 0.21154,z.025 = 1.96.


0.21154(1−0.21154)/(100+4), or (0.133, 0.290).



p(1− p)/(n+4).



SECTION 5.3 241

14. (a) Letn be the required sample size. Thenn satisfies the equation 0.05= 2.33√

p(1− p)/(n+4).


The equation becomes 0.05= 2.33√

0.5(1−0.5)/(n+4). Solving forn yieldsn = 539.

(b) X = 30,n = 200, p = (30+2)/(200+4)= 0.15686,z.01 = 2.33.


0.15686(1−0.15686)/(200+4), or (0.0975, 0.2162).

(c) Letn be the required sample size. Thenn satisfies the equation 0.05= 2.33√

p(1− p)/(n+4).


15. (a)X = 61,n = 189, p = (61+2)/(189+4)= 0.32642,z.05 = 1.645.


0.32642(1−0.32642)/(189+4), or (0.271, 0.382).

(b) Letn be the required sample size. Thenn satisfies the equation 0.03= 1.645√

p(1− p)/(n+4).


(c) Letn be the required sample size. Thenn satisfies the equation 0.05= 1.645√

p(1− p)/(n+4).


The equation becomes 0.03= 1.645√

0.5(1−0.5)/(n+4). Solving forn yieldsn = 748.

16. No, the data are not a simple random sample.

Section 5.3

1. (a) 1.796

(b) 2.447

(c) 63.657

(d) 2.048


242 CHAPTER 5

2. (a) 1.363

(b) 1.943

(c) 31.821

(d) 1.701

3. (a) 95%

(b) 98%

(c) 99%

(d) 80%

(e) 90%

4. False. The population must be approximately normal.

5. X = 5.900,s= 0.56921,n = 6, t6−1,.025= 2.571.


6), or (5.303, 6.497).

6. No. The maximum value is 39.920, yet the third quartile is only 8.103, and the interquartile range is 6.136.The distance between the third quartile and the maximum is thus more than 5 times the interquartile range. Themaximum value is an outlier, so it is not appropriate to assume that this is a sample from a normal population.

7. Yes it is appropriate, since there are no outliers.

X = 205.1267,s= 1.7174,n = 9, t9−1,.025= 2.306.


9), or (203.81, 206.45).

8. (a)X = 19.35,s= 0.577,n = 6, t6−1,.025= 2.571.


6), or (18.744, 19.956).


SECTION 5.3 243

(b) X = 19.35,s= 0.577,n = 6, t6−1,.01 = 3.365.


6), or (18.577, 20.143).

(c) No, the value 18.26 is an outlier.

9. (a) 0.3 0.35 0.4 0.45

(b) Yes, there are no outliers.

X = 0.3688,s= 0.0199797,n = 8, t8−1,.005= 3.499.


8), or (0.344, 0.394).

(c) 0.3 0.4 0.5

(d) No, the data set contains an outlier.

10. X = 86.560,s= 1.0213,n = 5, t5−1,.005= 4.604.


5), or (84.457, 88.663).

11. X = 13.040,s= 1.0091,n = 10,t10−1,.025= 2.262.


10), or (12.318, 13.762).

12. X = 7.220,s= 2.3285,n = 5, t5−1,.025= 2.776.


5), or (4.329, 10.111).

13. X = 1.250,s= 0.6245,n = 4, t4−1,.05 = 2.353.


4), or (0.515, 1.985).

14. (a) N−1 = 10−1= 9.

(b) X = 6.59635,s= 0.11213,n = 10,t10−1,.005= 3.250.


10), or (6.481, 6.712).

Alternatively, one may compute 6.59635±3.250(0.03546).


244 CHAPTER 5

15. (a) SE Mean is StDev/√

N, so 0.52640 = StDev/√

20, so StDev = 2.3541.

(b) X = 2.39374,s= 2.3541,n = 20,t20−1,.005= 2.861.

The lower limit of the 99% confidence interval is 2.39374−2.861(2.3541/√

20) = 0.888.

Alternatively, one may compute 2.39374−2.861(0.52640).

(c) X = 2.39374,s= 2.3541,n = 20,t20−1,.005= 2.861.

The upper limit of the 99% confidence interval is 2.39374+2.861(2.3541/√

20) = 3.900.

Alternatively, one may compute 2.39374+2.861(0.52640).

16. The value 85 comes from a normal population with standarddeviationσ = 8. Sinceσ is known, a 95%confidence interval is based onz.025 = 1.96. The confidence interval is 85±1.96(8), or (69.32, 100.68).

17. (a)X = 21.7, s= 9.4, n = 5, t5−1,.025= 2.776.


5), or (10.030, 33.370).

(b) No. The minimum possible value is 0, which is less than twosample standard deviations below the samplemean. Therefore it is impossible to observe a value that is two or more sample standard deviations below thesample mean. This suggests that the sample may not come from anormal population.

Section 5.4

1. X = 620,sX = 20,nX = 80,Y = 750,sY = 30,nY = 95,z.025 = 1.96.

The confidence interval is 750−620±1.96√

202/80+302/95, or (122.54, 137.46).

2. X = 126.31,sX = 18.10,nX = 87,Y = 162.41,sY = 28.49,nY = 132,z.025 = 1.96.

The confidence interval is 162.41−126.31±1.96√

18.102/87+28.492/132, or (29.9284, 42.2716).

3. X = 517.0, sX = 2.4, nX = 35,Y = 510.1,sY = 2.1, nY = 47,z.005 = 2.58.


2.42/35+2.12/47, or (5.589, 8.211).


SECTION 5.4 245

4. X = 1840,sX = 244,nX = 64,Y = 2475,sY = 760,nY = 88,z.025 = 1.96.


2442/64+7602/88, or (465.3, 804.7).

5. X = 26.50,sX = 2.37,nX = 39,Y = 37.14,sY = 3.66,nY = 142,z.025 = 1.96.


2.372/39+3.662/142, or (9.683, 11.597).

6. X = 3.72,sX = 0.51,nX = 75,Y = 4.82,sY = 0.42,nY = 100,z.025 = 1.96.


0.512/75+0.422/100, or (0.9582, 1.2418).

7. X = 0.58,sX = 0.16,nX = 80,Y = 0.41,sY = 0.18,nY = 60,z.01 = 2.33.


0.162/80+0.182/60, or (0.1017, 0.2383).

8. X = 1250,sX = 350,nX = 120,Y = 1400,sY = 250,nY = 90,z.01 = 2.33.


3502/120+2502/90, or (53.501, 246.499).

9. X = 242,sX = 20,nX = 47,Y = 220,sY = 31,nY = 42,z.025 = 1.96.


202/47+312/42, or (11.018, 32.982).

10. X = 1827,sX = 168,nX = 60,Y = 1658,sY = 225,nY = 180,z.025 = 1.96.


1682/60+2252/180, or (115.26, 222.74).

11. (a)X = 91.1, sX = 6.23,nX = 50,Y = 90.7, sY = 4.34,nY = 40,z.025 = 1.96.


6.232/50+4.342/40, or(−1.789,2.589).

(b) No. Since 0 is in the confidence interval, it may be regarded as being a plausible value for the mean differencein hardness.

12. The standard deviation of the difference between the means is√

σ2X/nX + σ2

Y/nY. EstimateσX ≈ sX = 6.23andσY ≈ sY = 4.34.


246 CHAPTER 5

If all 10 new welds are cooled at 10C, thennX = 60 andnY = 40.The standard deviation of the difference between the means is then

√6.232/60+4.342/40= 1.057.

If all 10 new welds are cooled at 30C, thennX = 50 andnY = 50.The standard deviation of the difference between the means is then

√6.232/50+4.342/50= 1.074.

If 5 of the new welds are cooled at 10C and 5 at 30C, thennX = 55 andnY = 45.The standard deviation of the difference between the means is then

√6.232/55+4.342/45= 1.060.

Therefore the confidence interval would be most precise if all 10 new welds were cooled at 10C.

13. It is not possible. The amounts of time spent in bed and spent asleep in bed are not independent.

14. X = 0.27,sX = 0.40,nX = 349,Y = 1.62,sY = 1.70,nY = 143,z.025 = 1.96.


0.402/349+1.702/143, or (1.068, 1.632).

Section 5.5

1. X = 20,nX = 100, pX = (20+1)/(100+2)= 0.205882,Y = 10,nY = 150, pY = (10+1)/(150+2)= 0.072368,z.05 = 1.645.

The confidence interval is 0.205882−0.072368± 1.645

√0.205882(1−0.205882)

100+2+

0.072368(1−0.072368)150+2

,

or (0.0591, 0.208).

2. X = 68,nX = 85, pX = (68+1)/(85+2)= 0.793103,Y = 105,nY = 120, pY = (105+1)/(120+2)= 0.868852,z.025 = 1.96.


√0.793103(1−0.793103)

85+2+

0.868852(1−0.868852)120+2

,

or (−0.180,0.0283).

3. X = 46,nX = 340, pX = (46+1)/(340+2)= 0.13743,Y = 21,nY = 85, pY = (21+1)/(85+2)= 0.25287,z.01 = 2.33.


√0.13743(1−0.13743)

340+2+

0.25287(1−0.25287)85+2

,

or (−0.232,0.00148).


SECTION 5.5 247

4. X = 110,nX = 1200,pX = (110+1)/(1200+2)= 0.092346,Y = 95,nY = 900, pY = (95+1)/(900+2)= 0.10643,z.005 = 2.58.


√0.092346(1−0.092346)

1200+2+

0.10643(1−0.10643)900+2

,

or (−0.0482,0.0201).

5. (a)X = 62,nX = 400, pX = (62+1)/(400+2)= 0.15672,Y = 12,nY = 100, pY = (12+1)/(100+2)= 0.12745,z.025 = 1.96.


√0.15672(1−0.15672)

400+2+

0.12745(1−0.12745)100+2

,

or (−0.0446,0.103).

(b) The width of the confidence interval is±1.96

√pX(1− pX)

nX +2+

pY(1− pY)

nY +2.

Estimate ˜pX = 0.15672 and ˜pY = 0.12745.

Then if 100 additional chips were sampled from the less expensive process,nX = 500 andnY = 100, so thewidth of the confidence interval would be approximately

±1.96

√0.15672(1−0.15672)

502+

0.12745(1−0.12745)102

= ±0.0721.

If 50 additional chips were sampled from the more expensive process,nX = 400 andnY = 150, so the width ofthe confidence interval would be approximately

±1.96

√0.15672(1−0.15672)

402+

0.12745(1−0.12745)152

= ±0.0638.

If 50 additional chips were sampled from the less expensive process and 25 additional chips were sampledfrom the more expensive process,nX = 450 andnY = 125, so the width of the confidence interval would beapproximately

±1.96

√0.15672(1−0.15672)

452+

0.12745(1−0.12745)127

= ±0.0670.

Therefore the greatest increase in precision would be achieved by sampling 50 additional chips from the moreexpensive process.

6. X = 112,nX = 143, pX = (112+1)/(143+2)= 0.77931,Y = 182,nY = 349, pY = (182+1)/(349+2)= 0.52137,z.025 = 1.96.


√0.77931(1−0.77931)

143+2+

0.52137(1−0.52137)349+2

,

or (0.173, 0.343).


248 CHAPTER 5

7. No. The sample proportions come from the same sample rather than from two independent samples.

8. X = 345,nX = 1500,pX = (345+1)/(1500+2)= 0.23036,Y = 195,nY = 1500,pY = (195+1)/(1500+2)= 0.13049,z.025 = 1.96.


√0.23036(1−0.23036)

1500+2+

0.13049(1−0.13049)1500+2

,

or (0.0726, 0.127).

9. X = 92,nX = 500, pX = (92+1)/(500+2)= 0.18526,Y = 65,nY = 500, pY = (65+1)/(500+2)= 0.13147,z.005 = 2.58.


√0.18526(1−0.18526)

500+2+

0.13147(1−0.13147)500+2

,

or (−0.0055,0.1131).

10. X = 2280,nX = 4644,pX = (2280+1)/(4644+2)= 0.49096,Y = 1982,nY = 4584,pY = (1982+1)/(4584+2)= 0.43240,z.05 = 1.645.


√0.49096(1−0.49096)

4644+2+

0.43240(1−0.43240)4584+2

,

or (0.0415, 0.0756).

11. No, these are not simple random samples.

Section 5.6

1. X = 1.5, sX = 0.25,nX = 7, Y = 1.0, sY = 0.15,nY = 5.

The number of degrees of freedom is

ν =

[0.252

7+

0.152

5

]2

(0.252/7)2

7−1+

(0.152/5)2

5−1

= 9, rounded down to the nearest integer.

t9,.005 = 3.250, so the confidence interval is 1.5−1.0± 3.250

√0.252

7+

0.152

5,

or (0.1234,0.8766).


SECTION 5.6 249

2. X = 6.83,sX = 1.72,nX = 5, Y = 3.32,sY = 1.17,nY = 7.


ν =

[1.722

5+

1.172

7

]2

(1.722/5)2

5−1+

(1.172/7)2

7−1



√1.722

5+

1.172

7,

or (1.3389,5.6811).

3. X = 1.13,sX = 0.0282843,nX = 5, Y = 1.174,sY = 0.0194936,nY = 5.


ν =

[0.02828432

5+

0.01949362

5

]2

(0.02828432/5)2

5−1+

(0.01949362/5)2

5−1



√0.02828432

5+

0.01949362

5,

or (−0.0978,0.00975).

4. X = 3.6, sX = 0.141421,nX = 4, Y = 3.2, sY = 0.182574,nY = 4.


ν =

[0.1414212

4+

0.1825742

4

]2

(0.1414212/4)2

4−1+

(0.1825742/4)2

4−1



√0.1414212

4+

0.1825742

4,

or (0.103, 0.697).

5. X = 73.1, sX = 9.1, nX = 10, Y = 53.9, sY = 10.7, nY = 10.


ν =

[9.12

10+

10.72

10

]2

(9.12/10)2

10−1+

(10.72/10)2

10−1



250 CHAPTER 5


√9.12

10+

10.72

10, or (7.798, 30.602).

6. X = 109.5, sX = 31.2, nX = 12, Y = 47.2, sY = 15.1, nY = 10.


ν =

[31.22

12+

15.12

10

]2

(31.22/12)2

12−1+

(15.12/10)2

10−1



√31.22

12+

15.12

10, or (32.523, 92.077).

7. X = 33.8, sX = 0.5, nX = 4, Y = 10.7, sY = 3.3, nY = 8.


ν =

[0.52

4+

3.32

8

]2

(0.52/4)2

4−1+

(3.32/8)2

8−1



√0.52

4+

3.32

8, or (20.278, 25.922).

8. X = 32.17,sX = 2.23,nX = 5, Y = 26.48,sY = 2.02,nY = 6.


ν =

[2.232

5+

2.022

6

]2

(2.232/5)2

5−1+

(2.022/6)2

6−1



√2.232

5+

2.022

6, or (1.942, 9.438).

9. X = 0.498,sX = 0.036,nX = 5, Y = 0.389,sY = 0.049,nY = 5.



SECTION 5.6 251

ν =

[0.0362

5+

0.0492

5

]2

(0.0362/5)2

5−1+

(0.0492/5)2

5−1



√0.0362

5+

0.0492

5, or (0.0447, 0.173).

10. X = 4.495714,sX = 0.177469,nX = 7, Y = 2.465000,sY = 0.507149,nY = 8.


ν =

[0.1774692

7+

0.5071492

8

]2

(0.1774692/7)2

7−1+

(0.5071492/8)2

8−1



√0.1774692

7+

0.5071492

8,

or (1.4763, 2.5851).

11. X = 482.7857,sX = 13.942125,nX = 14, Y = 464.7000,sY = 14.23789,nY = 9.


ν =

[13.9421252

14+

14.237892

9

]2

(13.9421252/14)2

14−1+

(14.237892/9)2

9−1



√13.9421252

14+

14.237892

9,

or (2.500, 33.671).

12. X = 62.71429,sX = 3.8607,nX = 7, Y = 60.40000,sY = 5.46097,nY = 10.


ν =

[3.86072

7+

5.460972

10

]2

(3.86072/7)2

7−1+

(5.460972/10)2

10−1



√3.86072

7+

5.460972

10,

or (−2.535,7.164).


252 CHAPTER 5

13. X = 229.54286,sX = 14.16885,nX = 7, Y = 143.95556,sY = 59.75699,nY = 9.


ν =

[14.168852

7+

59.756992

9

]2

(14.168852/7)2

7−1+

(59.756992/9)2

9−1



√14.168852

7+

59.756992

9,

or (38.931, 132.244).

14. X = 95,sX = 1, nX = 5, Y = 70,sY = 6, nY = 5.


ν =

[12

5+

62

5

]2

(12/5)2

5−1+

(62/5)2

5−1


t4,.025 = 2.776, so the confidence interval is 95−70± 2.776

√12

5+

62

5, or (17.448, 32.552).

15. X = 52,sX = 5, nX = 8, Y = 46,sY = 2, nY = 12.


ν =

[52

8+

22

12

]2

(52/8)2

8−1+

(22/12)2

12−1


t8,.01 = 2.896, so the confidence interval is 52−46± 2.896

√52

8+

22

12, or (0.614, 11.386).


SECTION 5.7 253

Section 5.7

1. D = 6.736667,sD = 6.045556,n = 9, t9−1,.025= 2.306.


9), or (2.090, 11.384).

2. D = 0.19375,sD = 0.120230,n = 8, t8−1,.025= 2.365.


8), or (0.0932, 0.2943).

3. The differences are−0.04,−0.05,−0.02,−0.02,0.01,−0.02,−0.05,0.01,−0.05,0.03.

D = −0.02,sD = 0.028674,n = 10,t10−1,.01 = 2.821.

The confidence interval is−0.02±2.821(0.028674/√

10), or (−0.0456,0.00558).

4. The differences are: 0.76,0.30,0.74,0.30,0.56,0.82.

D = 0.58,sD = 0.233581,n = 6, t6−1,.025= 2.571.


6), or (0.3348, 0.8252).

5. The differences are: 83,51,49,71,69,52,76,31.

D = 60.25,sD = 17.293682,n = 8, t8−1,.05 = 1.895.


8), or (48.664,71.836).

6. The differences are: 0.30,0.38,0.66,0.41,0.38,0.58,0.20,0.16,0.41,0.50.

D = 0.398,sD = 0.155835,n = 10,t10−1,.01 = 2.821.


10), or (0.259, 0.537).

7. The differences are: 8.4,8.6,10.5,9.6,10.7,10.8,10.7,11.3,10.7.

D = 10.144444,sD = 1.033333,n = 9, t9−1,.025= 2.306.


9), or (9.350, 10.939).

8. (a) Yes. This design involves two independent samples, soa confidence interval for the difference between the twomeans can be constructed by using expression (5.21) in Section 5.6.


254 CHAPTER 5

(b) It is likely to be less precise, because there is likely tobe a lot of variation in brake pad life from one car toanother.

9. (a) The differences are: 3.8,2.6,2.0,2.9,2.2,−0.2,0.5,1.3,1.3,2.1,4.8,1.5,3.4,1.4,1.1,1.9,−0.9,−0.3.

D = 1.74444,sD = 1.46095,n = 18,t18−1,.005= 2.898.


18), or (0.747, 2.742).

(b) The level 100(1−α)% can be determined from the equationt17,α/2(1.46095/√

18) = 0.5.

From this equation,t17,α/2 = 1.452. Thet table indicates that the value ofα/2 is between 0.05 and 0.10, andcloser to 0.10. Therefore the level 100(1−α)% is closest to 80%.

10. (a) Expression (5.21) in Section 5.6 should be used, because the design involves two independent samples.

(b) X = 23.93,sX = 1.79,nX = 18, Y = 22.19,sY = 1.95,nY = 18.


ν =

[1.792

18+

1.952

18

]2

(1.792/18)2

18−1+

(1.952/18)2

18−1



√1.792

18+

1.952

18, or (0.0349, 3.445).

(c) The confidence interval based on the two-sample design iswider because there is considerable variation inmileage between the cars.

Section 5.8

1. (a)X∗ ∼ N(8.5, 0.22), Y∗ ∼ N(21.2, 0.32)


(c) Answers will vary.


SECTION 5.8 255

(d) Yes,P is approximately normally distributed.

(e) Answers will vary.

2. (a)X∗ ∼ N(1.18, 0.022), Y∗ ∼ N(0.85, 0.022)



(d) Yes,W is approximately normally distributed.

(e) Answers will vary.

3. (a) Yes,A is approximately normally distributed.



4. (a) Yes,T is approximately normally distributed.



5. (a)N(0.27, 0.402/349) andN(1.62, 1.702/143). Since the values 0.27 and 1.62 are sample means, their variancesare equal to the population variances divided by the sample sizes.

(b) No,R is not approximately normally distributed.


256 CHAPTER 5


(d) It is not appropriate, sinceR is not approximately normally distributed.

6. (a) Using Method 1, the limits of the 95% confidence interval are the 2.5 and 97.5 percentiles of the bootstrapmeans,X.025 andX.975.

The 2.5 percentile is(Y25+Y26)/2 = 38.3818, and the 97.5 percentile is(Y975+Y976)/2 = 38.53865.


(b) Using Method 2, the 95% confidence interval is(2X−X.975, 2X−X.025).

X = 38.44727,X.025 = (Y25 +Y26)/2 = 38.3818, andX.975 = (Y975+Y976)/2 = 38.53865. The confidenceinterval is (38.3559, 38.5127).

(c) Using Method 1, the limits of the 90% confidence interval are the 5th and 95th percentiles of the bootstrapmeans,X.05 andX.95.

The 5th percentile is(Y50+Y51)/2 = 38.39135, and the 95th percentile is(Y950+Y951)/2 = 38.5227.


(d) Using Method 2, the 90% confidence interval is(2X−X.95, 2X−X.05).

X = 38.44727,X.05 =(Y50+Y51)/2= 38.39135, andX.95 = (Y950+Y951)/2= 38.5227. The confidence intervalis (38.3718, 38.5032).

7. (a,b,c) Answers will vary.

8. (a) 95% exactly (simulation results will be approximate).

(b) ≈ 91.5%

(c) ≈ 82.5%

9. (a) Coverage probability for traditional interval is≈ 0.89, mean length is≈ 0.585.

Coverage probability for the Agresti-Coull interval is≈ 0.98, mean length is≈ 0.51.

(b) Coverage probability for traditional interval is≈ 0.95, mean length is≈ 0.46.




(c) Coverage probability for traditional interval is≈ 0.92, mean length is≈ 0.305.


(d) The traditional method has coverage probability close to 0.95 forn= 17, but less than 0.95 for bothn = 10 andn = 40.

(e) The Agresti-Coull interval has greater coverage probability for sample sizes 10 and 40, and nearly equal for17.

(f) The Agresti-Coull method does.

10. (a) General coverage is≈ 0.95, pooled coverage is exactly 0.95 (simulation results will be approximate).

(b) General coverage is≈ 0.95, pooled coverage is≈ 0.94.

(c) General coverage is≈ 0.95, pooled coverage is≈ 0.999.

(d) General coverage is≈ 0.95, pooled coverage is≈ 0.66.

(e) No.

(f) iii.


1. The differences are 21,18,5,13,−2,10.

D = 10.833,sD = 8.471521,n = 6, t6−1,.025= 2.571.


6), or (1.942, 19.725).

2. X = 27.833333,sX = 8.328665,nX = 6, Y = 17.500000,sY = 4.370355,nY = 6.


ν =

[8.3286652

6+

4.3703552

6

]2

(8.3286652/6)2

6−1+

(4.3703552/6)2

6−1



258 CHAPTER 5


√8.3286652

6+

4.3703552

6,

or (1.252, 19.415).

3. X = 1919,nX = 1985,pX = (1919+1)/(1985+2)= 0.966281,Y = 4561,nY = 4988,pY = (4561+1)/(4988+2)= 0.914228,z.005 = 2.58.


√0.966281(1−0.966281)

1985+2+

0.914228(1−0.914228)4988+2

,

or (0.0374, 0.0667).

4. (a)X = 17.3, s= 1.2, n = 81,z.01 = 2.33.


81), or (16.989, 17.611).

(b) X = 17.3, s= 1.2, n = 81, so the upper confidence bound 17.5 satisfies 17.5 = 17.3+zα/2(1.2/√

81).



The level is 1−α = 1−2(0.0668)= 0.8664, or 86.64%.

(c) z.01 = 2.33. 2.33(1.2/√

n) = 0.1, son = 782.

5. X = 6.1, sX = 0.7, nX = 125,Y = 5.8, sY = 1.0, nY = 75,z.05 = 1.645.


0.72/125+1.02/75, or (0.084, 0.516).

6. The standard deviation for the difference between the means is√

σ2X/nX + σ2

Y/nY. EstimateσX ≈ sX = 0.7andσY ≈ sY = 1.0.

If 50 pieces of yarn are sampled from the old batch, thennX = 175 andnY = 75.The standard deviation for the difference between the meansis then

√0.72/175+1.02/75= 0.127.

If 50 pieces of yarn are sampled from the new batch, thennX = 125 andnY = 125.The standard deviation for the difference between the meansis then

√0.72/125+1.02/125= 0.109.

If 25 pieces of yarn are sampled from each batch, thennX = 150 andnY = 100.The standard deviation for the difference between the meansis then

√0.72/150+1.02/100= 0.115.

Therefore sampling 50 additional pieces from the new batch increases the precision the most.



7. (a)X = 13,n = 87, p = (13+2)/(87+4)= 0.16484,z.025 = 1.96.


0.16484(1−0.16484)/(87+4), or (0.0886, 0.241).



p(1− p)/(n+4).


8. X = 72.5, s= 5.8, n = 10,t10−1,.005= 3.250.


10), or (66.539, 78.461).

9. The higher the level, the wider the confidence interval. Therefore the narrowest interval, (4.20, 5.83), is the90% confidence interval, the widest interval, (3.57, 6.46),is the 99% confidence interval, and (4.01, 6.02) isthe 95% confidence interval.

10. Letn be the required sample size.


p(1− p)/(n+4).

Since there is no preliminary estimate of ˜p available, replace ˜p 0.5.


11. X = 7.909091,sX = 0.359039,nX = 11, Y = 8.00000,sY = 0.154919,nY = 6.


ν =

[0.3590392

11+

0.1549192

6

]2

(0.3590392/11)2

11−1+

(0.1549192/6)2

6−1



√0.3590392

11+

0.1549192

6,

or (−0.420,0.238).

12. (a)X = 1.69,s= 0.25,n = 80,z.025 = 1.96.


80), or (1.635, 1.745).


260 CHAPTER 5


Thenn satisfies the equation 0.02= 1.96σ/√

n.

Replaceσ with with s= 0.25. Solving forn yieldsn = 601.

13. Letn be the required sample size.

The 90% confidence interval based on 144 observations has width±0.35.

Therefore 0.35= 1.645σ/√

144, so 1.645σ = 4.2.

Now n satisfies the equation 0.2 = 1.645σ/√

n = 4.2/√

n. Solving forn yieldsn = 441.

14. (ii)±0.03. The original confidence has width±1.96σ/√

100. The new confidence interval has width±1.96σ/√

400=1.96σ/(2

√100). So the width of the new confidence interval is half that of theold.

15. (a) False. This a specific confidence interval that has already been computed. The notion of probability does notapply.

(b) False. The confidence interval specifies the location of the population mean. It does not specify the location ofa sample mean.

(c) True. This says that the method used to compute a 95% confidence interval succeeds in covering the true mean95% of the time.

(d) False. The confidence interval specifies the location of the population mean. It does not specify the location ofa future measurement.

16. (a) True. A 95% confidence interval is found by adding and subtracting 1.96(650/√

400) from the mean of 370.

(b) False. The confidence interval specifies the location of the population mean. It does not specify the proportionof the sample items that fall into any given interval.

(c) True. This describes the method to compute a 68% confidence interval, and this method produces an intervalthat covers the true value 68% of the times that it is used.

(d) False. So long as the sample mean is approximately normal, which will be the case when the sample is large,the method can be used.



(e) False. The confidence interval specifies the location of the population mean. It does not specify the proportionof the population that falls into any given interval.

17. (a)X = 37, and the uncertainty isσX = s/√

n = 0.1. A 95% confidence interval is 37±1.96(0.1),or (36.804, 37.196).

(b) Sinces/√

n = 0.1, this confidence interval is of the formX±1(s/√

n). The area to the left ofz= 1 is approxi-mately 0.1587. Thereforeα/2 = 0.1587, so the level is 1−α = 0.6826, or approximately 68%.

(c) The measurements come from a normal population.

(d) t9,.025 = 2.262. A 95% confidence interval is therefore 37±2.262(s/√

n) = 37±2.262(0.1),or (36.774, 37.226).

18. (a)X = 1500,s= 5, n = 10,t10−1,.025= 2.262.


10), or (1496.4, 1503.6).

(b) µnail =µbox100

(c) From part (a), we are 95% confident that the mean weight of abox of 100 nails is in the interval(1496.4, 1503.6). Therefore we are 95% confident that the mean weight of a single nail is in the interval(14.964, 15.036).

19. (a) SinceX is normally distributed with meannλ, it follows that for a proportion 1−α of all possible samples,−zα/2σX < X−nλ < zα/2σX.

Multiplying by −1 and addingX across the inequality yieldsX − zα/2σX < nλ < X + zα/2σX, which is thedesired result.

(b) Sincen is a constant,σX/n = σX/n =√

nλ/n =√

λ/n.

Thereforeσλ = σX/n.

(c) Divide the inequality in part (a) byn.


262 CHAPTER 5

(d) Substitute√

λ/n for σλ in part (c) to show that for a proportion 1−α of all possible samples,

λ−zα/2

√λ/n < λ < λ+zα/2

√λ/n.

The intervalλ±zα/2

√λ/n is therefore a level 1−α confidence interval forλ.

(e) n= 5, λ = 300/5= 60,σλ =√

60/5= 3.4641. A 95% confidence interval forλ is therefore 60±1.96(3.4641),or (53.210, 66.790).

20. (a)n = 1, λ = 64/1 = 64, σλ =√

64/1 = 8. A 95% confidence interval forλ is therefore 64± 1.96(8),or (48.32, 79.68).

(b) n = 4, λ = 256/4 = 64, σλ =√

64/4 = 4. A 95% confidence interval forλ is therefore 64± 1.96(4),or (56.16, 71.84).

(c) Letn be the required number of minutes. Thenn satisfies the equation 1= 1.96√

λ/n. Substitutingλ = 64 forλ and solving forn yieldsn = 245.86 minutes.

21. (a)µ= 1.6, σµ = 0.05,h = 15,σh = 1.0, τ = 25,στ = 1.0. V = τh/µ= 234.375

∂V∂µ

= −τh/µ2 = −146.484375,∂V∂h

= τ/µ= 15.625,∂V∂τ

= h/µ= 9.375

σV =

√(∂V∂µ

)2

σ2µ +

(∂V∂h

)2

σ2h +

(∂V∂τ

)2

σ2τ = 19.63862

(b) If the estimate ofV is normally distributed, then a 95% confidence interval forV is 234.375±1.96(19.63862)or (195.883, 272.867).

(c) The confidence interval is valid. The estimate ofV is approximately normal.






(b) Using Method 2, the 95% confidence interval is(2X−X.975, 2X−X.025).

X = 1385.1,X.025= (Y250+Y251)/2= 1305.5, andX.975= (Y9750+Y9751)/2= 1462.1. The confidence intervalis (1308.1, 1464.7).

(c) Using Method 1, the limits of the 99% confidence interval are the 0.5 and 99.5 percentiles of the bootstrapmeans,X.005 andX.995.



(d) Using Method 2, the 99% confidence interval is(2X−X.995, 2X−X.005).

X = 1385.1,X.005 = (Y50+Y51)/2 = 1283.4, andX.995 = (Y9950+Y9951)/2 = 1483.8. The confidence intervalis (1286.4, 1486.8).

23. (a) Answers may vary somewhat from the 0.5 and 99.5 percentiles in Exercise 22 parts (c) and (d).

(b) Answers may vary somewhat from the answer to Exercise 22 part (c).

(c) Answers may vary somewhat from the answer to Exercise 22 part (d).


264 CHAPTER 6

Chapter 6

Section 6.1

1. (a)X = 783,s= 120,n = 73. The null and alternate hypotheses areH0 : µ≤ 750 versusH1 : µ> 750.

z= (783−750)/(120/√

73) = 2.35. Since the alternate hypothesis is of the formµ > µ0, theP-value is thearea to the right ofz= 2.35.

ThusP = 0.0094.

(b) TheP-value is 0.0094, so ifH0 is true then the sample is in the most extreme 0.94% of its distribution.

2. (a)X = 1.102,s= 0.02,n = 65. The null and alternate hypotheses areH0 : µ≤ 1.1 versusH1 : µ> 1.1.

z= (1.102−1.1)/(0.02/√

65) = 0.81. Since the alternate hypothesis is of the formµ> µ0, theP-value is thearea to the right ofz= 0.81.

ThusP = 0.2090.


3. (a)X = 1.90,s= 21.20,n = 160. The null and alternate hypotheses areH0 : µ= 0 versusH1 : µ 6= 0.

z= (1.90−0)/(21.20/√

160) = 1.13. Since the alternate hypothesis is of the formµ 6= µ0, theP-value is thesum of the areas to the right ofz= 1.13 and to the left ofz= −1.13.

ThusP = 0.1292+0.1292= 0.2584.


4. (a)X = 11.98,s= 0.19,n = 100. The null and alternate hypotheses areH0 : µ= 12 versusH1 : µ 6= 12.

z= (11.98−12)/(0.19/√

100) = −1.05. Since the alternate hypothesis is of the formµ 6= µ0, theP-value isthe sum of the areas to the right ofz= 1.05 and to the left ofz= −1.05.

ThusP = 0.1469+0.1469= 0.2938.

(b) If the mean fill volume were 12 ounces, the probability of observing a sample mean as far from 12 as the valueof 11.98 that was actually observed would be 0.2938. Since 0.2938 is not a small probability, it is plausiblethat the mean fill volume is 12 ounces.


SECTION 6.1 265

5. (a)X = 51.2, s= 4.0, n = 110. The null and alternate hypotheses areH0 : µ≤ 50 versusH1 : µ> 50.

z= (51.2−50)/(4.0/√

110) = 3.15. Since the alternate hypothesis is of the formµ > µ0, theP-value is areato the right ofz= 3.15.

ThusP = 0.0008.

(b) If the mean tensile strength were 50 psi, the probabilityof observing a sample mean as large as the value of 51.2that was actually observed would be only 0.0008. Therefore we are convinced that the mean tensile strength isnot 50 psi or less, but is instead greater than 50 psi.

6. (a)X = 98,s= 20,n = 85. The null and alternate hypotheses areH0 : µ≥ 100 versusH1 : µ< 100.

z= (98−100)/(20/√

85) = −0.92. Since the alternate hypothesis is of the formµ< µ0, theP-value is area tothe left ofz= −0.92.

ThusP = 0.1788.

(b) If the mean chlorine concentration were 100µg/L, the probability of observing a sample mean as small as thevalue of 98 that was actually observed would be 0.1788. Since0.1788 is not a small probability, it is plausiblethat the mean chlorine concentration is 100µg/L or more.

7. (a)X = 715,s= 24,n = 60. The null and alternate hypotheses areH0 : µ≥ 740 versusH1 : µ< 740.

z= (715−740)/(24/√

60) = −8.07. Since the alternate hypothesis is of the formµ < µ0, theP-value is thearea to the left ofz= −8.07.

ThusP≈ 0.

(b) If the mean daily output were 740 tons or more, the probability of observing a sample mean as small as thevalue of 715 that was actually observed would be nearly 0. Therefore we are convinced that the mean dailyoutput is not 740 tons or more, but is instead less than 740 tons.

8. (a)X = 84.8, s= 0.5, n = 50. The null and alternate hypotheses areH0 : µ= 85 versusH1 : µ 6= 85.

z= (84.8−85)/(0.5/√

50) = −2.83. Since the alternate hypothesis is of the formµ 6= µ0, theP-value is thesum of the areas to the right ofz= 2.83 and to the left ofz= −2.83.

ThusP = 0.0023+0.0023= 0.0046.

(b) If the mean resistance were equal to 85 mΩ, the probability of observing a sample mean as far from 85 as thevalue of 84.8 that was actually observed would be only 0.0046. Therefore we are convinced that the mean dailyoutput is not equal to 85 mΩ.

9. (a)X = 39.6, s= 7, n = 58. The null and alternate hypotheses areH0 : µ≥ 40 versusH1 : µ< 40.


266 CHAPTER 6

z= (39.6−40)/(7/√

58) = −0.44. Since the alternate hypothesis is of the formµ< µ0, theP-value is the areato the left ofz= −0.44.

ThusP = 0.3300.

(b) If the mean air velocity were 40 cm/s, the probability of observing a sample mean as small as the value of 39.6that was actually observed would be 0.33. Since 0.33 is not a small probability, it is plausible that the mean airvelocity is 40 cm/s or more.

10. (a)X = 1356,s= 70,n = 100. The null and alternate hypotheses areH0 : µ≤ 1350 versusH1 : µ> 1350.

z= (1356−1350)/(70/√

100) = 0.86. Since the alternate hypothesis is of the formµ > µ0, theP-value is thearea to the right ofz= 0.86.

ThusP = 0.1949.

(b) If the mean compressive strength were 1350 kPa, the probability of observing a sample mean greater than orequal to the observed value of 1356 would be 0.195. Since 0.195 is not a small probability, it is plausible thatthe mean is 1350 kPa or less, and that the blocks do not meet thespecification.

11. (ii) 5. The null distribution specifies that the population mean, which is also the mean ofX, is the value on theboundary between the null and alternate hypotheses.

12. (iii) 4. TheP-value is the probability, under the assumption thatH0 is true, of observing a result as extremeas or more extreme than that actually observed. SinceP = 0.04, we expect to obtain a value ofX less than orequal to 8 approximately 4 times in 100.

13. X = 5.2 andσX = σ/√

n = 0.1. The null and alternate hypotheses areH0 : µ= 5.0 versusH1 : µ 6= 5.0.

z= (5.2−5.0)/0.1= 2.00. Since the alternate hypothesis is of the formµ 6= µ0, theP-value is the sum of theareas to the right ofz= 2.00 and to the left ofz= −2.00.

ThusP = 0.0228+0.0228= 0.0456.

14. (a) Two-tailed. The alternate hypothesis is of the formµ 6= µ0.

(b) H0 : µ= 73.5


SECTION 6.2 267

(c) P = 0.196

(d) X = 73.2461 ands/√

n = 0.1963. The null and alternate hypotheses areH0 : µ≥ 73.6 versusH1 : µ< 73.6.

z= (73.2461−73.6)/0.1963= −1.80. Since the alternate hypothesis is of the formµ< µ0, theP-value is thearea to the left ofz= −1.80.

ThusP = 0.0359.

(e) X = 73.2461, s/√

n = 0.1963, z.005 = 2.58. The confidence interval is 73.2461± 2.58(0.1963),or (72.74, 73.75).

15. (a) SE Mean =s/√

n = 2.00819/√

87= 0.2153.

(b) X = 4.07114. From part (a),s/√

n = 0.2153. The null and alternate hypotheses areH0 : µ ≤ 3.5 versusH1 : µ> 3.5.

z= (4.07114−3.5)/0.2153= 2.65.

(c) Since the alternate hypothesis is of the formµ> µ0, theP-value is the area to the right ofz= 2.65.

ThusP = 0.0040.

Section 6.2

1. P = 0.5. The larger theP-value, the more plausible the null hypothesis.

2. (a) True.

(b) False. We conclude thatH0 is plausible.

(c) True.

(d) False. We conclude thatH0 is plausible, and therefore thatH1 is plausible as well.


268 CHAPTER 6

3. (iv). A P-value of 0.01 means that ifH0 is true, then the observed value of the test statistic was in the mostextreme 1% of its distribution. This is unlikely, but not impossible.

4. (v). A P-value of 0.50 means that ifH0 is true, then the observed value of the test statistic was in the mostextreme 50% of its distribution. This is not at all unusual, so H0 is plausible. We can never conclude thatH1 isfalse, so therefore we conclude thatH1 is plausible as well.

5. (a) True. The result is statistically significant at any level greater than or equal to 2%.

(b) False.P > 0.01, so the result is not statistically significant at the 1% level.

(c) True. The null hypothesis is rejected at any level greater than or equal to 2%.

(d) False.P > 0.01, so the null hypothesis is not rejected at the 1% level.

6. (a) False.P≤ 0.05.

(b) True.

(c) True. The statement that the null hypothesis is rejectedat the 5% level means the same thing as the statementthat the result is statistically significant at the 5% level.

(d) True. The result is statistically significant at any level greater than or equal to 5%.

7. (a) No. TheP-value is 0.196, which is greater than 0.05.

(b) The sample mean is 73.2461. The sample mean is therefore closer to 73 than to 73.5. TheP-value for the nullhypothesisµ = 73 will therefore be larger than theP-value for the null hypothesisµ = 73.5, which is 0.196.Therefore the null hypothesisµ= 73 cannot be rejected at the 5% level.

8. H0 : µ≥ 5 is the best. IfH0 is rejected, we can be convinced thatµ< 5, and that the workplace is safe. IfH0 isnot rejected, then the workplace might not be safe.

9. (a)H0 : µ≤ 8. If H0 is rejected, we can conclude thatµ> 8, and that the new battery should be used.

(b) H0 : µ≤ 60,000. If H0 is rejected, we can conclude thatµ> 60,000, and that the new material should be used.


SECTION 6.2 269

(c) H0 : µ= 10. If H0 is rejected, we can conclude thatµ 6= 10, and that the flow meter should be recalibrated.

10. (a) There is uncertainty in the mean measurement. It is possible to get a mean measurement less than 4 even whenthe true value is 4 or more.

(b) H0 : µ≥ 4 versusH1 : µ< 4. If H0 is not rejected, then it is plausible that the mean radon concentration may be4 pCi/L or more.

(c) X = 3.72,s= 1.93,n = 75. The null and alternate hypotheses areH0 : µ≥ 4 versusH1 : µ< 4.

z= (3.72−4)/(1.93/√

75) = −1.26. Since the alternate hypothesis is of the formµ < µ0, theP-value is thearea to the left ofz= −1.26. ThusP = 0.1038.

The null hypothesis has some plausibility. Radon abatementis recommended.

11. (a) (ii) The scale is out of calibration. IfH0 is rejected, we conclude thatH0 is false, soµ 6= 10.

(b) (iii) The scale might be in calibration. IfH0 is not rejected, we conclude thatH0 is plausible, soµ might beequal to 10.

(c) No. The scale is in calibration only ifµ = 10. The strongest evidence in favor of this hypothesis wouldoccurif X = 10. But since there is uncertainty inX, we cannot be sure even then thatµ= 10.

12. (a) No.P = 0.30 is not small. Both the null and alternate hypotheses are therefore plausible.

(b) No, we cannot conclude that the null hypothesis is true. The alternate hypothesis is also plausible.

13. No, she cannot conclude that the null hypothesis is true,only that it is plausible.

14. (i) greater than 0.05. Since the value 1.4 is inside the 95% confidence interval, theP-value for testingH0 : µ= 1.4 versusH1 : µ 6= 1.4 will be greater than 0.05.

15. (i) H0 : µ= 1.2. For either of the other two hypotheses, theP-value would be 0.025.

16. (iii) 0.01 0.01.


270 CHAPTER 6

17. (a) Yes. The value 3.5 is greater than the upper confidencebound of 3.45. Quantities greater than the upperconfidence bound will haveP-values less than 0.05. ThereforeP < 0.05.

(b) No, we would need to know the 99% upper confidence bound to determine whetherP < 0.01.

18. (a) No, we would need to know the 99% lower confidence boundin order to determine whetherP < 0.01.

(b) Yes, since 50 is less than the 98% lower confidence bound, we know thatP < 0.02. ThereforeP < 0.05.

19. Yes, we can compute theP-value exactly. Since the 95% upper confidence bound is 3.45,we know that3.40+ 1.645s/

√n = 3.45. Therefores/

√n = 0.0304. Thez-score is(3.40− 3.50)/0.0304= −3.29. The

P-value is 0.0005, which is less than 0.01.

20. No, the confidence bound itself tells us only thatP< 0.02. To determine theP-value more precisely, we wouldneed to know the standard deviation of the sample mean. This is approximately equal tos/

√n wheres is the

sample standard deviation andn is the sample size. We know thats= 5, but we do not known.

Section 6.3

1. X = 25,n = 300, p = 25/300= 0.083333.

The null and alternate hypotheses areH0 : p≤ 0.05 versusH1 : p > 0.05.

z= (0.083333−0.5)/√

0.05(1−0.05)/300= 2.65.

Since the alternate hypothesis is of the formp > p0, theP-value is the area to the right ofz= 2.65,

soP = 0.0040. The claim is rejected.

2. No.n = 100 andp0 = 0.05, sonp0 = 5, which is less than 10.

3. X = 25,n = 400, p = 25/400= 0.0625.


z= (0.0625−0.05)/√

0.05(1−0.05)/400= 1.15.


soP= 0.1251. The company cannot conclude that more than 5% of their subscribers would pay for the channel.


SECTION 6.3 271

4. n = 304, p = 0.62.


z= (0.62−0.5)/√

0.5(1−0.5)/304= 4.18.


soP≈ 0. It can be concluded that more than half of patients prefer aphysician with high technical skills.

5. X = 330,n = 600, p = 330/600= 0.55.


z= (0.55−0.50)/√

0.50(1−0.50)/600= 2.45.


soP = 0.0071. It can be concluded that more than 50% of all likely voters favor the proposal.

6. X = 18,n = 50, p = 18/50= 0.36.

The null and alternate hypotheses areH0 : p≥ 0.50 versusH1 : p < 0.50.

z= (0.36−0.50)/√

0.50(1−0.50)/50= −1.98.

Since the alternate hypothesis is of the formp < p0, theP-value is the area to the left ofz= −1.98,

soP = 0.0239. It can be concluded that fewer than half of the incinerators meet the standard.

7. X = 470,n = 500, p = 470/500= 0.94.


z= (0.94−0.95)/√

0.95(1−0.95)/500= −1.03.


soP = 0.1515. The claim cannot be rejected.

8. X = 12,n = 300, p = 12/300= 0.04.


z= (0.04−0.08)/√

0.08(1−0.08)/300= −2.55.


soP = 0.0054. The machine can be qualified.

9. X = 15,n = 400, p = 15/400= 0.0375.


z= (0.0375−0.05)/√

0.05(1−0.05)/400= −1.15.


272 CHAPTER 6


soP = 0.1251. It cannot be concluded that the new process has a lower defect rate.

10. X = 37,n = 50, p = 37/50= 0.74.


z= (0.74−0.60)/√

0.60(1−0.60)/50= 2.02.


soP = 0.0217. It can be concluded that more than 60% of the measurements are satisfactory.

11. X = 17,n = 75, p = 17/75= 0.22667.


z= (0.22667−0.40)/√

0.40(1−0.40)/75= −3.06.


soP = 0.0011. It can be concluded that less than 40% of the fuses have burnout amperages greater than 15 A.

12. (a) One-tailed. The alternate hypothesis is of the formp < p0.

(b) H0 : p≥ 0.4

(c) Yes,P = 0.001, which is less than 0.02.

(d) The sample proportion is 0.304167. Therefore theP-value for the null hypothesisp ≥ 0.45 will be less thantheP-value for the null hypothesisp≥ 0.40, which is 0.001. It follows that the null hypothesisp≥ 0.45 canbe rejected at the 2% level.

(e) X = 73,n = 240, p = 73/240= 0.304167.


z= (0.304167−0.25)/√

0.25(1−0.25)/240= 1.94.

Since the alternate hypothesis is of the formp > p0, theP-value is the area to the right ofz= 1.94.

ThusP = 0.0262.

(f) X = 73,n = 240, p = (73+2)/(240+4)= 0.307377,z.05 = 1.645.


0.307377(1−0.307377)/(240+4), or (0.2588, 0.3560).


SECTION 6.4 273

13. (a) Sample p =p = 345/500= 0.690.

(b) The null and alternate hypotheses areH0 : p ≥ 0.7 versusH1 : µ < 0.7. n = 500. From part (a),p = 0.690.z= (0.690−0.700)/

√0.7(1−0.7)/500= −0.49.

(c) Since the alternate hypothesis is of the formp < p0, theP-value is the area to the left ofz= −0.49.

ThusP = 0.3121.

Section 6.4

1. (a)X = 100.01,s= 0.0264575,n = 3. There are 3−1= 2 degrees of freedom.

The null and alternate hypotheses areH0 : µ= 100 versusH1 : µ 6= 100.

t = (100.01−100)/(0.0264575/√

3) = 0.6547.

Since the alternate hypothesis is of the formµ 6= µ0, theP-value is the sum of the areas to the right oft = 0.6547and to the left oft = −0.6547.

From thet table, 0.50< P < 0.80. A computer package givesP = 0.580.

We conclude that the scale may well be calibrated correctly.

(b) Thet-test cannot be performed, because the sample standard deviation cannot be computed from a sample ofsize 1.

2. (a) Reject the assumptions: the value 221.03 is an outlier.

(b) The assumptions appear to be met.

(c) Reject the assumptions: the values are in increasing order, so this does not appear to be a simple randomsample.

3. (a)H0 : µ≤ 35 versusH1 : µ> 35

(b) X = 39,s= 4, n = 6. There are 6−1= 5 degrees of freedom.

From part (a), the null and alternate hypotheses areH0 : µ≤ 35 versusH1 : µ> 35.

t = (39−35)/(4/√

6) = 2.449.

Since the alternate hypothesis is of the formµ> µ0, theP-value is the area to the right oft = 2.449.



274 CHAPTER 6

(c) Yes, theP-value is small, so we conclude thatµ> 35.

4. (a)X = 20.75,s= 3.93,n = 26. There are 26−1= 25 degrees of freedom.

The null and alternate hypotheses areH0 : µ≤ 20 versusH1 : µ> 20.

t = (20.75−20)/(3.93/√

25) = 0.973.



We cannot conclude that the mean concentration is greater than 20 mg/kg.

(b) X = 20.75,s= 3.93,n = 26. There are 26−1= 25 degrees of freedom.

The null and alternate hypotheses areH0 : µ≥ 25 versusH1 : µ< 25.

t = (20.75−25)/(3.93/√

25) = −5.514.

Since the alternate hypothesis is of the formµ< µ0, theP-value is the area to the left oft = −5.514.

From thet table,P < 0.0005. A computer package givesP = 4.96×10−6.

We conclude that the mean concentration is less than 25 mg/kg.

5. (a)X = 61.3, s= 5.2, n = 12. There are 12−1= 11 degrees of freedom.


t = (61.3−60)/(5.2/√

12) = 0.866.



We cannot conclude that the mean concentration is greater than 60 mg/L.

(b) X = 61.3, s= 5.2, n = 12. There are 12−1= 11 degrees of freedom.


t = (61.3−65)/(5.2/√

12) = −2.465.



We conclude that the mean concentration is less than 65 mg/L.

6. X = 90.3, s= 1.290349,n = 5. There are 5−1= 4 degrees of freedom.


t = (90.3−90)/(1.290349/√

5) = 0.5199.




SECTION 6.4 275

We cannot conclude that the mean octane rating is greater than 90%.

7. (a)3.8 4 4.2

(b) Yes, the sample contains no outliers.

X = 4.032857,s= 0.061244,n = 7. There are 7−1= 6 degrees of freedom.


t = (4.032857−4)/(0.061244/√

7) = 1.419.

Since the alternate hypothesis is of the formµ 6= µ0, theP-value is the sum of the areas to the right oft = 1.419and to the left oft = −1.419.


It cannot be concluded that the mean thickness differs from 4mils.

(c)3.9 4 4.1 4.2 4.3

(d) No, the sample contains an outlier.

8. (a)H0 : µ≤ 85 versusH1 : µ> 85



t = (90.55−85)/(2.901551/√

6) = 4.685.



(c) Yes, theP-value is small, so we conclude thatµ> 85.

9. X = 1.88,s= 0.21,n = 18. There are 18−1= 17 degrees of freedom.


t = (1.88−2)/(0.21/√

18) = −2.424.



We can conclude that the mean warpage is less than 2 mm.


276 CHAPTER 6



t = (7.22−8)/(2.328519/√

5) = −0.7490.


From thet table,P≈ 0.25. A computer package givesP = 0.248.

We cannot conclude that the mean amount is less than 8%.


The null and alternate hypotheses areH0 : µ≥ 2.5 versusH1 : µ< 2.5.

t = (1.25−2.5)/(0.624500/√

4) = −4.003.



We can conclude that the mean amount absorbed is less than 2.5%.

12. (a) One-tailed. The alternate hypothesis is of the formµ> µ0.

(b) H0 : µ≤ 5.5

(c) Yes, theP-value is 0.002, which is less than 0.01.

(d) X = 5.92563,n = 5, s/√

n = 0.07046. There are 5−1= 4 degrees of freedom.


t = (5.92563−6.5)/0.07046= −8.152.



(e) X = 5.92563,n = 5, s/√

n = 0.07046. There are 5−1= 4 degrees of freedom.

t4,.005 = 4.604. The confidence interval is 5.92563±4.604(0.07046), or (5.6012, 6.2500).

13. (a) StDev = (SE Mean)√

N = 1.8389√

11= 6.0989.

(b) t10,.025 = 2.228. The lower 95% confidence bound is 13.2874−2.228(1.8389)= 9.190.

(c) t10,.025 = 2.228. The upper 95% confidence bound is 13.2874+2.228(1.8389)= 17.384.

(d) t = (13.2874−16)/1.8389= −1.475.


SECTION 6.5 277

Section 6.5

1. X = 8.5, sX = 1.9, nX = 58,Y = 11.9, sY = 3.6, nY = 58.

The null and alternate hypotheses areH0 : µX −µY ≥ 0 versusH1 : µX −µY < 0.

z= (8.5−11.9−0)/√

1.92/58+3.62/58= −6.36. Since the alternate hypothesis is of the formµX −µY < ∆,theP-value is the area to the left ofz= −6.36.

ThusP≈ 0.

We can conclude that the mean hospital stay is shorter for patients receiving C4A-rich plasma.

2. X = 20.95,sX = 14.5, nX = 482,Y = 22.79,sY = 15.6, nY = 614.


z = (20.95− 22.79− 0)/√

14.52/482+15.62/614= −2.02. Since the alternate hypothesis is of the formµX −µY < ∆, theP-value is the area to the left ofz= −2.02.

ThusP = 0.0217.

We can conclude that the mean weight of male spotted flounder is greater than that of females.

3. X = 5.92,sX = 0.15,nX = 42,Y = 6.05,sY = 0.16,nY = 37.

The null and alternate hypotheses areH0 : µX −µY = 0 versusH1 : µX −µY 6= 0.

z= (5.92−6.05−0)/√

0.152/42+0.162/37=−3.71. Since the alternate hypothesis is of the formµX −µY 6= ∆,theP-value is the sum of the areas to the right ofz= 3.71 and to the left ofz= −3.71.

ThusP = 0.0001+0.0001= 0.0002.

We can conclude that the mean dielectric constant differs between the two types of asphalt.

4. X = 27.2, sX = 1.2, nX = 40,Y = 28.1, sY = 2.0, nY = 40.


z= (27.2−28.1−0)/√

1.22/40+2.02/40=−2.44. Since the alternate hypothesis is of the formµX −µY < ∆,theP-value is the area to the left ofz= −2.44.

ThusP = 0.0073.

We can conclude that the mean mileage is greater with premiumfuel.

5. X = 40,sX = 12,nX = 75,Y = 42,sY = 15,nY = 100.

The null and alternate hypotheses areH0 : µX −µY > 0 versusH1 : µX −µY ≤ 0.

z= (40−42−0)/√

122/75+152/100= −0.98. Since the alternate hypothesis is of the formµX −µY ≤ ∆,theP-value is the area to the left ofz= −0.98.

ThusP = 0.1635.


278 CHAPTER 6

We cannot conclude that the mean reduction from drug B is greater than the mean reduction from drug A.

6. (a) The 60 counts may not be independent if they are taken inconsecutive time periods.

(b) X = 73.8, sX = 5.2, nX = 60,Y = 76.1, sY = 4.1, nY = 60.


z= (73.8−76.1−0)/√


ThusP = 0.0036.

We can conclude that machine 2 is faster.

7. (a)X = 7.79,sX = 1.06,nX = 80,Y = 7.64,sY = 1.31,nY = 80.

Hereµ1 = µX andµ2 = µY. The null and alternate hypotheses areH0 : µX −µY ≤ 0 versusH1 : µX −µY > 0.

z= (7.79−7.64−0)/√

1.062/80+1.312/80= 0.80. Since the alternate hypothesis is of the formµX−µY > ∆,theP-value is the area to the right ofz= 0.80.

ThusP = 0.2119.

We cannot conclude that the mean score on one-tailed questions is greater.

(b) The null and alternate hypotheses areH0 : µX −µY = 0 versusH1 : µX −µY 6= 0.

Thez-score is computed as in part (a):z= 0.80.

Since the alternate hypothesis is of the formµX −µY 6= ∆, theP-value is the sum of the areas to the right ofz= 0.80 and to the left ofz= 0.80.

ThusP = 0.2119+0.2119= 0.4238.

We cannot conclude that the mean score on one-tailed questions differs from the mean score on two-tailedquestions.

8. (a)X = 495.6, sX = 19.4, nX = 50,Y = 481.2, sY = 14.3, nY = 50.

The null and alternate hypotheses areH0 : µX −µY ≤ 0 versusH1 : µX −µY > 0.

z= (495.6−481.2−0)/√

19.42/50+14.32/50= 4.22. Since the alternate hypothesis is of the formµX −µY > ∆,theP-value is the area to the right ofz= 4.22.

ThusP≈ 0.

We can conclude that the mean speed for the new chips is greater.

(b) X = 495.6, sX = 19.4, nX = 50,Y = 391.2, sY = 17.2, nY = 60.


z = (495.6− 391.2− 100)/√

19.42/50+17.22/60 = 1.25. Since the alternate hypothesis is of the formµX −µY > ∆, theP-value is the area to the right ofz= 1.25.


SECTION 6.5 279

ThusP = 0.1056.

The data do not provide convincing evidence for this claim.

9. (a)X = 625,sX = 40,nX = 100,Y = 640,sY = 50,nY = 64.


z= (625−640−0)/√

402/100+502/64= −2.02. Since the alternate hypothesis is of the formµX −µY < ∆,theP-value is the area to the right ofz= −2.02.

ThusP = 0.0217.

We can conclude that the second method yields the greater mean daily production.

(b) The null and alternate hypotheses areH0 : µX −µY ≥−10 versusH1 : µX −µY < −10.

z= (625−640+10)/√

402/100+502/64=−0.67. Since the alternate hypothesis is of the formµX−µY < ∆,theP-value is the area to the right ofz= −0.67.

ThusP = 0.2514.

We cannot conclude that the mean daily production for the second method exceeds that of the first by morethan 10 tons.

10. X = 91.1, sX = 6.23,nX = 50,Y = 90.7, sY = 4.34,nY = 40.


z= (91.1−90.7−0)/√

6.232/50+4.342/40= 0.36. Since the alternate hypothesis is of the formµX−µY > ∆,theP-value is the area to the right ofz= 0.36.

ThusP = 0.3594. We cannot conclude that the mean hardness of welds cooled at 10C is greater than that ofwelds cooled at 30C.

11. X1 = 4387,s1 = 252,n1 = 75,X2 = 4260,s2 = 231,n2 = 75.

The null and alternate hypotheses areH0 : µ1−µ2 ≤ 0 versusH1 : µ1−µ2 > 0.

z= (4387−4260−0)/√

2522/75+2312/75= 3.22. Since the alternate hypothesis is of the formµX−µY > ∆,theP-value is the area to the right ofz= 3.22.

ThusP = 0.0006. We can conclude that new power supplies outlast old power supplies.

12. (a) One-tailed. The alternate hypothesis is of the formµX −µY < ∆.

(b) H0 : µX −µY ≥ 0

(c) Yes, theP-value is equal to 0.047, which is less than 0.050.


280 CHAPTER 6

(d) z= (3.94−4.43−0)/√


ThusP = 0.0446, and the result is similar to that of thet test.

Note that if the denominator√

0.232+0.182 is used forz, thenz= −1.68 andP = 0.0465. The result is stillsimilar to that of thet test.

(e) X = 3.94,sX = 2.65,nX = 135,Y = 4.43,sY = 2.38,nY = 180,z.005 = 2.58.


2.652/135+2.382/180, or(−1.235,0.255).

Alternatively, the confidence interval can be computed as 3.94−4.43±2.58√

0.232+0.182, or(−1.244,0.264).

13. (a) (i) StDev = (SE Mean)√

N = 1.26√

78= 11.128.

(ii) SE Mean = StDev/√

N = 3.02/√

63= 0.380484.

(b) z= (23.3−20.63−0)/√

1.262+0.3804842 = 2.03. Since the alternate hypothesis is of the formµX −µY 6= ∆,theP-value is the sum of the areas to the right ofz= 2.03 and to the left ofz= −2.03.

ThusP = 0.0212+0.0212= 0.0424, and the result is similar to that of thet test.

(c) X = 23.3, sX/√

nX = 1.26,Y = 20.63,sY/√

nY = 0.380484,z.01 = 2.33.


1.262+0.3804842, or (−0.3967,5.7367).

Section 6.6

1. (a)H0 : pX − pY ≥ 0 versusH1 : pX − pY < 0

(b) X = 960, nX = 1000, pX = 960/1000= 0.960, Y = 582, nY = 600, pY = 582/600= 0.970,

p = (960+582)/(1000+600)= 0.96375.

The null and alternate hypotheses areH0 : pX − pY ≥ 0 versusH1 : pX − pY < 0.

z=0.960−0.970√

0.96375(1−0.96375)(1/1000+1/600)= −1.04.

Since the alternate hypothesis is of the formpX − pY < 0, theP-value is the area to the left ofz= −1.04.

ThusP = 0.1492.

(c) SinceP = 0.1492, we cannot conclude that machine 2 is better. Thereforemachine 1 should be used.

2. (a)H0 : pX − pY ≥ 0 versusH1 : pX − pY < 0


SECTION 6.6 281

(b) X = 150, nX = 180, pX = 150/180= 0.83333, Y = 233, nY = 270, pY = 233/270= 0.86296,

p = (150+233)/(180+270)= 0.85111.


z=0.83333−0.86296√

0.85111(1−0.85111)(1/180+1/270)−0.86.


ThusP = 0.1949.

(c) No. SinceP= 0.1949, we cannot conclude that the proportion from vendor B isgreater than that from vendor A.

3. X = 133, nX = 400, pX = 133/400= 0.3325, Y = 50, nY = 100, pY = 50/100= 0.5,

p = (133+50)/(400+100)= 0.366.

The null and alternate hypotheses areH0 : pX − pY = 0 versusH1 : pX − pY 6= 0.

z=0.3325−0.5√

0.366(1−0.366)(1/400+1/100)= −3.11.

Since the alternate hypothesis is of the formpX − pY 6= 0, theP-value is the sum of the areas to the right ofz= 3.11 and to the left ofz= −3.11.

ThusP = 0.0009+0.0009= 0.0018.

We can conclude that the response rates differ between public and private firms.

4. X = 48, nX = 60, pX = 48/60= 0.8, Y = 44, nY = 60, pY = 44/60= 0.73333,

p = (48+44)/(60+60)= 0.76667.

The null and alternate hypotheses areH0 : pX − pY ≤ 0 versusH1 : pX − pY > 0.

z=0.8−0.73333√

0.76667(1−0.76667)(1/60+1/60)= 0.86.

Since the alternate hypothesis is of the formpX − pY > 0, theP-value is the area to the right ofz= 0.86.

ThusP = 0.1949.

We cannot conclude that the model with one hidden layer has the greater success rate.

5. X = 57, nX = 100, pX = 57/100= 0.57, Y = 135, nY = 200, pY = 135/200= 0.675,

p = (57+135)/(100+200)= 0.64.


z=0.57−0.675√

0.64(1−0.64)(1/100+1/200)= −1.79.


282 CHAPTER 6


ThusP = 0.0367.

We can conclude that awareness of the benefit increased afterthe advertising campaign.

6. X = 19, nX = 161, pX = 19/161= 0.11801, Y = 22, nY = 95, pY = 22/95= 0.23158,

p = (19+22)/(161+95)= 0.16016.


z=0.11801−0.23158√

0.16016(1−0.16016)(1/161+1/95)= −2.39.


ThusP = 0.0084.

We can conclude that the larger particles are more likely to contain coliform bacteria.

7. X = 20, nX = 1200, pX = 20/1200= 0.016667, Y = 15, nY = 1500, pY = 15/1500= 0.01,

p = (20+15)/(1200+1500)= 0.012963.


z=0.016667−0.01√

0.012963(1−0.012963)(1/1200+1/1500)= 1.52.


ThusP = 0.0643.

The evidence suggests that heavy packaging reduces the proportion of damaged shipments, but may not beconclusive.

8. X = 260, nX = 300, pX = 260/300= 0.86667, Y = 370, nY = 400, pY = 370/400= 0.925,

p = (260+370)/(300+400)= 0.9.


z=0.86667−0.925√

0.9(1−0.9)(1/300+1/400)= −2.55.


ThusP = 0.0054.

We can conclude that the proportion of items rated acceptable or better is greater on Wednesday than onMonday.

9. X = 22, nX = 41, pX = 22/41= 0.53659, Y = 18, nY = 31, pY = 18/31= 0.58065,


SECTION 6.6 283

p = (22+18)/(41+31)= 0.55556.

The null and alternate hypotheses areH0 : pX − pY = 0 versusH1 : pX − pY 6= 0.

z=0.53659−0.58065√

0.55556(1−0.55556)(1/41+1/31)= −0.37.

Since the alternate hypothesis is of the formpX − pY 6= 0, theP-value is the sum of the areas to the right ofz= 0.37 and to the left ofz= −0.37.

ThusP = 0.3557+0.3557= 0.7114.

We cannot conclude that the proportion of wells that meet thestandards differs between the two areas.

10. X = 76, nX = 100, pX = 76/100= 0.76, Y = 128, nY = 200, pY = 128/200= 0.64,

p = (76+128)/(100+200)= 0.68.


z=0.76−0.64√

0.68(1−0.68)(1/100+1/200)= 2.10.


ThusP = 0.0179.

We can conclude that drug A is more effective than drug B.

11. X = 285, nX = 500, pX = 285/500= 0.57, Y = 305, nY = 600, pY = 305/600= 0.50833,

p = (285+305)/(500+600)= 0.53636.


z=0.57−0.50833√

0.53636(1−0.53636)(1/500+1/600)= 2.04.


ThusP = 0.0207.

We can conclude that the proportion of voters favoring the proposal is greater in county A than in county B.

12. (a)X = 4243, nX = 82,486, pX = 4243/82,486= 0.05144, Y = 10,701, nY = 219,170, pY = 10,701/219,170= 0.04883,

p = (4243+10,701)/(82,486+219,170)= 0.04954.


z=0.05144−0.04883√

0.04954(1−0.04954)(1/82,486+1/219,170)= 2.95.



284 CHAPTER 6

ThusP = 0.0016.

We can conclude that accidents involving drivers aged 15–24are more likely to occur in driveways than acci-dents involving drivers aged 25–64.

(b) No, the observed difference is only 0.2%. Although this is statistically significant, it is not practically signifi-cant.

13. No, because the two samples are not independent.

14. (a) Two-tailed. The alternate hypothesis is of the formp1− p2 6= 0.

(b) H0 : p1− p2 = 0


15. (a) 101/153= 0.660131.

(b) 90(0.544444) = 49.

(c) X1 = 101, n1 = 153, p1 = 101/153= 0.660131, X2 = 49, n2 = 90, p2 = 49/90= 0.544444,

p = (101+49)/(153+90)= 0.617284.

z=0.660131−0.544444√

0.617284(1−0.617284)(1/153+1/90)= 1.79.

(d) Since the alternate hypothesis is of the formpX − pY 6= 0, theP-value is the sum of the areas to the right ofz= 1.79 and to the left ofz= −1.79.

ThusP = 0.0367+0.0367= 0.0734.

Section 6.7

1. (a)X = 3.05,sX = 0.34157,nX = 4, Y = 1.8, sY = 0.90921,nY = 4.


ν =

[0.341572

4+

0.909212

4

]2

(0.341572/4)2

4−1+

(0.909212/4)2

4−1



SECTION 6.7 285

t3 = (3.05−1.8−0)/√

0.341572/4+0.909212/4 = 2.574.


Since the alternate hypothesis is of the formµX −µY > ∆, theP-value is the area to the right oft = 2.574.


We can conclude that the mean strength of crayons made with dye B is greater than that made with dye A.

(b) X = 3.05,sX = 0.34157,nX = 4, Y = 1.8, sY = 0.90921,nY = 4.


ν =

[0.341572

4+

0.909212

4

]2

(0.341572/4)2

4−1+

(0.909212/4)2

4−1


t3 = (3.05−1.8−1)/√

0.341572/4+0.909212/4 = 0.5148.




We cannot conclude that the mean strength of crayons made with dye B exceeds that of those made with dye Aby more than 1 J.

2. X = 255.8, sX = 8.216325,nX = 6, Y = 270.7375,sY = 11.902933,nY = 8.


ν =

[8.2163252

6+

11.9029332

8

]2

(8.2163252/6)2

6−1+

(11.9029332/8)2

8−1


t11 = (255.8−270.7375−0)/√

8.2163252/6+11.9029332/8 = −2.776.


Since the alternate hypothesis is of the formµX −µY 6= ∆, theP-value is the sum of the areas to the right oft = 2.776 and to the left oft = −2.776.


We can conclude that the dissolve times differ between the two shapes.

3. X = 1.93,sX = 0.84562,nX = 7, Y = 3.05,sY = 1.5761,nY = 12.



286 CHAPTER 6

ν =

[0.845622

7+

1.57612

12

]2

(0.845622/7)2

7−1+

(1.57612/12)2

12−1


t16 = (1.93−3.05−0)/√

0.845622/7+1.57612/12= −2.014.




The null hypothesis is suspect.

4. (a)X = 47.79,sX = 2.19,nX = 14, Y = 47.15,sY = 2.65,nY = 40.


ν =

[2.192

14+

2.652

40

]2

(2.192/14)2

14−1+

(2.652/40)2

40−1


t27 = (47.79−47.15−0)/√

2.192/14+2.652/40= 0.8891.




We cannot conclude that the time to perform a lift of low difficulty is less with the video system than with thetagman system.

(b) X = 69.33,sX = 6.26,nX = 12, Y = 58.5, sY = 5.59,nY = 24.


ν =

[6.262

12+

5.592

24

]2

(6.262/12)2

12−1+

(5.592/24)2

24−1


t19 = (69.33−58.5−0)/√

6.262/12+5.592/24= 5.067.




We can conclude that the time to perform a lift of moderate difficulty is less with the video system than withthe tagman system.


SECTION 6.7 287

(c) X = 109.71,sX = 17.02,nX = 17, Y = 84.52,sY = 13.51,nY = 29.


ν =

[17.022

17+

13.512

29

]2

(17.022/17)2

17−1+

(13.512/29)2

29−1


t27 = (109.71−84.52−0)/√

17.022/17+13.512/29= 5.215.




We can conclude that the time to perform a lift of high difficulty is less with the video system than with thetagman system.

5. X = 2.1062,sX = 0.029065,nX = 5, Y = 2.0995,sY = 0.033055,nY = 5.


ν =

[0.0290652

5+

0.0330552

5

]2

(0.0290652/5)2

5−1+

(0.0330552/5)2

5−1


t7 = (2.1062−2.0995−0)/√

0.0290652/5+0.0330552/5 = 0.3444.




We cannot conclude that the calibration has changed from thefirst to the second day.

6. X = 66.1667, sX = 19.630758, nX = 6, Y = 28.2, sY = 17.767949, nY = 5.

Since the population standard deviations are assumed to be equal, estimate their common value with the pooledstandard deviation

sp =

√(nX −1)s2

X +(nY −1)s2Y

nX +nY −2= 18.825613.

The number of degrees of freedom isν = nX −nY −2 = 6+5−2= 9.

t9 =66.1667−28.2

18.825613√

1/6+1/5= 3.331.



288 CHAPTER 6



We can conclude that the weights differ.

7. X = 53.0, sX = 1.41421,nX = 6, Y = 54.5, sY = 3.88587,nY = 6.


ν =

[1.414212

6+

3.885872

6

]2

(1.414212/6)2

6−1+

(3.885872/6)2

6−1


t6 = (53.0−54.5−0)/√

1.414212/6+3.885872/6 = −0.889.


Since the alternate hypothesis is of the formµX −µY < ∆, theP-value is the area to the left oft = −0.889.


We cannot conclude that the mean cost of the new method is lessthan that of the old method.

8. (a) No. Even though the sample standard deviations are equal, the population variances may be quite different.

(b) X = 0.51,sX = 0.09,nX = 20, Y = 0.62,sY = 0.09,nY = 14.


ν =

[0.092

20+

0.092

14

]2

(0.092/20)2

20−1+

(0.092/14)2

14−1


t28 = (0.51−0.62−0)/√

0.092/20+0.092/14= −3.507.




We can conclude that the mean mass ratios differ between summer and winter.

9. X = 22.1, sX = 4.09,nX = 11, Y = 20.4, sY = 3.08,nY = 7.



SECTION 6.7 289

ν =

[4.092

11+

3.082

7

]2

(4.092/11)2

11−1+

(3.082/7)2

7−1


t15 = (22.1−20.4−0)/√

4.092/11+3.082/7 = 1.002.




We cannot conclude that the mean compressive stress is greater for no. 1 grade lumber than for no. 2 grade.

10. X = 6.02,sX = 0.02,nX = 10, Y = 6, sY = 0.01,nY = 5.


ν =

[0.022

10+

0.012

5

]2

(0.022/10)2

10−1+

(0.012/5)2

5−1


t12 = (6.02−6−0)/√

0.022/10+0.012/5 = 2.582.




We can conclude that the mean measurements differ between the two methods.

11. X = 2.465000,sX = 0.507149,nX = 8, Y = 4.495714,sY = 0.177469,nY = 7.


ν =

[0.5071492

8+

0.1774692

7

]2

(0.5071492/8)2

8−1+

(0.1774692/7)2

7−1


t8 = (2.465000−4.495714−0)/√

0.5071492/8+0.1774692/7 = −10.608.




We can conclude that the dissolution rates differ.


290 CHAPTER 6

12. X = 3.6, sX = 0.14142,nX = 4, Y = 3.2, sY = 0.18257,nY = 4.


ν =

[0.141422

4+

0.182572

4

]2

(0.141422/4)2

4−1+

(0.182572/4)2

4−1


t5 = (3.6−3.2−0.1)/√

0.141422/4+0.182572/4 = 2.598.

The null and alternate hypotheses areH0 : µX −µY ≤ 0.1 versusH1 : µX −µY > 0.1.



We can conclude that more than 0.1µg is absorbed between 30 and 60 minutes after exposure.

13. (a)X = 109.5, sX = 31.2, nX = 12, Y = 47.2, sY = 15.1, nY = 10.


ν =

[31.22

12+

15.12

10

]2

(31.22/12)2

12−1+

(15.12/10)2

10−1


t16 = (109.5−47.2−0)/√

31.22/12+15.12/10= 6.111.




We can conclude that mean mutation frequency for men in theirsixties is greater than for men in their twenties.

(b) X = 109.5, sX = 31.2, nX = 12, Y = 47.2, sY = 15.1, nY = 10.


ν =

[31.22

12+

15.12

10

]2

(31.22/12)2

12−1+

(15.12/10)2

10−1


t16 = (109.5−47.2−25)/√

31.22/12+15.12/10= 3.659.



From thet table,P ≈ 0.001. A computer package givesP = 0.0011. We can conclude that mean mutationfrequency for men in their sixties is greater than for men in their twenties by more than 25 sequences perµg.


SECTION 6.7 291

14. X = 464.700,sX = 14.2379,nX = 9, Y = 482.786,sY = 13.9421,nY = 14.


ν =

[14.23792

9+

13.94212

14

]2

(14.23792/9)2

9−1+

(13.94212/14)2

14−1


t16 = (464.700−482.786−0)/√

14.23792/9+13.94212/14= −2.997.


Since the alternate hypothesis is of the formµX −µY < ∆, theP-value is the area to the left oft = −2.997.


We can conclude that the mean breaking strength is greater for hockey sticks made from Composite B.

15. X = 62.714,sX = 3.8607,nX = 7, Y = 60.4, sY = 5.461,nY = 10.


ν =

[3.86072

7+

5.4612

10

]2

(3.86072/7)2

7−1+

(5.4612/10)2

10−1


t14 = (62.714−60.4−0)/√

3.86072/7+5.4612/10= 1.024.




We cannot conclude that the mean permeability coefficient at60C differs from that at 61C.

16. (a) One-tailed. The alternative hypothesis is of the form µX −µY > ∆.

(b) H0 : µX −µY ≤ 0


17. (a) SE Mean = StDev/√

N = 0.482/√

6 = 0.197.

(b) StDev = (SE Mean)√

N = 0.094√

13= 0.339.

(c) X−Y = 1.755−3.239= −1.484.


292 CHAPTER 6

(d) t =1.755−3.239√

0.4822/6+0.0942= −6.805.

Section 6.8

1. D = 4.2857,sD = 1.6036,n = 7. There are 7−1= 6 degrees of freedom.

The null and alternate hypotheses areH0 : µD = 0 versusH1 : µD 6= 0.

t = (4.2857−0)/(1.6036/√

7) = 7.071.

Since the alternate hypothesis is of the formµD 6= ∆, theP-value is the sum of the areas to the right oft = 7.071and to the left oft = −7.071.

From thet table,P < 0.001. A computer package givesP = 0.00040.

We can conclude that there is a difference in latency betweenmotor point and nerve stimulation.



t = (0.31545−0)/(0.36445/√

11) = 2.871.



We can conclude that there is a difference in the mean impedance ratio between the standing and sitting position.

3. The differences are 4,−1,0,7,3,4,−1,5,−1,5.

D = 2.5, sD = 2.9907,n = 10. There are 10−1= 9 degrees of freedom.


t = (2.5−0)/(2.9907/√

10) = 2.643.



We can conclude that the etch rates differ between center andedge.

4. The differences are 0.0195,−0.0120,0.0131,0.1041,0.0211,0.0434,0.0991,0.0458,−0.0249,0.0539.

D = 0.03631,sD = 0.042323,n = 10. There are 10−1= 9 degrees of freedom.



SECTION 6.8 293

t = (0.03631−0)/(0.042323/√

10) = 2.713.



There is fairly strong evidence that the mean emissions differ between the two temperatures.

5. The differences are−7,−21,4,−16,2,−9,−20,−13.

D = −10,sD = 9.3808,n = 8. There are 8−1= 7 degrees of freedom.


t = (−10−0)/(9.3808/√

8) = −3.015.



We can conclude that the mean amount of corrosion differs between the two formulations.

6. The differences are 3.1,−1.2,0.4,−0.4,−3.1,−7.7.

D = −1.48333,sD = 3.6625,n = 6. There are 6−1= 5 degrees of freedom.


t = (−1.48333−0)/(3.6625/√

6) = −0.9921.



We cannot conclude that the mean speeds of the two processorsdiffer.

7. The differences are 35,57,14,29,31.

D = 33.2, sD = 15.498,n = 5.

There are 5−1= 4 degrees of freedom.

The null and alternate hypotheses areH0 : µD ≤ 0 versusH1 : µD > 0.

t = (33.2−0)/(15.498/√

5) = 4.790.

Since the alternate hypothesis is of the formµD > ∆, theP-value is the area to the right oft = 4.790.


We can conclude that the mean strength after 6 days is greaterthan the mean strength after 3 days.

8. The differences are−0.04,−0.05,−0.02,−0.02,0.01,−0.02,−0.05,0.01,−0.05,0.03.


294 CHAPTER 6

D = 0.02,sD = 0.028674,n = 10.



t = (0.02−0)/(0.028674/√

10) = 2.206.



We can conclude that scale 2 reads heavier, on the average, than scale 1.

9. The differences are: 0.76,0.30,0.74,0.30,0.56,0.82.

D = 0.58,sD = 0.233581,n = 6. There are 6−1= 5 degrees of freedom.


t = (0.58−0)/(0.233581/√

6) = 6.082.



We can conclude that there is a difference in concentration between the shoot and the root.

10. (a) The differences are: 8.4,8.6,10.5,9.6,10.7,10.8,10.7,11.3,10.7.

D = 10.1444sD = 1.0333,n = 9. There are 9−1= 8 degrees of freedom.


t = (10.1444−0)/(1.0333/√

9) = 29.452.



We can conclude that mean lifetime of rear brake pads is greater than that of front brake pads.

(b) D = 10.1444sD = 1.0333,n = 9. There are 9−1= 8 degrees of freedom.


t = (10.1444−10)/(1.0333/√

9) = 0.419.



We cannot conclude that mean lifetime of rear brake pads is exceeds that of front brake pads by more than10,000 miles.

11. (a) The differences are 5.0,4.6,1.9,2.6,4.4,3.2,3.2,2.8,1.6,2.8.


SECTION 6.8 295

Let µR be the mean number of miles per gallon for taxis using radial tires, and letµB be the mean number ofmiles per gallon for taxis using bias tires. The appropriatenull and alternate hypotheses areH0 : µR−µB ≤ 0versusH1 : µR−µB > 0.


t = (3.21−0)/(1.1338/√

10) = 8.953.



We can conclude that the mean number of miles per gallon is higher with radial tires.

(b) The appropriate null and alternate hypotheses areH0 : µR−µB ≤ 2 vs.H1 : µR−µB > 2.


t = (3.21−2)/(1.1338/√

10) = 3.375.


From thet table, 0.001< P < 0.005. A computer package givesP = 0.0041. We can conclude that the meanmileage with radial tires is more than 2 miles per gallon higher than with bias tires.

12. (a) One-tailed. The alternate hypothesis is of the formµD > µ0.

(b) H0 : µX −µY ≤ 0

(c) No,P = 0.038, which is greater than 0.01.

(d) D = 33.6316,sD/√

n = 17.1794,n = 12,t12−1,.01 = 2.718.

The confidence interval is 33.6316±2.718(17.1794), or (−13.062,80.325).

13. (a) SE Mean = StDev/√

N = 2.9235/√

7 = 1.1050.

(b) StDev = (SE Mean)√

N = 1.0764√

7 = 2.8479.

(c) µD = µX −µY = 12.4141−8.3476= 4.0665.

(d) t = (4.0665−0)/1.19723= 3.40.


296 CHAPTER 6

Section 6.9

1. (a) The signed ranks are Signedx x−14 Rank

20 6 121 7 25 −9 −3

24 10 425 11 532 18 643 29 7


The sum of the positive signed ranks isS+ = 25. n = 7.

Since the alternate hypothesis is of the formµ > µ0, theP-value is the area under the signed rank probabilitymass function corresponding toS+ ≥ 25.

From the signed-rank table,P = 0.0391.

We can conclude that the mean concentration is greater than 14 g/L.

(b) The signed ranks are Signedx x−30 Rank

32 2 125 −5 −224 −6 −321 −9 −420 −10 −543 13 65 −25 −7



Since the alternate hypothesis is of the formµ < µ0, theP-value is the area under the signed rank probabilitymass function corresponding toS+ ≤ 7.

From the signed-rank table,P > 0.1094.

We cannot conclude that the mean concentration is less than 30 g/L.


SECTION 6.9 297

(c) The signed ranks are Signedx x−18 Rank

20 2 121 3 224 6 325 7 45 −13 −5

32 14 643 25 7



Since the alternate hypothesis is of the formµ 6= µ0, theP-value is twice the area under the signed rank proba-bility mass function corresponding toS+ ≥ 23.

From the signed-rank table,P = 2(0.0781) = 0.1562.

We cannot conclude that the mean concentration differs from18 g/L.


41.01 0.01 141.37 0.37 241.42 0.42 340.50 −0.50 −441.70 0.70 541.83 0.83 6.541.83 0.83 6.542.68 1.68 8




From the signed-rank table,P = 0.0273.

We can conclude that the mean thickness is greater than 41 mm.


298 CHAPTER 6

(b) The signed ranks are Signedx x−41.8 Rank

41.83 0.03 1.541.83 0.03 1.541.70 −0.10 −341.42 −0.38 −441.37 −0.43 −541.01 −0.79 −642.68 0.88 740.50 −1.30 −8





We cannot conclude that the mean thickness is less than 41.8 mm.


41.83 −0.17 −1.541.83 −0.17 −1.541.70 −0.30 −341.42 −0.58 −441.37 −0.63 −542.68 0.68 641.01 −0.99 −740.50 −1.50 −8



Since the alternate hypothesis is of the formµ 6= µ0, theP-value is twice the area under the signed rank proba-bility mass function corresponding toS+ ≤ 6.


We cannot conclude that the mean thickness differs from 42 mm.


SECTION 6.9 299


41.1 −3.9 −141.0 −4.0 −240.0 −5.0 −338.1 −6.9 −452.3 7.3 537.3 −7.7 −636.2 −8.8 −734.6 −10.4 −833.6 −11.4 −933.4 −11.6 −1059.4 14.4 1130.3 −14.7 −1230.1 −14.9 −1328.8 −16.2 −1463.0 18.0 1526.8 −18.2 −1667.8 22.8 1719.0 −26.0 −1872.3 27.3 1914.3 −30.7 −2078.6 33.6 2180.8 35.8 229.1 −35.9 −23

81.2 36.2 24


The sum of the positive signed ranks isS+ = 134.n = 24.

Sincen > 20, compute thez-score ofS+ and use thez table.

z=S+−n(n+1)/4√

n(n+1)(2n+1)/24= −0.46.

Since the alternate hypothesis is of the formµ< µ0, theP-value is the area to the left ofz= −0.46. From theztable,P = 0.3228.

We cannot conclude that the mean conversion is less than 45.


300 CHAPTER 6


30.1 0.1 130.3 0.3 228.8 −1.2 −326.8 −3.2 −433.4 3.4 533.6 3.6 634.6 4.6 736.2 6.2 837.3 7.3 938.1 8.1 1040.0 10.0 1119.0 −11.0 −12.541.0 11.0 12.541.1 11.1 1414.3 −15.7 −159.1 −20.9 −16

52.3 22.3 1759.4 29.4 1863.0 33.0 1967.8 37.8 2072.3 42.3 2178.6 48.6 2280.8 50.8 2381.2 51.2 24


The sum of the positive signed ranks isS+ = 249.5. n = 24.


z=S+−n(n+1)/4√

n(n+1)(2n+1)/24= 2.84.

Since the alternate hypothesis is of the formµ > µ0, theP-value is the area to the right ofz= 2.84. From theztable,P = 0.0023.

We can conclude that the mean conversion is greater than 30.


SECTION 6.9 301


52.3 −2.7 −159.4 4.4 263.0 8.0 367.8 12.8 441.1 −13.9 −541.0 −14.0 −640.0 −15.0 −738.1 −16.9 −872.3 17.3 937.3 −17.7 −1036.2 −18.8 −1134.6 −20.4 −1233.6 −21.4 −1333.4 −21.6 −1478.6 23.6 1530.3 −24.7 −1630.1 −24.9 −1780.8 25.8 1828.8 −26.2 −19.581.2 26.2 19.526.8 −28.2 −2119.0 −36.0 −2214.3 −40.7 −239.1 −45.9 −24




z=S+−n(n+1)/4√

n(n+1)(2n+1)/24= −2.27.

Since the alternate hypothesis is of the formµ 6= µ0, theP-value is the sum of the areas to the right ofz= 2.27and to the left ofz= −2.27.

From thez table,P = 0.0116+0.0116= 0.0232.

We can conclude that the mean conversion differs from 55.


302 CHAPTER 6


40.76 −0.24 −140.57 −0.43 −241.48 0.48 339.68 −1.32 −443.06 2.06 543.53 2.53 643.68 2.68 743.76 2.76 844.86 3.86 946.14 5.14 10




From the signed-rank table, 0.0137< P < 0.0244.

We can conclude that the mean thickness is greater than 41 mm.


43.06 0.06 143.53 0.53 243.68 0.68 343.76 0.76 441.48 −1.52 −544.86 1.86 640.76 −2.24 −740.57 −2.43 −846.14 3.14 939.68 −3.32 −10





We cannot conclude that the mean thickness is less than 43 mm.


SECTION 6.9 303


43.76 −0.24 −143.68 −0.32 −243.53 −0.47 −344.86 0.86 443.06 −0.94 −546.14 2.14 641.48 −2.52 −740.76 −3.24 −840.57 −3.43 −939.68 −4.32 −10



Since the alternate hypothesis is of the formµ 6= µ0, theP-value is twice the area under the signed rank proba-bility mass function corresponding toS+ ≤ 10.


We cannot conclude that the mean thickness differs from 44 mm.

5. The signed ranks are Signedx x−0 Rank

0.01 0.01 20.01 0.01 2

−0.01 −0.01 −20.03 0.03 40.05 0.05 5.5

−0.05 −0.05 −5.5−0.07 −0.07 −7−0.11 −0.11 −8−0.13 −0.13 −9

0.15 0.15 10



Since the alternate hypothesis is of the formµ 6= µ0, theP-value is twice the area under the signed rank proba-bility mass function corresponding toS+ ≤ 23.5.

From the signed-rank table,P > 2(0.1162) = 0.2324.

We cannot conclude that the gauges differ.


304 CHAPTER 6

6. The ranks of the combined samples are Value Rank Sample0.43 1 Y0.73 2 X0.93 3 Y1.12 4 X1.24 5 X1.91 6 Y2.56 7 Y2.93 8 X3.72 9 Y6.19 10 Y

11.00 11 Y


The test statisticW is the sum of the ranks corresponding to theX sample.

W = 19. The sample sizes arem= 4 andn = 7.

Since the alternate hypothesis is of the formµX − µY 6= ∆, theP-value is twice the area under the rank-sumprobability mass function corresponding toW ≤ 19.

From the rank-sum table,P > 2(0.0545) = 0.1090.

We cannot conclude that the mean rates differ.

7. The ranks of the combined samples are Value Rank Sample12 1 X13 2 X15 3 X18 4 Y19 5 X20 6 X21 7 X23 8 Y24 9 Y27 10 X30 11 Y32 12 Y35 13 Y40 14 Y




Since the alternate hypothesis is of the formµX − µY 6= ∆, theP-value is twice the area under the rank-sumprobability mass function corresponding toW ≤ 34.

From the rank-sum table,P = 2(0.0087) = 0.0174.


SECTION 6.9 305

We can conclude that the mean recovery times differ.

8. The ranks of the combined samples are Value Rank Sample1255 1.5 X1255 1.5 X1270 3 X1280 4 X1287 5 X1291 6 X1296 7 X1301 8 Y1302 9 X1306 10 X1310 11 X1314 12 X1318 13 X1321 14 Y1326 15 X1328 16 X1329 17 X1332 18 Y1341 19 Y1343 20 Y1349 21 Y1364 22 Y1372 23 Y1374 24 Y1376 25 Y1378 26 Y1387 27 Y1396 28 Y1397 29 Y1399 30 Y




Sincen andmare both greater than 8, compute thez-score ofW and use thez table.

z=W−m(m+n+1)/2√

mn(m+n+1)/12= −4.21.

Since the alternate hypothesis is of the formµX −µY < ∆, theP-value is the area to the left ofz= −4.21. Fromthez table,P≈ 0.

We can conclude that the mean strength is greater for blocks cured 6 days.


306 CHAPTER 6

9. The ranks of the combined samples are Value Rank Sample51 1 Y59 2 Y64 4 X64 4 Y64 4 Y66 6 X69 7 X72 8 X74 9.5 X74 9.5 Y75 11.5 X75 11.5 Y77 14 X77 14 X77 14 Y80 16.5 Y80 16.5 Y81 18.5 Y81 18.5 Y83 20.5 X83 20.5 Y85 22.5 X85 22.5 Y86 24 Y88 25 X91 26 X





z=W−m(m+n+1)/2√

mn(m+n+1)/12= 0.31.

Since the alternate hypothesis is of the formµX −µY 6= ∆, theP-value is the sum of the areas to the right ofz= 0.31 and to the left ofz= −0.31. From thez table,P = 0.3783+0.3783= 0.7566.

We cannot conclude that the mean scores differ.


SECTION 6.10 307

10. The ranks of the combined samples are Value Rank Sample15.7 1 Y15.9 2 Y

16 3 Y16.1 4 X16.2 5 Y16.3 6 Y16.4 7 Y16.7 8 X16.9 9 X17.2 10 X19.8 11 X




Since the alternate hypothesis is of the formµX −µY > ∆, theP-value is the area under the rank-sum probabilitymass function corresponding toW ≥ 42.

From the rank-sum table,P = 0.0152.

We can conclude that the mean time is less for route A.

Section 6.10

1. (a) Let p1 represent the probability that a randomly chosen rivet meets the specification, letp2 represent theprobability that it is too short, and letp3 represent the probability that it is too long. Then the null hypothesisis H0 : p1 = 0.90, p2 = 0.05, p3 = 0.05

(b) The total number of observation isn = 1000. The expected values arenp1, np2 andnp3, or 900, 50, and 50.

(c) The observed values are 860, 60, and 80.

χ2 = (860−900)2/900+(60−50)2/50+(80−50)2/50= 21.7778.

(d) There are 3− 1 = 2 degrees of freedom. From theχ2 table, P < 0.005. A computer package givesP = 1.9×10−5.

The true percentages differ from 90%, 5%, and 5%.

2. (a) Let pi j denote the probability that an observation is in columnj, given that it is in rowi. Then the nullhypothesis isH0 : For each columnj, p1 j = p2 j = p3 j .


308 CHAPTER 6

(b) The row totals areO1. = 400,O2. = 300,O3. = 500. The column totals areO.1 = 968,O.2 = 179,O.3 = 53.The grand total isO.. = 1200.

The expected values areEi j = Oi.O. j/O.., as shown in the following table.

322.6667 59.6667 17.6667242.0000 44.7500 13.2500403.3333 74.5833 22.0833

(c) χ2 = ∑3i=1∑3

j=1(Oi j −Ei j )2/Ei j = 7.033

(d) There are(3−1)(3−1) = 4 degrees of freedom. From theχ2 table,P > 0.10. A computer package givesP = 0.13.

There is no evidence that the proportions of rollers in the three categories differ among the machinists.

3. The row totals areO1. = 173 andO2. = 210. The column totals areO.1 = 181,O.2 = 99,O.3 = 31,O.4 = 11,O.5 = 61. The grand total isO.. = 383.


Net Excess Capacity

< 0% 0 – 10% 11 – 20% 21 – 30% > 30%Small 81.7572 44.7180 14.0026 4.9687 27.5535Large 99.2428 54.2820 16.9974 6.0313 33.4465

There are(2−1)(5−1)= 4 degrees of freedom.

χ2 = ∑2i=1∑5

j=1(Oi j −Ei j )2/Ei j = 12.945.

From theχ2 table, 0.01< P < 0.05. A computer package givesP = 0.012.

It is reasonable to conclude that the distributions differ.

4. χ2 = ∑9i=1(Oi −Ei)

2/Ei = 21.475.



The theoretical model does not explain the observed values well.

5. The row totals areO1. = 41,O2. = 39, andO3. = 412. The column totals areO.1 = 89,O.2 = 163,O.3 = 240.The grand total isO.. = 492.



SECTION 6.10 309

< 1 1 to< 5 ≥ 5Diseased 7.4167 13.583 20.000

Sensitized 7.0549 12.921 19.024Normal 74.528 136.50 200.98


χ2 = ∑3i=1∑3

j=1(Oi j −Ei j )2/Ei j = 10.829.


There is some evidence that the proportions of workers in thevarious disease categories differ among exposurelevels.

6. The row totals areO1. = 171 andO2. = 208. The column totals areO.1 = 14, O.2 = 74, O.3 = 40, O.4 = 41,O.5 = 39,O.6 = 36,O.7 = 62,O.8 = 73. The grand total isO.. = 379.


Percent of Full Operating Capacity Labor Force Currently Employed

> 100% 95 – 100% 90 – 94% 85 – 89% 80 – 84% 75 – 79% 70 – 74%< 70%Small 6.3166 33.3879 18.0475 18.4987 17.5963 16.2427 27.9736 32.9367Large 7.6834 40.6121 21.9525 22.5013 21.4037 19.7573 34.0264 40.0633


χ2 = ∑2i=1∑8

j=1(Oi j −Ei j )2/Ei j = 8.8837.

From theχ2 table,P > 0.10. A computer package givesP = 0.261.

There is no evidence that the distributions differ.

7. (a) The row totals areO1. = 37,O2. = 25, andO3. = 35. The column totals areO.1 = 27,O.2 = 35,O.3 = 35. Thegrand total isO.. = 97.


10.2990 13.3505 13.35056.9588 9.0206 9.02069.7423 12.6289 12.6289

(b) Theχ2 test is appropriate here, since all the expected values are greater than 5.


χ2 = ∑3i=1∑3

j=1(Oi j −Ei j )2/Ei j = 6.4808.


There is no evidence that the rows and columns are not independent.


310 CHAPTER 6

8. (a) The row totals areO1. = 40, O2. = 10, O3. = 50. The column totals areO.1 = 70, O.2 = 10, O.3 = 20. Thegrand total isO.. = 100.


28 4 87 1 2

35 5 10

(b) The chi-square test should not be performed, because some of the expected values are less than 5.

9. (iii) Both row totals and column totals in the observed table must be the same as the row and column totals,respectively, in the expected table.

10. It is possible to compute the expected values.

The grand total is given to be 150. Therefore the missing row and column totals must be the values that makethe grand total ,O.., come out to 150.

Therefore the row totals areO1. = 25,O2. = 10,O3. = 40,O4. = 75. The column totals areO.1 = 50,O.2 = 20,O.3 = 80.


8.33 3.33 13.333.33 1.33 5.33

13.33 5.33 21.3325.00 10.00 40.00

11. Let pi , i = 0,1, ...,9 be he probability that a random digit has the valuei. The null hypothesis isH0 : pi = 0.1for all i.

The total number of observations isn = 200. The expected values are equal tonpi = 200(0.1) = 20.

The test statistic isχ2 = (21−20)2/20+(17−20)2/20+(20−20)2/20+(18−20)2/20+(25−20)2/20+(16−20)2/20+(28−20)2/20+(19−20)2/20+(22−20)2/20+(14−20)2/20= 8.



There is no evidence that the random number generator is not working properly.

12. The row totals areO1. = 47,O2. = 63,O3. = 32,O4. = 25. The column totals areO.1 = 58,O.2 = 71,O.3 = 38.The grand total isO.. = 167.



SECTION 6.10 311

Morning Evening NightInfluenza 16.323 19.982 10.695Headache 21.880 26.784 14.335Weakness 11.114 13.605 7.281

Shortness of Breath 8.6826 10.629 5.689


χ2 = ∑4i=1∑3

j=1(Oi j −Ei j )2/Ei j = 17.572.


We can conclude that the proportions of workers with the various symptoms differ among shifts.

13. The row totals areO1. = 217 andO2. = 210. The column totals areO.1 = 32, O.2 = 15, O.3 = 37, O.4 = 38,O.5 = 45,O.6 = 48,O.7 = 46,O.8 = 42,O.9 = 34,O.10 = 36,O.11 = 28,O.12 = 26. The grand total isO.. = 427.


Month1 2 3 4 5 6 7 8 9 10 11 12

Known 16.26 7.62 18.80 19.31 22.87 24.39 23.38 21.34 17.28 18.30 14.23 13.21Unknown 15.74 7.38 18.20 18.69 22.13 23.61 22.62 20.66 16.72 17.70 13.77 12.79


χ2 = ∑2i=1∑12

j=1(Oi j −Ei j )2/Ei j = 41.33.

From theχ2 table,P < 0.005. A computer package givesP = 2.1×10−5.

We can conclude that the proportion of false alarms whose cause is known differs from month to month.

14. (a) 22

(b) 39.86

(c) C2

(d) C3

(e) No. TheP-value is 0.301, so the null hypothesis of independence cannot be rejected.

(f) No. It cannot be concluded that the null hypothesis is true.


312 CHAPTER 6

Section 6.11

1. ν1 = 7, ν2 = 20. From theF table, the upper 5% point is 2.51.

2. ν1 = 2, ν2 = 5. From theF table, the upper 1% point is 13.27.

3. (a) The upper 1% point of theF5,7 distribution is 7.46. Therefore theP-value is 0.01.

(b) TheP-value for a two-tailed test is twice the value for the corresponding one-tailed test.

ThereforeP = 0.02.

4. (a) The sample variance for day 1 isσ21 = 0.0192308. The sample variance for day 2 isσ2

2 = 0.148077. The samplevariance for day 3 isσ2

3 = 0.268590.

The sample sizes aren1 = n2 = n3 = 13.

The null and alternate hypotheses areH0 : σ22/σ2

1 ≤ 1 versusH1 : σ22/σ2

1 > 1.

The test statistic isF = σ22/σ2

1 = 7.7000. The numbers of degrees of freedom are 12 and 12.

TheP-value is the area to the right of 7.7000 under theF12,12 probability density function.

From theF table,P < 0.001. A computer package givesP = 0.00064.

We can conclude that the variability of the process on the second day is greater than on the first day.

(b) The null and alternate hypotheses areH0 : σ23/σ2

2 ≤ 1 versusH1 : σ23/σ2

2 > 1.




From theF table,P > 0.1. A computer package givesP = 0.158.

We cannot conclude that the variability of the process on thethird day is greater than on the second day.

5. The sample variance of the breaking strengths for composite A is σ21 = 202.7175. The sample size isn1 = 9.

The sample variance of the breaking strengths for compositeB is σ22 = 194.3829. The sample size isn2 = 14.


2 = 1 versusH1 : σ21/σ2

2 6= 1.



Since this is a two-tailed test, theP-value is twice the area to the right of 1.0429 under theF8,13 probabilitydensity function.



SECTION 6.12 313

We cannot conclude that the variance of the breaking strength varies between the composites.

6. The sample variance of the times for the first month isσ21 = 200.7562. The sample size isn1 = 7.

The sample variance of the times for the seventh month isσ22 = 3570.8978. The sample size isn2 = 9.


1 ≤ 1 versusH1 : σ22/σ2

1 > 1.




From theF table, 0.001< P < 0.01. A computer package givesP = 0.0012.

We can conclude that the time to freeze-up is more variable inthe seventh month than in the first month.

Section 6.12

1. (a) True.H0 is rejected at any level greater than or equal to 0.03.

(b) False. 2% is less than 0.03.

(c) False. 10% is greater than 0.03.

2. (a) LetX be the sample mean of the 60 specimens.

UnderH0, the population mean isµ = 7, and the population standard deviation is estimated with the samplestandard deviations= 0.5.

The null distribution ofX is therefore normal with mean 7 and standard deviation 0.5/√

60= 0.064550.

Since this is a two-tailed test, the rejection region for a 5%level test consists of the upper and lower 2.5% ofthe null distribution.

Thez-scores corresponding to the upper and lower 2.5% of the normal distribution arez= 1.96 andz=−1.96,respectively.

Therefore the rejection region consists of all values ofX greater than or equal to 7+1.96(0.064550)= 7.1265,along with all values ofX less than or equal to 7−1.96(0.064550)= 6.8735.

H0 will be rejected ifX ≥ 7.1265 or ifX ≤ 6.8735.

(b) Since this is a two-tailed test, the rejection region fora 10% level test consists of the upper and lower 5% ofthe null distribution.

Thez-scores corresponding to the upper and lower 5% of the normaldistribution arez= 1.645 andz=−1.645,respectively.

Therefore the rejection region consists of all values ofX greater than or equal to 7+1.645(0.064550)= 7.1062,


314 CHAPTER 6

along with all values ofX less than or equal to 7−1.645(0.064550)= 6.8938.

Since 6.87< 6.8938,H0 will be rejected at the 10% level.

(c) Since this is a two-tailed test, the rejection region fora 1% level test consists of the upper and lower 0.5% ofthe null distribution.

Thez-scores corresponding to the upper and lower 0.5% of the normal distribution arez= 2.58 andz=−2.58,respectively.

Therefore the rejection region consists of all values ofX greater than or equal to 7+2.58(0.064550)= 7.1665,along with all values ofX less than or equal to 7−2.58(0.064550)= 6.8335.

Since 6.87> 6.8335,H0 will not be rejected at the 1% level.

(d) UnderH0, thez-score of 7.2 is(7.2−7)/0.064550= 3.10.

Since this is a two-tailed test, the level is the sum of the areas to the right ofz= 3.10 and to the left ofz=−3.10.

Therefore the level isα = 0.0010+0.0010= 0.0020.

3. (a)H0 : µ≥ 90 versusH1 : µ< 90

(b) LetX be the sample mean of the 150 times.

UnderH0, the population mean isµ= 90, and the population standard deviation isσ = 5.

The null distribution ofX is therefore normal with mean 90 and standard deviation 5/√

150= 0.408248.

Since the alternate hypothesis is of the formµ< µ0, the rejection region for a 5% level test consists of the lower5% of the null distribution.

Thez-score corresponding to the lower 5% of the normal distribution isz= −1.645.

Therefore the rejection region consists of all values ofX less than or equal to 90−1.645(0.408248)= 89.3284.

H0 will be rejected ifX < 89.3284.

(c) This is not an appropriate rejection region. The rejection region should consist of values forX that will maketheP-value of the test less than or equal to a chosen threshold level. Therefore the rejection region must be ofthe formX ≤ x0. This rejection region is of the formX ≥ x0, and so it consists of values for which theP-valuewill be greater than some level.

(d) This is an appropriate rejection region.

UnderH0, thez-score of 89.4 is(89.4−90)/0.408248= −1.47.

Since the alternate hypothesis is of the formµ< µ0, the level is the area to the left ofz= −1.47.

Therefore the level isα = 0.0708.


SECTION 6.13 315

(e) This is not an appropriate rejection region. The rejection region should consist of values forX that will maketheP-value of the test less than a chosen threshold level. This rejection region contains values ofX greater than90.6, for which theP-value will be large.

4. (a) Type I error.H0 is true and was rejected.

(b) Correct decision.H0 is true and was not rejected.

(c) Type II error.H0 is false and was not rejected.

(d) Correct decision.H0 is false and was rejected.

5. (a)H0 says that the claim is true. ThereforeH0 is true, so rejecting it is a type I error.

(b) Correct decision.H0 is false and was rejected.

(c) Correct decision.H0 is true and was not rejected.

(d) Type II error.H0 is false and was not rejected.

6. The maximum probability of rejectingH0 when true is the levelα = 0.10.

7. The 1% level. The error in question is rejectingH0 when true, which is a type I error. When the level is smaller,it is less likely to rejectH0, and thus less likely to make a type I error.

Section 6.13

1. (a) True. This is the definition of power.

(b) True. WhenH0 is false, making a correct decision means rejectingH0.

(c) False. The power is 0.90, not 0.10.


316 CHAPTER 6

(d) False. The power is not the probability thatH0 is true.

2. (a) True. This is the definition of power.

(b) False. The power is the probability of not making a type IIerror, not the probability of making a type I error.

(c) False. The power does not specify the probability of a type I error.

(d) False. The power is the probability ofnot making a type II error, not the probability of making a type II error.

(e) True. The probability of making a type II error is equal to1 minus the power.

(f) False. The power is not the probability thatH0 is false.

3. increase. If the level increases, the probability of rejecting H0 increases, so in particular, the probability ofrejectingH0 when it is false increases.

4. increase. If the sample size increases, the probability of rejectingH0 when it is false increases.

5. (a)H0 : µ≥ 50,000 versusH1 : µ< 50,000.H1 is true, since the true value ofµ is 49,500.

(b) The level is the probability of rejectingH0 when it is true.

Under H0, X is approximately normally distributed with mean 50,000 andstandard deviationσX = 5000/

√100= 500.

The probability of rejectingH0 is P(X ≤ 49,400).

UnderH0, thez-score of 49,400 isz= (49,400−50,000)/500= −1.20.The level of the test is the area under the normal curve to the left of z= −1.20.

Therefore the level is 0.1151.

The power is the probability of rejectingH0 whenµ= 49,500.

X is approximately normally distributed with mean 49,500 andstandard deviationσX = 5000/

√100= 500.

The probability of rejectingH0 is P(X ≤ 49,400).

Thez-score of 49,400 isz= (49,400−49,500)/500= −0.20.The power of the test is thus the area under the normal curve tothe left ofz= −0.20.

Therefore the power is 0.4207.


SECTION 6.13 317

(c) Since the alternate hypothesis is of the formµ < µ0, the 5% rejection region will be the regionX ≤ x5, wherex5 is the 5th percentile of the null distribution.

Thez-score corresponding to the 5th percentile isz= −1.645.

Thereforex5 = 50,000−1.645(500)= 49,177.5.

The rejection region isX ≤ 49,177.5.

The power is thereforeP(X ≤ 49,177.5) whenµ= 49,500.

Thez-score of 49,177.5 isz= (49,177.5−49,500)/500= −0.645. We will usez= −0.65.

The power is therefore the area to the left ofz= −0.65.

Thus the power is 0.2578.

(d) For the power to be 0.80, the rejection region must beX ≤ x0 whereP(X ≤ x0) = 0.80 whenµ= 49,500.

Thereforex0 is the 80th percentile of the normal curve whenµ = 49,500. Thez-score corresponding to the80th percentile isz= 0.84.

Thereforex0 = 49,500+0.84(500)= 49,920.

Now compute the level of the test whose rejection region isX ≤ 49,920.

The level isP(X ≤ 49,920) whenµ= 50,000.

Thez-score of 49,920 isz= (49,920−50,000)/500= −0.16.

The level is the area under the normal curve to the left ofz= −0.16.

Therefore the level is 0.4364.

(e) Letn be the required number of tires.

The null distribution is normal withµ= 50,000 andσX = 5000/√

n. The alternate distribution is normal withµ= 49,500 andσX = 5000/

√n.

Let x0 denote the boundary of the rejection region.

Since the level is 5%, thez-score ofx0 is z= −1.645 under the null distribution.

Thereforex0 = 50,000−1.645(5000/√

n).

Since the power is 0.80, thez-score ofx0 is z= 0.84 under the alternate distribution.

Thereforex0 = 49,500+0.84(5000/√

n).

It follows that 50,000−1.645(5000/√

n) = 49,500+0.84(5000/√

n).


6. (a) The null distribution ofX is normal with meanµ= 12 and standard deviationσX = 2/√

100= 0.2.

The level isP(X ≤ 11.7), computed under the null distribution.

Thez-score of 11.7 isz= (11.7−12)/0.2= −1.50.

The level is the area to the left ofz= −1.50.

Thus the level is 0.0668.


318 CHAPTER 6

The alternate distribution ofX is normal with meanµ= 11.5 and standard deviationσX = 2/

√100= 0.2.

The power isP(X ≤ 11.7), computed under the alternate distribution.

Thez-score of 11.7 isz= (11.7−11.5)/0.2= 1.00.

The power is the area to the left ofz= 1.00.


(b) To make the power 0.90, the rejection region must beX ≤ x0 whereP(X ≤ x0) = 0.90 under the alternatedistribution.

Thez-score corresponding to the 90th percentile isz= 1.28.

Thereforex0 = 11.5+1.28(0.2)= 11.756.

The rejection region should beX ≤ 11.756.

The level is thenP(X ≤ 11.756) under the null distribution.

Thez-score of 11.756 isz= (11.756−12)/0.2= −1.22.



(c) To make the level 5%, the rejection region must beX ≤ x0 whereP(X ≤ x0) = 0.05 under the null distribution.


Thereforex0 = 12−1.645(0.2)= 11.671.

The rejection region should beX ≤ 11.671.

The power is thenP(X ≤ 11.671), computed under the alternate distribution.

Thez-score of 11.671 isz= (11.671−11.5)/0.2= 0.855. We will usez= 0.86.




The null distribution is normal withµ= 12 andσX = 2/√

n. The alternate distribution is normal withµ= 11.5andσX = 2/

√n.



Thereforex0 = 12−1.645(2/√

n).


Thereforex0 = 11.5+1.28(2/√

n).

It follows that 12−1.645(2/√

n) = 11.5+1.28(2/√

n).



SECTION 6.13 319

7. (ii). Since 7 is farther from the null mean of 10 than 8 is, the power against the alternativeµ = 7 will begreater than the power against the alternativeµ= 8.

8. (a) The null distribution ofX is Bin(250,0.10). Thus the null distribution is approximately normal with meanµ= 250(0.10) = 25 and standard deviationσ =

√250(0.1)(1−0.1)= 4.743416.

The level isP(X ≤ 18), computed under the null distribution.

To findP(X ≤ 18), use the continuity correction, find thez-score of 18.5.

Thez-score of 18.5 isz= (18.5−25)/4.743416= −1.37.



(b) The alternate distribution ofX is Bin(250,0.06). Thus the alternate distribution is approximately normal withmeanµ= 250(0.06) = 15 and standard deviationσ =

√250(0.06)(1−0.06)= 3.75500.

The power isP(X ≤ 18), computed under the alternate distribution.

To findP(X ≤ 18), use the continuity correction, find thez-score of 18.5.

Thez-score of 18.5 isz= (18.5−15)/3.75500= 0.93.



(c) No. To compute the level, use the standard deviation assumed by the null hypothesis, which is√

250(0.1)(0.9)=4.7434. To compute the power, use the standard deviation assumed by the assumed alternate hypothesis, whichis√

250(0.06)(0.94)= 3.7550.


The null distribution is approximately normal withµ= 0.1n andσ =√

n(0.1)(0.9). The alternate distributionis normal withµ= 0.06n andσ =

√n(0.06)(0.94). Let x0 denote the boundary of the rejection region.


Thereforex0 = 0.1n−1.645√

n(0.1)(0.9).


Thereforex0 = 0.06n+1.28√

n(0.06)(0.94).

It follows that 0.1n−1.645√

n(0.1)(0.9) = 0.06n+1.28√

n(0.06)(0.94).

Dividing by√

n yields the equation 0.1√

n−1.645√

(0.1)(0.9) = 0.06√

n+1.28√

(0.06)(0.94).


9. (a) Two-tailed. The alternate hypothesis is of the formp 6= p0.


320 CHAPTER 6

(b) p = 0.5

(c) p = 0.4

(d) Less than 0.7. The power for a sample size of 150 is 0.691332, and the power for a smaller sample size of 100would be less than this.

(e) Greater than 0.6. The power for a sample size of 150 is 0.691332, and the power for a larger sample size of200 would be greater than this.

(f) Greater than 0.65. The power against the alternativep = 0.4 is 0.691332, and the alternativep = 0.3 is fartherfrom the null thanp = 0.4. So the power against the alternativep = 0.3 is greater than 0.691332.

(g) It’s impossible to tell from the output. The power against the alternativep = 0.45 will be less than the poweragainstp = 0.4, which is 0.691332. But we cannot tell without calculatingwhether the power will be less than0.65.

10. (a) One-tailed. The alternate hypothesis is of the formµ> µ0.

(b) 4. The alternate mean is equal to the null plus the difference, and the difference is 1, so, if the null mean is 3,the alternate is 4.

(c) Greater than 0.85. The power is 0.857299 when the sample size is 18, so it will be greater than this when thesample size is 25.

(d) Less than 0.90. The power against a difference of 1 is 0.857299. The power against a smaller difference of 0.5will be less than this.

(e) Less than 0.85. The sample size of 18 is the smallest that will produce power greater than or equal to the targetpower of 0.85.

11. (a) Two-tailed. The alternate hypothesis is of the formµ1−µ2 6= ∆.

(b) Less than 0.9. The sample size of 60 is the smallest that will produce power greater than or equal to the targetpower of 0.9.

(c) Greater than 0.9. The power is greater than 0.9 against a difference of 3, so it will be greater than 0.9 againstany difference greater than 3.


SECTION 6.14 321

Section 6.14

1. (a) There are six tests, so the Bonferroni-adjustedP-values are found by multiplying the originalP-values by 6.For the setting whose originalP-value is 0.002, the Bonferroni-adjustedP-value is therefore 0.012. Since thisvalue is small, we can conclude that this setting reduces theproportion of defective parts.

(b) The Bonferroni-adjustedP-value is 6(0.03) = 0.18. Since this value is not so small, we cannot conclude thatthis setting reduces the proportion of defective parts.

2. (iii). The Bonferroni-adjustedP-value is 5(0.03) = 0.15, which is not small enough to conclude that thisprocess is better than the current process. However, it might be. The best thing to do is to run it again, andsee if itsP-value remains small. It is of less interest to rerun processes whoseP-values were big the first timearound, since there is no evidence that any of them are betterthan the current process.

3. The originalP-value must be 0.05/20= 0.0025.

4. (a) For additive A:X = 12.938,s= 1.1168,n = 10. t = (12.938−12)/(1.1168/√

10) = 2.655.

TheP value is the area to the left oft = 2.655.


From thet table, the area to the right is between 0.01 and 0.025.

Therefore theP-value is between 0.975 and 0.99. A computer package givesP = 0.987.

For additive B:X = 10.890,s= 1.2395,n = 10. t = (10.890−12)/(1.2395/√

10) = −2.832.

TheP value is the area to the left oft = −2.832.



For additive C:X = 11.721,s= 2.3433,n = 10. t = (11.721−12)/(2.3433/√

10) = −0.3768.




For additive D:X = 10.473,s= 1.3996,n = 10. t = (10.473−12)/(1.3996/√

10) = −3.4503.



322 CHAPTER 6



For additive E:X = 11.524,s= 1.0892,n = 10. t = (11.524−12)/(1.0892/√

10) = −1.3822.




(b) (i) At least one of the new additives results in an improvement. The Bonferroni-adjustedP-value for method Dis 0.018. This is obtained by multiplying the unadjustedP-value, which is 0.0036, by 5. Since the Bonferroni-adjustedP-value is small, we may conclude that additive D results in animprovement.

Alternatively, thet table shows thatP< 0.005. Therefore the Bonferroni-adjustedP-value is less than 5(0.005)=0.025. Again the adjustedP-value is small, so we may conclude that additive D results inan improvement.

5. (a) No. LetX represent the number of times in 200 days thatH0 is rejected. If the mean burn-out amperage isequal to 15 A every day, the probability of rejectingH0 is 0.05 each day, soX ∼ Bin(200,0.05).

The probability of rejectingH0 10 or more times in 200 days is thenP(X ≥ 10), which is approximately equalto 0.5636. So it would not be unusual to rejectH0 10 or more times in 200 trials ifH0 is always true.

Alternatively, note that if the probability of rejectingH0 is 0.05 each day, the mean number of times thatH0

will be rejected in 200 days is(200)(0.05) = 10. Therefore observing 10 rejections in 200 days is consistentwith the hypothesis that the mean burn-out amperage is equalto 15 A every day.

(b) Yes. LetX represent the number of times in 200 days thatH0 is rejected. If the mean burn-out amperage isequal to 15 A every day, the probability of rejectingH0 is 0.05 each day, soX ∼ Bin(200,0.05).

The probability of rejectingH0 20 or more times in 200 days is thenP(X ≥ 20) which is approximately equalto 0.0010.

So it would be quite unusual to rejectH0 20 times in 200 trials ifH0 is always true.

We can conclude that the mean burn-out amperage differed from 15 A on at least some of the days.

Section 6.15

1. (a)X = 22.10,σX = 0.34,Y = 14.30,σY = 0.32,V =√

X2 +Y2 = 26.323.∂V∂X

= X/√

X2 +Y2 = 0.83957,∂V∂Y

= Y/√

X2+Y2 = 0.54325,


SECTION 6.15 323

σV =

√(∂V∂X

)2

σ2X +

(∂V∂Y

)2

σ2Y = 0.3342.

V = 26.323,σV = 0.3342

(b) V = 26.323,σV = 0.3342. The null and alternate hypotheses areH0 : µV ≤ 25 versusH1 : µV > 25.

z= (26.323−25)/0.3342= 3.96.

Since the alternate hypothesis is of the formµ> µ0, theP-value is the area to the right ofz= 3.96.

ThusP≈ 0.

(c) Yes,V is approximately normally distributed.

2. (a) p1 = 0.42,σp1 = 0.049, p2 = 0.23,σp2 = 0.043, p = p1p2 = 0.0966.

∂p∂p1

= p2 = 0.23,∂V∂p2

= p1 = 0.42, σp =

√(∂p∂p1

)2

σ2p1

+

(∂p∂p2

)2

σ2p2

= 0.0213.

p = 0.0966,σp = 0.0213.

(b) p = 0.0966,σp = 0.0213. The null and alternate hypotheses areH0 : p≥ 25 versusH1 : p < 25.

z= (0.0966−0.10)/0.0213= −0.16.

Since the alternate hypothesis is of the formµ> µ0, theP-value is the area to the left ofz= −0.16.

ThusP = 0.4364.

(c) The distribution ofp differs somewhat from normal.



The confidence interval is (38.3818, 38.53865). Values outside this interval can be rejected at the 5% level.

Therefore null hypotheses (ii) and (iv) can be rejected.

(b) Using Method 1, the limits of the 90% confidence interval are the 5th and 95th percentiles of the bootstrapmeans,X.05 andX.95.

The 5th percentile is(Y50+Y51)/2 = 38.39135, and the 95th percentile is(Y950+Y951)/2 = 38.5227.

The confidence interval is (38.39135, 38.5227). Values outside this interval can be rejected at the 10% level.

Therefore null hypotheses (i), (ii), and (iv) can be rejected.

4. (a) Answers may vary.

(b) Answers may vary.


324 CHAPTER 6

(c) No, because the simulated samples differ somewhat each time the bootstrap is performed.

5. No, the value 103 is an outlier.

6. (a) There are 16 outcomes in both groups together. UnderH0, any group of eight is equally likely to comprise the

values for regular gasoline. Therefore the total number of possible outcomes is

(168

)= 12,870.

(b) R= 25.8500,P = 31.1375.

(c) P≈ 0.013.

(d) R= 25.85,sR = 5.7249,nR = 8, S= 31.1375,sS = 3.0156,nS = 8.


ν =

[5.72492

8+

3.01562

8

]2

(5.72492/8)2

8−1+

(3.01562/8)2

8−1


t10 = (25.85−31.1375)/√

5.72492/8+3.01562/8 = −2.3113.

The null and alternate hypotheses areH0 : µR−µS≥ 0 versusH1 : µR−µS< 0.

Since the alternate hypothesis is of the formµR−µS< ∆, theP-value is the area to the left oft = −2.3113.


This result is not reliable, because the data contain outliers.

7. (a)s2A = 200.28,s2

B = 39.833,s2A/s2

B = 5.02.

(b) No, theF-test requires the assumption that the data are normally distributed. These data contain an outlier(103), so theF-test should not be used.

(c) P≈ 0.37

8. P≈ 0.19



9. (a) The test statistic ist =X−7

s/√

7. H0 will be rejected if|t| > 2.447.

(b) The power is approximately 0.60.

10. (a) The power of the two-sample test is approximately 0.46, and the power of the paired test is approximately 0.41.The two-sample test has slightly greater power.

(b) The power of the two-sample test is approximately 0.46, and the power of the paired test is approximately 0.97.The paired test has much greater power.


1. This requires a test for the difference between two means.The data are unpaired. Letµ1 represent the pop-ulation mean annual cost for cars using regular fuel, and letµ2 represent the population mean annual costfor cars using premium fuel. Then the appropriate null and alternate hypotheses areH0 : µ1− µ2 ≥ 0 versusH1 : µ1− µ2 < 0. The test statistic is the difference between the sample mean costs between the two groups.Thez table should be used to find theP-value.

2. The data are paired, so this requires a paired test for the mean difference. Letµ1 represent the population meanpulse recovery rate when using the old breathing style, and letµ2 represent the population mean pulse recoveryrate when using the new breathing style. The appropriate null and alternate hypotheses areH0 : µ1− µ2 ≤ 0versusH1 : µ1−µ2 > 0. For each swimmer, letDi denote the difference in recovery rates between the old andnew styles, and letD ands denote the sample mean and standard deviation, respectively, of these differences.The test statistic isD/(s/

√15). Thet table should be used to find theP-value. There are 14 degrees of freedom.

3. This requires a test for a population proportion. Letp represent the population proportion of defective partsunder the new program. The appropriate null and alternate hypotheses areH0 : p≥ 0.10 versusH1 : p < 0.10.The test statistic is the sample proportion of defective parts. Thez table should be used to find theP-value.

4. This requires anF test. Letσ21 represent the population variance for the crushing strength of the old material,

and letσ22 represent the population variance for the crushing strength of the new material. Then the appropriate

null and alternate hypotheses areH0 : σ21/σ2

2 ≤ 1 versus H1 : σ21/σ2

2 > 1. The test statistic is the ratios21/s2

2of the two sample variances. TheF table should be used to find theP-value. There are 15 and 19 degrees offreedom.


326 CHAPTER 6

5. (a)H0 : µ≥ 16 versusH1 : µ< 16



t = (15.887−16)/(0.13047/√

10) = −2.739.

(c) Since the alternate hypothesis is of the formµ< µ0, theP-value is the area to the left oft = −2.739.


We conclude that the mean fill weight is less than 16 oz.

6. (a)H0 : p = 0.20 versusH1 : p 6= 0.20.

(b) X = 192,n = 1280,p = 192/1280= 0.15.

z= (0.15−0.20)/√

0.20(1−0.20)/1280= −4.47.

(c) Since the alternate hypothesis is of the formp 6= p0, theP-value is the sum of the areas to the right ofz= 4.47and to the left ofz= −4.47.

ThusP≈ 0+0= 0.

We conclude that choices for the SAT are not generated at random.

7. (a)H0 : µ1−µ2 = 0 versusH1 : µ1−µ2 6= 0

(b) D = 1608.143,sD = 2008.147,n = 7. There are 7−1= 6 degrees of freedom. The null and alternate hypothe-ses areH0 : µD = 0 versusH1 : µD 6= 0.

t = (1608.143−0)/(2008.147/√

7) = 2.119.

(c) Since the alternate hypothesis is of the formµD 6= µ0, theP-value is the sum of the areas to the right oft = 2.119and to the left oft = −2.119.


The null hypothesis is suspect, but one would most likely notfirmly conclude that it is false.

8. s1 = 3.06,s2 = 1.41. The sample sizes aren1 = 21 andn2 = 21.


2 ≤ 1 versusH1 : σ21/σ2

2 > 1.


1 = 3.062/1.412 = 4.7098.

The numbers of degrees of freedom are 20 and 20.




From theF table,P < 0.001. A computer package givesP = 0.00053.

We can conclude that the measurements made on the base metal are more variable than those made on the weldmetal.

9. X = 5.6, s= 1.2, n = 85. The null and alternate hypotheses areH0 : µ≤ 5 versusH1 : µ> 5.

z= (5.6−5)/(1.2/√

85) = 4.61.

Since the alternate hypothesis is of the formµ> µ0, theP-value is the area to the right ofz= 4.61.

ThusP≈ 0.



t = (0.17−0)/(1.0122/√

10) = 0.5311.

Since the alternate hypothesis is of the formµD 6= µ0, the P-value is the sum of the areas to the right oft = 0.5311 and to the left oft = −0.5311.


We cannot conclude that there is a difference in the mean noise levels between acceleration and decelerationlanes.

11. (a) The null distribution ofX is normal with meanµ= 100 and standard deviationσX = 0.1/√

100 = 0.01.

Since the alternate hypothesis is of the formµ 6= µ0, the rejection region will consist of both the upper andlower 2.5% of the null distribution.

Thez-scores corresponding to the boundaries of upper and lower 2.5% arez= 1.96 andz=−1.96, respectively.

Therefore the boundaries are 100+1.96(0.01)= 100.0196 and 100−1.96(0.01)= 99.9804.

RejectH0 if X ≥ 100.0196 or ifX ≤ 99.9804.

(b) The null distribution ofX is normal with meanµ= 100 and standard deviationσX = 0.1/√

100 = 0.01.

Since the alternate hypothesis is of the formµ 6= µ0, the rejection region will consist of both the upper andlower 5% of the null distribution.

Thez-scores corresponding to the boundaries of upper and lower 5% arez= 1.645 andz= −1.645, respec-tively.


RejectH0 if X ≥ 100.01645 or ifX ≤ 99.98355.

(c) Yes


328 CHAPTER 6

(d) No

(e) Since this is a two-tailed test, there are two critical points, equidistant from the null mean of 100. Since onecritical point is 100.015, the other is 99.985.

The level of the test is the sumP(X ≤ 99.985) + P(X ≥ 100.015), computed under the null distribution.

The null distribution is normal with meanµ= 100 and standard deviationσX = 0.01.

Thez-score of 100.015 is(100.015−100)/0.01= 1.5. Thez-score of 99.985 is(99.985−100)/0.01= −1.5.

The level of the test is therefore 0.0668+0.0668= 0.1336, or 13.36%.

12. (a) First find the rejection region.

The null distribution ofX is normal with meanµ= 4 and standard deviationσX = 0.20/√

100 = 0.02.

Since the level is 5%, and the alternate hypothesis is of the formµ> µ0, the rejection region must be the upper5% of the null distribution. Thez-score of the boundary of the upper 5% of the normal distribution isz= 1.645.

The boundary is therefore 4+1.645(0.02)= 4.0329.

The rejection region is thereforeX ≥ 4.0329.

The power isP(X ≥ 4.0329), computed under the alternate distribution.

The alternate distribution is normal with meanµ= 4.04 and standard deviationσX = 0.02.

Thez-score of 4.0329 is(4.0329−4.04)/0.02= −0.355. We will usez= −0.36.

The power is the area to the right ofz= −0.36, which is 0.6406.


The null distribution is normal withµ= 4 andσX = 0.2/√

n. The alternate distribution is normal withµ= 4.04andσX = 0.2/

√n.


Since the level is 5%, thez-score ofx0 is z= 1.645 under the null distribution.

Thereforex0 = 4+1.645(0.2/√

n).

Since the power is 0.95, thez-score ofx0 is z= −1.645 under the alternate distribution.

Thereforex0 = 4.04−1.645(0.2/√

n).

It follows that 4+1.645(0.2/√

n) = 4.04−1.645(0.2/√

n).


(c) Letx0 be the boundary of the rejection region.

Since the power is 0.90,P(X ≥ x0) = 0.90, computed under the alternate distribution.

Thereforex0 is the 10th percentile of the alternate distribution.

The alternate distribution is normal withµ= 4.04 andσX = 0.2/√

100= 0.02.


Thereforex0 = 4.04−1.28(0.02)= 4.0144.

The level isP(X ≥ 4.0144), computed under the null distribution.



Thez-score of 4.0144 is(4.0144−4)/0.02= 0.72.

The level is the area to the right ofz= 0.72, which is 0.2358, or 23.58%.

(d) The power isP(X ≥ 4.02), computed under the alternate hypothesis.


Thez-score of 4.02 is(4.02−4.04)/0.02= −1.00.


13. (a) The null hypothesis specifies a single value for the mean: µ = 3. The level, which is 5%, is therefore theprobability that the null hypothesis will be rejected whenµ= 3. The machine is shut down ifH0 is rejected atthe 5% level. Therefore the probability that the machine will be shut down whenµ= 3 is 0.05.

(b) First find the rejection region.

The null distribution ofX is normal with meanµ= 3 and standard deviationσX = 0.10/√

50= 0.014142.

Since the alternate hypothesis is of the formµ 6= µ0, the rejection region will consist of both the upper andlower 2.5% of the null distribution.

The z-scores corresponding to the boundaries of upper and lower 2.5% arez= 1.96 andz= −1.96, respec-tively.


H0 will be rejected ifX ≥ 3.0277 or ifX ≤ 2.9723.

The probability that the equipment will be recalibrated is therefore equal toP(X ≥ 3.0277) +P(X ≤ 2.9723), computed under the assumption thatµ= 3.01.

Thez-score of 3.0277 is(3.0277−3.01)/0.014142= 1.25.

Thez-score of 2.9723 is(2.9723−3.01)/0.014142= −2.67.

ThereforeP(X ≥ 3.0277) = 0.1056, andP(X ≤ 2.9723) = 0.0038.

The probability that the equipment will be recalibrated is equal to 0.1056 + 0.0038 = 0.1094.

14. The row totals areO1. = 1811 andO2. = 200. The column totals areO.1 = 539, O.2 = 526, O.3 = 495,O.4 = 451. The grand total isO.. = 2011.


Line

1 2 3 4Pass 485.395 473.688 445.771 406.147Fail 53.605 52.312 49.229 44.853


χ2 = ∑2i=1∑4

j=1(Oi j −Ei j )2/Ei j = 4.676.


330 CHAPTER 6


We cannot conclude that the failure rates differ.

15. X = 37,n = 37+458= 495, p = 37/495= 0.074747.


z= (0.074747−0.10)/√

0.10(1−0.10)/495= −1.87.

Since the alternate hypothesis is of the formp < p0, the P-value is the area to the left ofz = −1.87, soP = 0.0307.

Since there are four samples altogether, the Bonferroni-adjustedP-value is 4(0.0307) = 0.1228.

We cannot conclude that the failure rate on line 3 is less than0.10.

16. (a) Both samples have a mean of 131 and a variance of 107,210.

(b) The ranks of the combined samples are Value Rank Sample−738 1 Y

0 2 X2 3 X3 4 X4 5 X

10 6 X20 7 X40 8 X

100 9 X162 10 Y222 11 Y242 12 Y252 13 Y258 14 Y259 15 Y260 16 Y262 17 Y

1000 18 X





z=W−m(m+n+1)/2√

mn(m+n+1)/12= −2.08.



Since the alternate hypothesis is of the formµX −µY 6= ∆, theP-value is the sum of the areas to the right ofz= 2.08 and to the left ofz= −2.08.

From thez table,P = 0.0188+0.0188= 0.0376.

It seems reasonable to conclude that the population means are unequal.

(c) No, theX sample is skewed to the right, while theY sample is skewed to the left. It does not seem reasonableto assume that these samples came from populations of the same shape.

17. (a) Both samples have a median of 20.

(b) The ranks of the combined samples are Value Rank Sample−10 1 Y−9 2 Y−8 3 Y−7 4 Y−6 5 Y−5 6 Y−4 7 Y

1 8 X2 9 X3 10 X4 11 X5 12 X6 13 X7 14 X

20 15.5 X20 15.5 Y21 17 Y22 18 Y23 19 Y24 20 Y25 21 Y26 22 Y27 23 Y40 24 X50 25 X60 26 X70 27 X80 28 X90 29 X

100 30 X

The null and alternate hypotheses areH0 : median ofX − median ofY = 0 versusH1 : median ofX − median ofY 6= 0.


W = 281.5. The sample sizes arem= 15 andn = 15.


332 CHAPTER 6


z=W−m(m+n+1)/2√

mn(m+n+1)/12= 2.03.

Since the alternate hypothesis is of the form median ofX−median ofY 6= ∆, theP-value is the sum of the areasto the right ofz= 2.03 and to the left ofz= −2.03.

From thez table,P = 0.0212+0.0212= 0.0424.

The P-value is fairly small, and so it provides reasonably strongevidence that the population medians aredifferent.

(c) No, theX sample is heavily skewed to the right, while theY sample is strongly bimodal. It does not seemreasonable to assume that these samples came from populations of the same shape.

18. (a)X = 15.95,sX = 0.20976,nX = 16, Y = 16.00,sY = 0.18974,nY = 16.


ν =

[0.209762

16+

0.189742

16

]2

(0.209762/16)2

16−1+

(0.189742/16)2

16−1


t29 = (15.95−16.00−0)/√

0.209762/16+0.189742/16= −0.7071.




SinceP > 0.05, we cannot conclude at the 5% level that the means are different.

(b) The sample variance for the old process isσ21 = 0.044. The sample variance for the new process isσ2

2 = 0.036.The sample sizes aren1 = n2 = 16.


2 ≤ 1 versusH1 : σ21/σ2

2 > 1.





SinceP > 0.05, we cannot conclude at the 5% level that the variance of thenew process is less than that of theold process.

19. (a) LetµA be the mean thrust/weight ratio for Fuel A, and letµB be the mean thrust/weight ratio for Fuel B. Theappropriate null and alternate hypotheses areH0 : µA−µB ≤ 0 versusH1 : µA−µB > 0.

(b) A = 54.919,sA = 2.5522,nA = 16, B = 53.019,sB = 2.7294,nB = 16.




ν =

[2.55222

16+

2.72942

16

]2

(2.55222/16)2

16−1+

(2.72942/16)2

16−1


t29 = (54.919−53.019−0)/√

2.55222/16+2.72942/16= 2.0339.

The null and alternate hypotheses areH0 : µA−µB ≤ 0 versusH1 : µA−µB > 0.

Since the alternate hypothesis is of the formµA−µB > ∆, theP-value is the area to the right oft = 2.0339.


SinceP≤ 0.05, we can conclude at the 5% level that the means are different.

20. (a)X = 1.59,s= 0.25,n = 80. The null and alternate hypotheses areH0 : µ≤ 1.50 versusH1 : µ> 1.50.

z= (1.59−1.50)/(0.25/√

80) = 3.22. Since the alternate hypothesis is of the formµ> µ0, theP-value is thearea to the right ofz= 3.22.

ThusP = 0.0006. SinceP≤ 0.05,H0 is rejected.

(b) First find the rejection region.

The null distribution ofX is normal with meanµ= 1.5 and standard deviationσX = 0.33/√

80= 0.036895.

Since the level is 5%, and the alternate hypothesis is of the formµ> µ0, the rejection region must be the upper5% of the null distribution. Thez-score of the boundary of the upper 5% of the normal distribution isz= 1.645.

The boundary is therefore 1.5+1.645(0.036895)= 1.5607.

The rejection region is thereforeX ≥ 1.5607.

The power isP(X ≥ 1.5607), computed under the alternate distribution.


Thez-score of 1.5607 is(1.5607−1.6)/0.036895= −1.07.



The null distribution ofX is normal with meanµ= 1.5 and standard deviationσX = 0.33/√

n.

The alternate distribution is normal with meanµ= 1.6 and standard deviationσX = 0.33/√

n.


Since the level is 5%, thez-score ofx0 is z= 1.645 under the null distribution.

Thereforex0 = 1.5+1.645(0.33/√

n).

Since the power is 0.99, thez-score ofx0 is z= −2.33 under the alternate distribution.

Thereforex0 = 1.6−2.33(0.33/√

n).

It follows that 1.5+1.645(0.33/√

n) = 1.6−2.33(0.33/√

n). Solving forn yieldsn = 173.


334 CHAPTER 6

21. (a) Yes.

(b) The conclusion is not justified. The engineer is concluding thatH0 is true because the test failed to reject.

22. X = 347,n = 510, p = 347/510= 0.68039.


z= (0.68039−0.48)/√

0.48(1−0.48)/510= 9.06.


soP≈ 0.

We can conclude that the survey method tends to oversample males.

23. The row totals areO1. = 214 andO2. = 216. The column totals areO.1 = 65,O.2 = 121,O.3 = 244. The grandtotal isO.. = 430.


Numbers of SkeletonsSite 0–4 years 5–19 years 20 years or more

Casa da Moura 32.349 60.219 121.43Wandersleben 32.651 60.781 122.57


χ2 = ∑2i=1∑3

j=1(Oi j −Ei j )2/Ei j = 2.1228.


We cannot conclude that the age distributions differ between the two sites.

24. The row totals areO1. = 127,O2. = 123,O3. = 135. The column totals areO.1 = 102,O.2 = 39, O.3 = 48,O.4 = 88,O.5 = 57,O.6 = 51. The grand total isO.. = 385.


Years of EducationState 0 1–4 5–6 7–9 10–11 12 or moreHaryana 33.647 12.865 15.834 29.029 18.803 16.823Bihar 32.587 12.460 15.335 28.114 18.210 16.294Uttar Pradesh 35.766 13.675 16.831 30.857 19.987 17.883


χ2 = ∑3i=1∑6

j=1(Oi j −Ei j )2/Ei j = 63.301.

From theχ2 table,P < 0.005. A computer package givesP = 8.6×10−10.



We can conclude that the educational levels differ among thethree states.


336 CHAPTER 7

Chapter 7

Section 7.1

1. x = y = 4, ∑ni=1(xi −x)2 = 28, ∑n

i=1(yi −y)2 = 28, ∑ni=1(xi −x)(yi −y) = 23.

r =∑n

i=1(xi −x)(yi −y)√∑n

i=1(xi −x)2√

∑ni=1(yi −y)2

= 0.8214.

2. (a)y has been replaced withy+ 3, x is unchanged. This involves only adding a constant, which does not changethe correlation coefficient.

(b) x has been replaced with 10x+ 1, y has been replaced withy+ 3. This involves only adding a constant andmultiplying by a constant, neither of which changes the correlation coefficient.

(c) x has been replaced with 10y+ 33, y has been replaced with 2x+ 2. This involves only adding a constant,multiplying by a constant, and interchangingx andy, none of which changes the correlation coefficient.

3. (a) The correlation coefficient is appropriate. The points are approximately clustered around a line.

(b) The correlation coefficient is not appropriate. The relationship is curved, not linear.

(c) The correlation coefficient is not appropriate. The plotcontains outliers.

4. (a) True. This is a result of the plot being clustered around a line with positive slope.

(b) False. Since the plot is clustered around a line with negative slope, below average values of one variable willtend to be associated with above average values of the other.

(c) False. The correlation coefficient describes the shape of the plot, not its location relative to the axes.


SECTION 7.1 337

5.

The heights and weights for the men (dots) are on thewhole greater than those for the women (xs). There-fore the scatterplot for the men is shifted up and tothe right. The overall plot exhibits a higher correlationthan either plot separately. The correlation betweenheights and weights for men and women taken togetherwill be more than 0.6.

6. (a) Letx represent velocity and lety represent acceleration.

x = 1.662, y = 7.018, ∑ni=1(xi −x)2 = 2.23928, ∑n

i=1(yi −y)2 = 6.96808,

∑ni=1(xi −x)(yi −y) = −3.17098.

r =∑n


i=1(xi −x)2√

∑ni=1(yi −y)2

= −0.8028.

(b)

0.5 1 1.5 2 2.5 3 3.54

4.5

5

5.5

6

6.5

7

7.5

8

8.5

Velocity

Acce

lera

tio

n

(c) No, the point (2.92, 5.00) is an outlier.

(d) No effect. Converting units from meters to centimeters and from seconds to minutes involves multiplying by aconstant, which does not change the correlation coefficient.

7. (a) Letx represent temperature,y represent stirring rate, andz represent yield.

Thenx = 119.875, y = 45, z= 75.590625, ∑ni=1(xi −x)2 = 1845.75,


338 CHAPTER 7

∑ni=1(yi −y)2 = 1360, ∑n

i=1(zi −z)2 = 234.349694, ∑ni=1(xi −x)(yi −y) = 1436,

∑ni=1(xi −x)(zi −z) = 481.63125, ∑n

i=1(yi −y)(zi −z) = 424.15.

The correlation coefficient between temperature and yield is

r =∑n

i=1(xi −x)(zi −z)√∑n

i=1(xi −x)2√

∑ni=1(zi −z)2

= 0.7323.

The correlation coefficient between stirring rate and yieldis

r =∑n

i=1(yi −y)(zi −z)√∑n

i=1(yi −y)2√

∑ni=1(zi −z)2

= 0.7513.

The correlation coefficient between temperature and stirring rate is

r =∑n


i=1(xi −x)2√

∑ni=1(yi −y)2

= 0.9064.

(b) No, the result might be due to confounding, since the correlation between temperature and stirring rate is farfrom 0.

(c) No, the result might be due to confounding, since the correlation between temperature and stirring rate is farfrom 0.

8. (a) Letx represent temperature,y represent stirring rate, andz represent yield.

Thenx = 126.5, y = 45, z= 75.59, ∑ni=1(xi −x)2 = 2420, ∑n

i=1(yi −y)2 = 2000,

∑ni=1(zi −z)2 = 225.986, ∑n

i=1(xi −x)(yi −y) = 0, ∑ni=1(xi −x)(zi −z) = 0.66,

∑ni=1(yi −y)(zi −z) = 518.1.

The correlation coefficient between temperature and yield is

r =∑n

i=1(xi −x)(zi −z)√∑n

i=1(xi −x)2√

∑ni=1(zi −z)2

= 0.0009.

The correlation coefficient between stirring rate and yieldis

r =∑n

i=1(yi −y)(zi −z)√∑n

i=1(yi −y)2√

∑ni=1(zi −z)2

= 0.7707.

The correlation coefficient between temperature and stirring rate is

r =∑n


i=1(xi −x)2√

∑ni=1(yi −y)2

= 0.

(b) Yes, the correlation between temperature and yield is nearly 0, and there is no confounding of temperature withstirring rate since their correlation is 0.


SECTION 7.1 339

(c) Yes, the correlation between stirring rate and yield is fairly large and positive, and there is no confounding oftemperature with stirring rate since their correlation is 0.

(d) This design is better because temperature and stirring rate are not confounded.

9. (a)x = 105.63, y = 38.84, ∑ni=1(xi −x)2 = 2.821, ∑n

i=1(yi −y)2 = 47.144,

∑ni=1(xi −x)(yi −y) = −5.902, n = 10.

r =∑n


i=1(xi −x)2√

∑ni=1(yi −y)2

= −0.51178.

W =12

ln1+ r1− r

= −0.56514,σW =√

1/(10−3) = 0.377964.

A 95% confidence interval forµW is W±1.96σW, or (−1.30595,0.17567).

A 95% confidence interval forρ is

(e2(−1.30595)−1

e2(−1.30595) +1,

e2(0.17567)−1

e2(0.17567) +1

), or (−0.8632, 0.1739).

(b) The null and alternate hypotheses areH0 : ρ ≥ 0.3 versusH1 : ρ < 0.3.

r = −0.51178, W =12

ln1+ r1− r

= −0.56514, σW =√

1/(10−3) = 0.377964.

UnderH0, takeρ = 0.3, soµW =12

ln1+0.31−0.3

= 0.30952.

The null distribution ofW is therefore normal with mean 0.30952 and standard deviation 0.377964.

Thez-score of−0.56514 is(−0.56514−0.30952)/0.377964= −2.31. Since the alternate hypothesis is of theform ρ < ρ0, theP-value is the area to the left ofz= −2.31.

ThusP = 0.0104.

We conclude thatρ < 0.3.

(c) r = −0.51178,n = 10,U = r√

n−2/√

1− r2 = −1.6849.

UnderH0, U has a Student’st distribution with 10−2= 8 degrees of freedom.

Since the alternate hypothesis is of the formρ 6= ρ0, theP-value is the sum of the areas to the right oft = 1.6849and to the left oft = −1.6849.


It is plausible thatρ = 0.

10. (a)r = 0.10,n = 400,U = r√

n−2/√

1− r2 = 2.00504.

UnderH0, U has a Student’st distribution with 400−2= 398 degrees of freedom. Since the number of degreesof freedom is large, use thez table.


340 CHAPTER 7

Since the alternate hypothesis isH1 : ρ > 0, theP-value is the area to the right ofz= 2.01.

ThusP = 0.0222.

We conclude thatρ > 0.

(b) No, we can only conclude thatρ > 0. We cannot conclude thatρ is large.

11. r = −0.509,n = 23,U = r√

n−2/√

1− r2 = −2.7098.

UnderH0, U has a Student’st distribution with 23−2= 21 degrees of freedom.

Since the alternate hypothesis isH1 : ρ 6= 0, theP-value is the sum of the areas to the right oft = 2.7098 and tothe left oft = −2.7098.


We conclude thatρ 6= 0.

12. x = 0 and∑ni=1(xi −x)2 = 10.

y = [−2+(−1)+0+1+y]/5= (y−2)/5, soy = 5y+2.

Express∑ni=1(xi −x)(yi −y) , ∑n

i=1(yi −y)2 , andr in terms ofy:

∑ni=1(xi −x)(yi −y) = −2(−2−y)+ (−1)(−1−y)+0(0−y)+1(1−y)+2(5y+2−y) = 10y+10.

∑ni=1(yi −y)2 = (−2−y)2 +(−1−y)2 +(0−y)2 +(1−y)2 +(5y+2−y)2 = 20y2 +20y+10.

Now r =10y+10√

10√

20y2 +20y+10=

y+1√y2 +2y+1

, so(1−2r2)y2 +(2−2r2)y+(1− r2) = 0.

For any given value ofr, substituter into (1−2r2)y2+(2−2r2)y+(1− r2) = 0 and solve fory, then computey = 5y+2.

(a) Substituter = 1 to obtain−y2 = 0, soy = 0, andy = 5(0)+2= 2.

(b) Substituter = 0 to obtainy2 +2y+1 = 0, soy = −1, andy = 5(−1)+2= −3.

(c) Substituter = 0.5 to obtain(1/2)y2 + (3/2)y+ (3/4) = 0. There are two roots,y = −0.633975 andy =−2.366025. The two corresponding values ofy arey = −1.169873 andy = −9.830127. Ify = −1.169873thenr = 0.5, If y = −9.830127 thenr = −0.5.

(d) From part (c),y = −9.830127.


SECTION 7.2 341

(e) For the correlation to be equal to−1, the points would have to lie on a straight line with negative slope. Thereis no value fory for which this is the case.

Section 7.2

1. (a) 245.82+1.13(65)= 319.27 pounds

(b) The difference iny predicted from a one-unit change inx is the slopeβ1 = 1.13. Therefore the change in thenumber of lbs of steam predicted from a change of 5C is 1.13(5) = 5.65 pounds.

2. (a)−196.32+2.42(102.7)= 52.21 ksi.

(b) The difference iny predicted from a one-unit change inx is the slopeβ1 = 2.42. Therefore the change in tensilestrength predicted from a change of 3 in the Brinell hardnessis 2.42(3) = 7.26 ksi.

3. r2 = 1− ∑ni=1(yi − yi)

2

∑ni=1(yi −y)2 = 1− 1450

9615= 0.8492.

4. r2 = 1− ∑ni=1(yi − yi)

2

∑ni=1(yi −y)2 = 1− 33.9

181.2= 0.8129.

5. (a)−0.2967+0.2738(70)= 18.869 in.

(b) Letx be the required height. Then 19= −0.2967+0.2738x, sox = 70.477 in.

(c) No, some of the men whose points lie below the least-squares line will have shorter arms.

6. n = 40, ∑ni=1(xi −x)2 = 98,775, ∑n

i=1(yi −y)2 = 19.10, x = 26.36, y = 0.5188,∑n

i=1(xi −x)(yi −y) = 826.94.


342 CHAPTER 7

(a) r =∑n


i=1(xi −x)2√

∑ni=1(yi −y)2

= 0.602052.

(b) The error sum of squares is∑ni=1(yi − yi)

2 = (1− r2)∑ni=1(yi −y)2 = 12.177.

The regression sum of squares is∑ni=1(yi −y)2−∑n

i=1(yi − yi)2 = 19.10−12.177= 6.923.

The total sum of squares is∑ni=1(yi −y)2 = 19.10.

(c) β1 =∑n

i=1(xi −x)(yi −y)

∑ni=1(xi −x)2 = 0.008371956 andβ0 = y− β1x = 0.298115.

The equation of the least-squares line isy = 0.298115+0.008371956x

(d) 0.298115+0.008371956(40)= 0.633 mm

(e) Letx be the required temperature. Then 0.5 = 0.298115+0.008371956x, sox = 24.1C.

(f) No. If the actual amount of warping happens to come out above the value predicted by the least-squares line, itcould exceed 0.5 mm.

7. β1 = rsy/sx = (0.85)(1.9)/1.2= 1.3458.β0 = y− β1x = 30.4−1.3458(8.1)= 19.499.

The equation of the least-squares line isy = 19.499+1.3458x.

8. (a)

40 50 60 70 80 90 100100

200

300

400

500

600

700

800

900

Life

tim

e (

ho

urs

)

Temperature (° C)

The linear model is appropriate.


SECTION 7.2 343

(b) x = 67.5, y = 492.833333, ∑ni=1(xi −x)2 = 3575.0, ∑n

i=1(yi −y)2 = 650887.667,∑n

i=1(xi −x)(yi −y) = −45085.0.

β1 =∑n


∑ni=1(xi −x)2 = −12.61119 andβ0 = y− β1x = 1344.089.

The equation of the least-squares line isy = 1344.089−12.61119x.

(c) The fitted values are found by computing, for eachi, the quantityyi = β0 + β1xi . The residuals are found bycomputing, for eachi, the quantityei = yi − yi. The fitted values and residuals are

(839.64,11.36), (776.59,−141.59), (713.53,50.47), (650.47,57.53), (587.42,−118.42), (524.36,136.64),(461.31,124.69), (398.25,−27.25), (335.19,1.81), (272.14,−27.14), (209.08,−80.08), (146.03,11.97)

(d) By −12.61119(5) = −63.06 hours.

(e) 1344.089−12.61119(73)= 423.47 hours.

(f) No, the value 120C is outside the range of temperatures in the experiment.

(g) Letx be the required temperature. Then 500= 1344.089−12.61119x, sox = 66.93C.

9. (a)

5 10 15 20 25 304

5

6

7

8

Weight

Mile

ag

e


(b) x = 19.5, y = 5.534286, ∑ni=1(xi −x)2 = 368.125, ∑n

i=1(yi −y)2 = 9.928171,∑n

i=1(xi −x)(yi −y) = −57.1075.

β1 =∑n


∑ni=1(xi −x)2 = −0.1551307 andβ0 = y− β1x = 8.55933.



344 CHAPTER 7

(c) By 0.1551307(5) = 0.776 miles per gallon.

(d) 8.55933−0.1551307(15)= 6.23 miles per gallon.

(e) miles per gallon per ton

(f) miles per gallon

10. (a)

1 1.2 1.4 1.6 1.8 20

5

10

15

20

25

Density

Pe

rce

nt A

sh


(b) x = 1.39, y = 10.774, ∑ni=1(xi −x)2 = 0.05325, ∑n

i=1(yi −y)2 = 294.72752,

∑ni=1(xi −x)(yi −y) = 3.9272.

β1 =∑n


∑ni=1(xi −x)2 = 73.75023 andβ0 = y− β1x = −91.7388.

The equation of the least-squares line isy = −91.7388+73.75023x.

(c) 73.75023(0.1) = 7.375%

(d) −91.7388+73.75023(1.40)= 11.51%

(e) The fitted values are the valuesyi = β0 + β1xi , for each valuexi . They are shown in the following table.

Fitted Value

x y y = β0 + β1x1.25 1.93 0.449

1.325 4.63 5.9801.375 8.95 9.6681.45 15.05 15.1991.55 23.31 22.574


SECTION 7.2 345

(f) The residuals are the valuesei = yi − yi for eachi. They are shown in the following table.

Fitted Value Residual

x y y = β0 + β1x e= y− y1.25 1.93 0.449 1.481

1.325 4.63 5.980 −1.3501.375 8.95 9.668 −0.7181.45 15.05 15.199 −0.1491.55 23.31 22.574 0.736

The point whose residual has the largest magnitude is (1.25,1.93).

(g) r =∑n


i=1(xi −x)2√

∑ni=1(yi −y)2

= 0.991318.

(h) The error sum of squares is∑ni=1(yi − yi)

2 = (1− r2)∑ni=1(yi −y)2 = 5.10.

The regression sum of squares is∑ni=1(yi −y)2−∑n

i=1(yi − yi)2 = 289.63.

The total sum of squares is∑ni=1(yi −y)2 = 294.73.

(i) The quotient of the regression sum of squares divided by the total sum of squares is 289.63/294.73= 0.9827.This is the square of the correlation coefficient.

To see this, note that∑n

i=1(yi −y)2−∑ni=1(yi − yi)

2

∑ni=1(yi −y)2 = 1− ∑n

i=1(yi − yi)2

∑ni=1(yi −y)2 = 1− (1− r2) = r2.

11. (a)

4 4.5 5 5.5 67

7.5

8

8.5

9

Concentration

Dry

ing

Tim

e


(b) x = 4.9, y = 8.11, ∑ni=1(xi −x)2 = 3.3, ∑n

i=1(yi −y)2 = 2.589, ∑ni=1(xi −x)(yi −y) = −2.75.

β1 =∑n


∑ni=1(xi −x)2 = −0.833333 andβ0 = y− β1x = 12.19333.


346 CHAPTER 7


(c) The fitted values are the valuesyi = β0 + β1xi , and the residuals are the valuesei = yi − yi , for each valuexi .They are shown in the following table.

Fitted Value Residual

x y y = β0 + β1x e= y− y4.0 8.7 8.860 −0.1604.2 8.8 8.693 0.1074.4 8.3 8.527 −0.2274.6 8.7 8.360 0.3404.8 8.1 8.193 −0.0935.0 8.0 8.027 −0.0275.2 8.1 7.860 0.2405.4 7.7 7.693 0.0075.6 7.5 7.527 −0.0275.8 7.2 7.360 −0.160

(d) −0.833333(0.1)= −0.0833. Decrease by 0.0833 hours.

(e) 12.19333−0.833333(4.4)= 8.53 hours.

(f) No, because 7% is outside the range of concentrations present in the data.

(g) Letx be the required concentration. Then 8.2 = 12.19333−0.833333x, sox = 4.79%.

(h) We cannot specify such concentration. According to the least-squares line, a concentration of 7.43% wouldresult in a drying time of 6 hours. However, since 7.43% is outside the range of the data, this prediction isunreliable.

12. β1 = rsy/sx = 0.7(100/2) = 35. β0 = y− β1x = 1350−35(5)= 1175.

The equation of the least-squares line isy = 1175+35x.

13. β1 = rsy/sx = 0.5(10/0.5) = 10. β0 = y− β1x = 50−10(3) = 20.

The equation of the least-squares line isy = 20+10x.


SECTION 7.2 347

14. (a)

0 2 4 6 8 100

2

4

6

8

10

12

x

y

(b) No, becausex andy are not linearly related.

(c)

0 2 4 6 8 10−2

−1

0

1

2

3

x

z

(d) x = 5.5, z= 0.41, ∑ni=1(xi −x)2 = 82.5, ∑n

i=1(zi −z)2 = 18.2494, ∑ni=1(xi −x)(zi −z) = 38.55.

β1 =∑n

i=1(xi −x)(zi −z)

∑ni=1(xi −x)2 = 0.4672727 andβ0 = z− β1x = −2.16.

The equation of the least-squares line isz= −2.16+0.4672727x.

The predicted value ofzwhenx = 4.5 is z= −2.16+0.4672727(4.5)= −0.057273.

This is an appropriate method of prediction, becausex andzare linearly related.

(e) y = ez = e−0.057273= 0.944. Sincez is a reasonable predictor ofz, and sincey= ez, ez is a reasonable predictorof y.

15. (iii) equal to $34,900. Since 70 inches is equal tox, the predictedy value,y will be equal toy = 34,900.

To see this, note thaty = β0 + β1x = (y− β1x)+ β1x = y.


348 CHAPTER 7

16. (a) Letx represent time andy = T1 represent temperature.

x = 10, y = 64.428571, ∑ni=1(xi −x)2 = 770, ∑n

i=1(yi −y)2 = 17111.14286,

∑ni=1(xi −x)(yi −y) = 3597.

β1 =∑n


∑ni=1(xi −x)2 = 4.671428 andβ0 = y− β1x = 17.714286.


(b) Letx represent time andy = T2 represent temperature.

x = 10, y = 63.761905, ∑ni=1(xi −x)2 = 770, ∑n

i=1(yi −y)2 = 16999.80952,

∑ni=1(xi −x)(yi −y) = 3576.

β1 =∑n


∑ni=1(xi −x)2 = 4.644156 andβ0 = y− β1x = 17.320346.


(c) Letx represent time andy = T3 represent temperature.

x = 10, y = 63.857143, ∑ni=1(xi −x)2 = 770, ∑n

i=1(yi −y)2 = 18480.57143,

∑ni=1(xi −x)(yi −y) = 3710.

β1 =∑n


∑ni=1(xi −x)2 = 4.818182 andβ0 = y− β1x = 15.675325.


(d) β0= (17.714286 + 17.320346 + 15.675325)/3 = 16.9033.

β1 = (4.671428+4.644156+4.818182)/3= 4.7113.


(e) Letx represent time andy = (T1 +T2+T3)/3 represent temperature.

x = 10, y = 64.015873, ∑ni=1(xi −x)2 = 770, ∑n

i=1(yi −y)2 = 17321.21693,

∑ni=1(xi −x)(yi −y) = 3267.666667.

β1 =∑n


∑ni=1(xi −x)2 = 4.7113 andβ0 = y− β1x = 16.9033.

(f) The results in parts (d) and (e) are the same.


SECTION 7.3 349

Section 7.3

1. (a)x= 2.40, y= 12.18, ∑ni=1(xi −x)2 = 52.00, ∑n

i=1(yi −y)2 = 498.96, ∑ni=1(xi −x)(yi −y) = 160.27, n= 25.

β1 =∑n


∑ni=1(xi −x)2 = 3.082115 andβ0 = y− β1x = 4.782923.

(b) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.990000. s2 =(1− r2)∑n

i=1(yi −y)2

n−2= 0.216939.

(c) s=√

0.216939= 0.465767. sβ0= s

√1n

+x2

∑ni=1(xi −x)2 = 0.18085.

sβ1=

s√∑n

i=1(xi −x)2= 0.064590.

There aren−2 = 23 degrees of freedom.t23,.025= 2.069.

Therefore a 95% confidence interval forβ0 is 4.782923±2.069(0.18085), or (4.409, 5.157).

The 95% confidence interval forβ1 is 3.082115±2.069(0.064590), or (2.948, 3.216).

(d) β1 = 3.082115, sβ1= 0.064590, n = 25. There are 25−2= 23 degrees of freedom.

The null and alternate hypotheses areH0 : β1 ≤ 3 versusH1 : β1 > 3.

t = (3.082115−3)/0.064590= 1.271.

Since the alternate hypothesis is of the formβ1 > b, theP-value is the area to the right oft = 1.271.


We cannot conclude that the claim is false.

(e) β0 = 4.782923, sβ0= 0.18085, n = 25. There are 25−2= 23 degrees of freedom.

The null and alternate hypotheses areH0 : β0 ≥ 5.5 versusH1 : β0 < 5.5.

t = (4.782923−5.5)/0.18085= −3.965.

Since the alternate hypothesis is of the formβ0 < b, theP-value is the area to the left oft = −3.965.


We can conclude that the claim is false.

(f) x = 1.5, y = 4.782923+3.082115(1.5)= 9.4060955.

sy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 = 0.109803. There are 23 degrees of freedom.t23,.005= 2.807.

Therefore a 99% confidence interval for the mean response is 9.4060955±2.807(0.109803),or (9.098, 9.714).


350 CHAPTER 7

(g) x = 1.5, y = 4.782923+3.082115(1.5)= 9.4060955.

spred = s

√

1+1n

+(x−x)2


Therefore a 99% prediction interval is 9.4060955±2.807(0.478535), or (8.063, 10.749).

(h) The confidence interval is more useful, since it is concerned with the true length of the spring, while theprediction interval is concerned with the next measurementof the length.

2. (a)n−2= 6−2= 4

(b) β1 = 0.30714, sβ1= 0.02063,t4,.025 = 2.776, so a 95% confidence interval is 0.30714± 2.776(0.02063),

or (0.250, 0.364).

(c) β0 = 1.4381, sβ0= 0.62459, t4,.025 = 2.776, so a 95% confidence interval is 1.4381± 2.776(0.62459),

or (−0.296, 3.172).

(d) The null and alternate hypotheses areH0 : β1 = 0.40 versusH1 : β1 6= 0.40.

β1 = 0.30714,sβ1= 0.02063,t = (0.30714−0.40)/0.02063= −4.501.

Since the alternate hypothesis is of the formβ1 6= b, theP-value is the sum of the areas to the right oft = 4.501and to the left oft = −4.501.


The claim is not plausible.

(e) x = 0, The mean response isβ0 + β1(0) = β0.

The null and alternate hypotheses areH0 : β0 ≤ 1.0 versusH1 : β0 > 1.0.

β0 = 1.4381,sβ0= 0.62459,t = (1.4381−1.0)/0.62459= 0.701.

Since the alternate hypothesis is of the formβ0 > b, theP-value is the area to the right oft = 0.701.


The claim is plausible.

3. (a) The slope is 0.84451; the intercept is 44.534.

(b) Yes, theP-value for the slope is≈ 0, so horsepower is related to NOx.


SECTION 7.3 351

(c) 44.534+0.84451(10)= 52.979 mg/s.

(d) r =√

r2 =√

0.847= 0.920.

(e) Sincen = 123 is large, use thez table to construct a confidence interval.

z.05 = 1.645, so a 90% confidence interval is 78.31±1.645(2.23), or (74.6, 82.0).

(f) No. A reasonable range of predicted values is given by the95% prediction interval,which is (29.37, 127.26).

4. (a) 0.75269

(b) β1 = 0.75269,sβ1= 0.02912. Sincen = 123 is large, use thez table to construct a confidence interval.

z.025 = 1.96. A 95% confidence interval is 0.75269±1.96(0.02912), or (0.696, 0.810).

(c) The null and alternate hypotheses areH0 : β1 ≥ 0.80 versusH1 : β1 < 0.80.

β1 = 0.75269,sβ1= 0.02912,z= (0.75269−0.80)/0.02912= −1.62.

Since the alternate hypothesis is of the formβ1 < b, theP-value is the area to the left ofz= −1.62.

From thez table,P = 0.0526.

5. (a)H0 : βC−βE = 0

(b) βC andβE are independent and normally distributed with meansβC andβE, respectively, and estimated standarddeviationssβC

= 0.03267 andsβE= 0.02912.

Sincen = 123 is large, the estimated standard deviations can be treated as good approximations to the truestandard deviations.

βC = 0.84451 andβE = 0.75269.

The test statistic isz= (βC− βE)/√

s2βC

+s2βE

= (0.84451−0.75269)/√

0.032672+0.029122 = 2.10.

Since the alternate hypothesis is of the formβC−βE 6= 0, theP-value is the sum of the areas to the right ofz= 2.10 and to the left ofz= −2.10.

ThusP = 0.0179+0.0179= 0.0358.

We can conclude that the effect of horsepower differs in the two cases.


352 CHAPTER 7

6. (a)x = 3.733, y = 65.985, ∑ni=1(xi −x)2 = 6.28701, ∑n

i=1(yi −y)2 = 1360.86625,∑n

i=1(xi −x)(yi −y) = 69.29655, n = 10.

β1 =∑n


∑ni=1(xi −x)2 = 11.022179 andβ0 = y− β1x = 24.839206.

The least-squares line isy = 24.839206+11.022179x.

(b) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.561259. s=

√(1− r2)∑n

i=1(yi −y)2

n−2= 8.639063.

(c) s= 8.639063.sβ1=

s√∑n

i=1(xi −x)2= 3.445439.

There aren−2 = 8 degrees of freedom.t8,.025 = 2.306.


(d) x = 4, y = 24.839206+11.022179(4)= 68.9279.sy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 = 2.882640.

t8,.025 = 2.306. A 95% confidence interval for the mean response is 68.9279±2.306(2.882640)or (62.28, 75.58).

(e) Lety4 = β0 + β1(4) denote the mean withdrawal strength of nails 4 mm in diameter.

The null and alternate hypotheses areH0 : y4 ≤ 60 versusH1 : y4 > 60.

y = β0 + β1(4) = 68.9279,sy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 = 2.882640.

t = (68.9279−60)/2.882640= 3.097. There are 10−2= 8 degrees of freedom.

Since the alternate hypothesis isy4 > 60, theP-value is the area to the right oft = 3.097.


We may conclude that the mean withdrawal strength of nails 4 mm in diameter is greater than 60 N/mm.

(f) y = β0 + β1(4) = 68.9279,spred = s

√

1+1n

+(x−x)2

∑ni=1(xi −x)2 = 9.107306.

t8,.025 = 2.306. A 95% prediction interval is 68.9279±2.306(9.107306) or (47.93, 89.93).

7. (a)x = 20.888889, y = 62.888889, ∑ni=1(xi −x)2 = 1036.888889, ∑n

i=1(yi −y)2 = 524.888889,∑n

i=1(xi −x)(yi −y) = −515.111111, n = 9.

β1 =∑n


∑ni=1(xi −x)2 = −0.496785 andβ0 = y− β1x = 73.266181.


SECTION 7.3 353

The least-squares line isy = 73.266181−0.496785x

(b) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.487531, s=

√(1− r2)∑n

i=1(yi −y)2

n−2= 6.198955,

sβ0= s

√1n

+x2

∑ni=1(xi −x)2 = 4.521130, sβ1

=s√

∑ni=1(xi −x)2

= 0.192510.

There are 9−2= 7 degrees of freedom.t7,.025 = 2.365.


The 95% confidence interval forβ1 is−0.496785±2.365(0.192510), or (−0.952,−0.0415).

(c) y = 60.846550,spred = s

√

1+1n

+(x−x)2

∑ni=1(xi −x)2 = 6.582026.


(d) The standard error of prediction isspred = s

√

1+1n

+(x−x)2

∑ni=1(xi −x)2 .

Given two values forx, the one that is farther fromx will have the greater value ofspred, and thus the widerprediction interval. Sincex = 20.888889, the prediction interval forx = 30 will be wider than the one forx = 15.

8. (a)x = 4.259875 y = 0.907458, ∑ni=1(xi −x)2 = 3.770927, ∑n

i=1(yi −y)2 = 0.402126,∑n

i=1(xi −x)(yi −y) = 1.138876, n = 24.

β1 =∑n


∑ni=1(xi −x)2 = 0.302015 andβ0 = y− β1x = −0.379088.

The least-squares line isy = −0.379088+0.302015x

(b) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.855348, s=

√(1− r2)∑n

i=1(yi −y)2

n−2= 0.0514199,

sβ0= s

√1n

+x2

∑ni=1(xi −x)2 = 0.113286, sβ1

=s√

∑ni=1(xi −x)2

= 0.0264794.


Therefore a 95% confidence interval forβ0 is−0.379088±2.074(0.113286), or (−0.614,−0.144).

The 95% confidence interval forβ1 is 0.302015±2.074(0.0264794), or (0.247,0.357).


354 CHAPTER 7

(c) By 0.302015(0.5) = 0.1510 volts.

(d) y = 0.919577,spred = s

√

1+1n

+(x−x)2

∑ni=1(xi −x)2 = 0.0524910.


(e) Lety4.5 = β0 + β1(4.5) denote the mean potential of molecules whose model value is 4.5.

The null and alternate hypotheses areH0 : y4.5 ≤ 0.93 versusH1 : y4.5 > 0.93.

y = β0 + β1(4.5) = 0.979980,sy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 = 0.0122717.

t = (0.979980−0.93)/0.0122717= 4.073. There are 24−2= 22 degrees of freedom.

Since the alternate hypothesis isy4.5 > 0.93, theP-value is the area to the right oft = 4.073.


We may conclude that the mean potential of molecules whose model value is 4.5 is greater than 0.93 volts.

9. (a)x = 0.762174, y = 1.521957, ∑ni=1(xi −x)2 = 4.016183, ∑n

i=1(yi −y)2 = 2.540524,∑n

i=1(xi −x)(yi −y) = −3.114696, n = 46.

β1 =∑n


∑ni=1(xi −x)2 = −0.775536 andβ0 = y− β1x = 2.113050.

The least-squares line isy = 2.113050−0.775536x

(b) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.950812, s=

√(1− r2)∑n

i=1(yi −y)2

n−2= 0.053293,

sβ0= s

√1n

+x2

∑ni=1(xi −x)2 = 0.021738, sβ1

=s√

∑ni=1(xi −x)2

= 0.026593.

There are 46−2= 44 degrees of freedom.t44,.025≈ 2.015.


The 95% confidence interval forβ1 is−0.775536±2.015(0.026593), or (−0.829,−0.722).

(c) y = β0 + β1(1.7) = 0.79464.

(d) y = 0.79464,sy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 = 0.0261477.

There are 46−2= 44 degrees of freedom.t44,.025≈ 2.015.


SECTION 7.3 355

Therefore a 95% confidence interval is 0.79464±2.015(0.0261477), or (0.742, 0.847).

(e) y = 0.79464,spred = s

√

1+1n

+(x−x)2

∑ni=1(xi −x)2 = 0.0593616.

t44,.025≈ 2.015. A 95% prediction interval is 0.79464±2.015(0.0593616) or (0.675, 0.914).

10. (a) If the standard deviations is held constant, the width of a confidence interval is proportional to 1/√

∑ni=1(xi −x)2.

Let SA, SB, andSC denote the values of∑ni=1(xi −x)2 for Engineers A, B, and C, respectively.

ThenSA = 22.8, SB = 2SA = 45.6, andSC = 4SA = 91.2.

Let WA, WB, andWC denote the widths of the confidence intervals for Engineers A, B, and C, respectively.

ThenWA/WB ≈√

SB/SA =√

2.

The answer is approximate because the values ofswill in fact vary somewhat among the engineers.

(b) WA/WC ≈√

SC/SA = 2.

The answer is approximate because the values ofswill in fact vary somewhat among the engineers.

(c) If the standard deviations is held constant, the width of a confidence interval is proportional to

√1n

+(x−x)2

∑ni=1(xi −x)2 .

The value ofx is 2.5.

The values ofx arexA = 2.8, xB = 2.8, xC = 5.6.

The values ofn arenA = 5, nB = 10,nC = 5.

Let WA, WB, andWC denote the widths of the confidence intervals for Engineers A, B, and C, respectively.

Then WA, WB, WC are proportional to

√15

+(2.5−2.8)2

22.8= 0.45,

√110

+(2.5−2.8)2

45.6= 0.32,

√15

+(2.5−5.6)2

91.2= 0.55 respectively.

These results are approximate because the values ofswill in fact vary somewhat among the engineers.

B is most likely to be the shortest; C is most likely to be the longest.

11. The width of a confidence interval is proportional tosy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 .

Sinces, n, x, and∑ni=1(xi −x)2 are the same for each confidence interval, the width of the confidence interval

is an increasing function of the differencex−x.

x = 1.51966. The value 1.5 is closest tox and the value 1.8 is the farthest fromx.


356 CHAPTER 7

Therefore the confidence interval at 1.5 would be the shortest, and the confidence interval at 1.8 would be thelongest.

12. The width of a confidence interval is proportional tosy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 .

Sinces, n, x, and∑ni=1(xi −x)2 are the same for each confidence interval, the width of the confidence interval

is an increasing function of the differencex−x.

x = 2.40. The value 2.45 is closest tox and the value 2.15 is the farthest fromx.

Therefore the confidence interval at 2.45 would be the shortest, and the confidence interval at 2.15 would bethe longest.

13. s2 = (1− r2)∑ni=1(yi −y)2/(n−2) = (1−0.9111)(234.19)/(17−2)= 1.388.

14. r2 satisfies the equations2 = (1− r2)∑ni=1(yi −y)2/(n−2).

Therefore 0.0670= (1− r2)(57.313)/(25−2), sor2 = 0.9731.

15. (a)t = 1.71348/6.69327= 0.256.

(b) n = 25, so there aren−2 = 23 degrees of freedom. TheP-value is for a two-tailed test, so it is equal to thesum of the areas to the right oft = 0.256 and to the left oft = −0.256.

ThusP = 0.40+0.40= 0.80.

(c) sβ1satisfies the equation 3.768= 4.27473/sβ1

, sosβ1= 1.13448.

(d) n = 25, so there aren−2 = 23 degrees of freedom. TheP-value is for a two-tailed test, so it is equal to thesum of the areas to the right oft = 3.768 and to the left oft = −3.768.

ThusP = 0.0005+0.0005= 0.001.

16. (a)β0 satisfies the equation 0.688= β0/0.43309, soβ0 = 0.29797.

(b) n = 20, so there aren−2 = 18 degrees of freedom. TheP-value is for a two-tailed test, so it is equal to thesum of the areas to the right oft = 0.688 and to the left oft = −0.688.


SECTION 7.4 357

ThusP = 0.25+0.25= 0.50.

(c) t = 0.18917/0.065729= 2.878.

(d) n = 20, so there aren−2 = 18 degrees of freedom. TheP-value is for a two-tailed test, so it is equal to thesum of the areas to the right oft = 2.878 and to the left oft = −2.878.

ThusP = 0.005+0.005= 0.010.

17. (a)y = 106.11+0.1119(4000)= 553.71.

(b) y = 106.11+0.1119(500)= 162.06.

(c) Below. For values ofx near 500, there are more points below the least squares estimate than above it.

(d) There is a greater amount of vertical spread on the right side of the plot than on the left.

Section 7.4

1. (a) lny = −0.4442+0.79833lnx

(b) y = eln y = e−0.4442+0.79833(ln2500) = 330.95.

(c) y = eln y = e−0.4442+0.79833(ln1600) = 231.76.

(d) The 95% prediction interval for lny is given as (3.9738, 6.9176).

The 95% prediction interval fory is therefore(e3.9738,e6.9176), or (53.19, 1009.89).


358 CHAPTER 7

2. (a)

6 7 8 9

−2

−1

0

1

2

Fitted Value

Re

sid

ua

l

The least-squares line isy= 5.576+0.0142x. The linear modelis not appropriate. The point (276.02, 9.36) is highly influential.

(b)

3 4 5 6 7 8 9 10−0.4

−0.2

0

0.2

0.4

Fitted Value

Re

sid

ua

l

The least-squares line isy = 6.148+ 0.629lnx. The linearmodel appears to be appropriate.

(c)

5 6 7 8 9 10

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Fitted Value

Re

sid

ua

l

The least-squares line isy = 5.172+0.286x. The linear modelis not appropriate. The point (276.02, 9.36) is highly influential.

(d) The most appropriate model isy = 6.148+0.629lnx. The prediction isy = 6.148+0.629ln50= 8.609.

(e) The 95% confidence interval is (8.025, 9.192).


SECTION 7.4 359

3. (a)y = 20.162+1.269x

(b)

50 55 60 65 70 75

−10

−5

0

5

10

15

Fitted Value

Re

sid

ua

l

There is no apparent pattern to the residual plot. The linearmodel looks fine.

(c)

0 5 10 15 20 25−15

−10

−5

0

5

10

15

Order of Observations

Re

sid

ua

l

The residuals increase over time. The linear model is not ap-propriate as is. Time, or other variables related to time, must beincluded in the model.

4. (a)

0 10 20 30 40 500

5

10

15

20

25

30

Scaled Distance (m/kg0.5)

PP

V (

mm

/se

c)

The relationship is not linear.


360 CHAPTER 7

(b)

−0.5 0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

Fitted Value

Re

sid

ua

l

The least-squares line is lny = 5.278−1.465lnx, wherey rep-resents PPV andx represents Scaled Distance.

(c) The predicted value of lny is lny = 5.278− 1.465ln20= 0.88925. Therefore the predicted value ofy isy = eln y = e0.88925= 2.433. The 95% prediction interval is (0.608, 9.735).

5. (a)y = −235.32+0.695x.

(b)

0 1000 2000 3000 4000 5000−1500

−1000

−500

0

500

1000

1500

Fitted Value

Re

sid

ua

l

The residual plot shows a pattern, with positive residuals at thehigher and lower fitted values, and negative residuals in themid-dle. The model is not appropriate.

(c) lny = −0.0745+0.925lnx.


SECTION 7.4 361

(d)

4.5 5 5.5 6 6.5 7 7.5 8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

Fitted Value

Re

sid

ua

l

The residual plot shows no obvious pattern. The model is ap-propriate.

(e) The log model is more appropriate. The 95% prediction interval is(197.26, 1559.76).

6. (a)y = 36.632−3.293x

(b)

−20 −10 0 10 20 30−20

−15

−10

−5

0

5

10

15

20

Fitted Value

Re

sid

ua

l

(c) y = −5.366+85.439(1/x)


362 CHAPTER 7

(d)

0 10 20 30 40 50−10

−5

0

5

10

Fitted Value

Re

sid

ua

l

(e) The liney = −5.366+85.439(1/x) fits better. The 95% confidence interval is (8.743,14.701).

7. (a)

1 1.5 2 2.5 3 3.5

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

Fitted Value

Re

sid

ua

l


(b)

0.5 1 1.5 2 2.5 3

−0.6

−0.4

−0.2

0

0.2

0.4

Fitted Value

Re

sid

ua

l

The least-squares line isy = 0.199+1.207lnx.


SECTION 7.4 363

(c)

0 0.5 1 1.5

−1.4

−1.2

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

Fitted Value

Re

sid

ua

l

The least-squares line is lny = −0.0679+0.137x.

(d)

1 1.2 1.4 1.6 1.8 2

−0.6

−0.4

−0.2

0

0.2

Fitted Value

Re

sid

ua

l

The least-squares line is√

y = 0.956+0.0874x.

(e) The modely = 0.199+1.207lnx fits best. Its residual plot shows the least pattern.

(f)

0 10 20 30 40 50−1

−0.5

0

0.5

1

The residuals show no pattern with time.

(g) The modely = 0.199+1.207lnx fits best. The predicted value isy = 0.199+1.207ln5.0 = 2.14.


364 CHAPTER 7

(h) The 95% prediction interval is (1.689, 2.594).

8. (a)R2 = 0.464+0.571R1

(b)

1 1.5 2 2.5 3 3.5 4 4.5

−0.5

0

0.5

Fitted Value

Re

sid

ua

l

The linear model is not appropriate. The residual plot has astrong pattern.

(c)

1.6 1.8 2 2.2

−0.3

−0.2

−0.1

0

0.1

Fitted Value

Re

sid

ua

l

For R1 < 4, the least squares line isR2 =1.233+ 0.264R1. There is a pattern to theresidual plot, so the linear model does notseem appropriate.

2.5 3 3.5 4 4.5

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

Fitted Value

Re

sid

ua

l

For R1 ≥ 4, the least squares line isR2 =−0.190+0.710R1. The residual plot does nothave much pattern, so the linear model seemsappropriate.

(d) The linear model seems to work well whenR1 ≥ 4, but less well whenR1 < 4. It might make sense to use alinear model to predictR2 whenR1 ≥ 4, but to fit a non-linear model to predictR2 from R1 whenR1 < 4.

9. (a) The model is log10y = β0 + β1 log10x+ ε. Note that the natural log (ln) could be used in place of log10, but


SECTION 7.4 365

common logs are more convenient since partial pressures areexpressed as powers of 10.

(b)

−2 −1.5 −1 −0.5−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

Fitted Value

Re

sid

ua

l

The least-squares line is log10y = −3.277−0.225log10x. The linear model appears to fitquite well.

(c) The theory says that the coefficientβ1 of log10x in the linear model is equal to−0.25. The estimated value is

β = −0.225. We determine whether the data are consistent with the theory by testing the hypothesesH0 : β1 =−0.25 vs.H1 : β1 6= −0.25. The value of the test statistic ist = 0.821. There are 21 degrees of freedom, so0.20< P < 0.50 (a computer package givesP = 0.210). We do not rejectH0, so the data are consistent withthe theory.

10. (a)

100 200 300 400 500 600 700150

200

250

300

350

400

450

Ester Level (mg/l)

Acid

Le

ve

l (m

g/l)

Outlier

(b) β0 = 352.32, sβ0= 52.774,β1 = −0.2495,sβ1

= 0.1951.

(c) t14 = −1.28, 0.20< P < 0.50. A computer package givesP = 0.221.

(d) β0 = 359.23, sβ0= 91.076,β1 = −0.2834,sβ1

= 0.4108.

(e) t13 = −0.690,P≈ 0.50. A computer package givesP = 0.502.


366 CHAPTER 7

(f) Not useful. TheP-values for the slope are not small, whether or not the outlier is included. Therefore it isplausible that the slope is equal to 0, so a linear model is nothelpful for prediction.

11. (a)y = 2049.87−4.270x

(b) (12, 2046) and (13, 1954) are outliers. The least-squares line with (12, 2046) deleted isy = 2021.85−2.861x.The least-squares line with (13, 1954) deleted isy= 2069.30−5.236x. The least-squares line with both outliersdeleted isy = 2040.88−3.809x.

(c) The slopes of the least-squares lines are noticeably affected by the outliers. They ranged from−2.861 to−5.236.

12. (a)y = 457.270059+87.555863x

0 2000 4000 6000 8000

−5000

0

5000

Fitted Value

Re

sid

ua

l

(b) y = −504.001756+1165.271152lnx

−2000 −1000 0 1000 2000 3000 4000 5000

−2000

−1000

0

1000

2000

3000

4000

5000

6000

Fitted Value

Re

sid

ua

l

(c) lny = 5.258333+0.799002lnx

4 5 6 7 8 9

−1

−0.5

0

0.5

1

Fitted Value

Res

idua

l

(d) The residual plot shows that the model lny = 5.258+0.799lnx fits best.



(e) Lety be the predicted flow when the discharge is 50 km3/yr. Then lny= 5.258333+0.799002ln50= 8.38405.

So y = e8.38405= 4376.7 m3/s

(f) The 95% prediction interval for lny is (7.282961,9.485138).

The 95% prediction interval fory is (e7.282961,e9.485138), or (1455.3,13162.6).

13. The equation becomes linear upon taking the log of both sides:lnW = β0 + β1 lnL, whereβ0 = lna andβ1 = b.

14. lny = β0 + β1 lnx1 + β2 lnx2, whereβ0 = lna, β1 = −b, andβ2 = −c.

15. (a) A physical law.

(b) It would be better to redo the experiment. If the results of an experiment violate a physical law, then somethingwas wrong with the experiment, and you can’t fix it by transforming variables.


1. (a)x = 1.48, y = 1.466, ∑ni=1(xi −x)2 = 0.628, ∑n

i=1(yi −y)2 = 0.65612,∑n

i=1(xi −x)(yi −y) = 0.6386, n = 5.

β1 =∑n


∑ni=1(xi −x)2 = 1.016879 andβ0 = y− β1x = −0.038981.

(b) 0

(c) The molar absorption coefficientM.

(d) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.989726, s=

√(1− r2)∑n

i=1(yi −y)2

n−2= 0.0474028,


368 CHAPTER 7

sβ0= s

√1n

+x2

∑ni=1(xi −x)2 = 0.0910319.

The null and alternate hypotheses areH0 : β0 = 0 versusH1 : β0 6= 0.

There aren−2= 3 degrees of freedom.t = (−0.038981−0)/0.0910319= −0.428.



The data are consistent with the Beer-Lambert law.

2. (a)y = 4.9x2, so the relationship betweenx andy is not linear. The correlation coefficient is not appropriate.

(b) y = 4.9w, so the relationship betweenw andy is linear. The correlation coefficient is appropriate.

(c) v =√

4.9x, so the relationship betweenv andx is linear. The correlation coefficient is appropriate.

(d) v =√

4.9√

w, so the relationship betweenv andw is not linear. The correlation coefficient is not appropriate.

(e) lny= ln4.9+2lnx, so the relationship between lny and lnx is linear. The correlation coefficient is appropriate.

3. (a)

40 50 60 70 80 90 10040

50

60

70

80

90

100

(b) x = 71.101695, y = 70.711864, ∑ni=1(xi −x)2 = 10505.389831, ∑n

i=1(yi −y)2 = 10616.101695,

∑ni=1(xi −x)(yi −y) = −7308.271186, n = 59.

β1 =∑n


∑ni=1(xi −x)2 = −0.695669 andβ0 = y− β1x = 120.175090.



Ti+1 = 120.175090−0.695669Ti.

(c) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.478908, s=

√(1− r2)∑n

i=1(yi −y)2

n−2= 9.851499,

sβ1=

s√∑n

i=1(xi −x)2= 0.096116.

There aren−2 = 57 degrees of freedom.t57,.025≈ 2.002.


(d) y = 120.175090−0.695669(70)= 71.4783 minutes.

(e) y = 71.4783,sy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 = 1.286920.

There are 59−2= 57 degrees of freedom.t57,.01 ≈ 2.394.

Therefore a 98% confidence interval is 71.4783±2.394(1.286920), or (68.40, 74.56).

(f) y = 71.4783,spred = s

√

1+1n

+(x−x)2

∑ni=1(xi −x)2 = 9.935200.

t57,.005≈ 2.6649. A 95% prediction interval is 71.4783±2.6649(9.935200) or (45.00, 97.95).

4. (a)

55 60 65 70 75 80 85 90−30

−20

−10

0

10

20

30

Fitted Value

Re

sid

ua

l

There appears to be some heteroscedasticity, with alarger spread in residuals for smaller fitted values.


370 CHAPTER 7

(b)

0 10 20 30 40 50 60−30

−20

−10

0

10

20

30

Order of Observations

Re

sid

ua

l

No violations of the standard assumptions are indi-cated.

5. (a)x = 50, y = 47.909091, ∑ni=1(xi −x)2 = 11000, ∑n

i=1(yi −y)2 = 9768.909091,

∑ni=1(xi −x)(yi −y) = 10360, n = 11.

β1 =∑n


∑ni=1(xi −x)2 = 0.941818 andβ0 = y− β1x = 0.818182.

(b) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.998805, s=

√(1− r2)∑n

i=1(yi −y)2

n−2= 1.138846.

β0 = 0.818182, sβ0= s

√1n

+x2

∑ni=1(xi −x)2 = 0.642396.


There are 11−2= 9 degrees of freedom.t = (0.818182−0)/0.642396= 1.274.



It is plausible thatβ0 = 0.

(c) The null and alternate hypotheses areH0 : β1 = 1 versusH1 : β1 6= 1.

β1 = 0.941818, sβ1=

s√∑n

i=1(xi −x)2= 0.010858.

There are 11−2= 9 degrees of freedom.t = (0.941818−1)/0.010858= −5.358.



We can conclude thatβ1 6= 1.



(d) Yes, since we can conclude thatβ1 6= 1, we can conclude that the machine is out of calibration.

Since two coefficients were tested, some may wish to apply theBonferroni correction, and multiply theP-valuefor β1 by 2. The evidence thatβ1 6= 1 is still conclusive.

(e) x = 20, y = β0 + β1(20) = 19.65455.

sy = s

√1n

+(x−x)2



(f) x = 80, y = β0 + β1(80) = 76.163636.

sy = s

√1n

+(x−x)2



(g) No, when the true value is 20, the result of part (e) shows that a 95% confidence interval for the mean of themeasured values is (18.58, 20.73). Therefore it is plausible that the mean measurement will be 20, so that themachine is in calibration.

6. (a)x = 4.036667, y = 51.3333, ∑ni=1(xi −x)2 = 21.150533, ∑n

i=1(yi −y)2 = 3411.333333,

∑ni=1(xi −x)(yi −y) = 268.166667, n = 6.

β1 =∑n


∑ni=1(xi −x)2 = 12.678955 andβ0 = y− β1x = 0.152617.

y = 0.152617+12.678955x

(b) y = β0 + β1(3.00) = 38.189.

(c) No, because no values as large as 8 were used for the independent variable.

7. (a)x = 18.142857, y = 0.175143, ∑ni=1(xi −x)2 = 418.214286, ∑n

i=1(yi −y)2 = 0.0829362,

∑ni=1(xi −x)(yi −y) = 3.080714, n = 14.

β1 =∑n


∑ni=1(xi −x)2 = 0.00736635 andβ0 = y− β1x = 0.0414962.


372 CHAPTER 7

y = 0.0414962+0.00736635x

(b) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.273628, s=

√(1− r2)∑n

i=1(yi −y)2

n−2= 0.0708535.

β1 = 0.00736635,sβ1=

s√∑n

i=1(xi −x)2= 0.00346467.


Therefore a 95% confidence interval for the slope is 0.00736635±2.179(0.00346467),or(−0.00018, 0.01492).

(c) x = 20, y = β0 + β1(20) = 0.188823.

sy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 = 0.0199997.

There are 12 degrees of freedom.t12,.025= 2.179.


(d) x = 20, y = β0 + β1(20) = 0.188823.

spred = s

√

1+1n

+(x−x)2

∑ni=1(xi −x)2 = 0.073622.

There are 12 degrees of freedom.t12,.05 = 1.782.


8. (a)x = 146.875,y = 1.21625,∑ni=1(xi −x)2 = 74249.75, ∑n

i=1(yi −y)2 = 0.588575,

∑ni=1(xi −x)(yi −y) = −182.9175, n = 16.

β1 =∑n


∑ni=1(xi −x)2 = −0.00246354 andβ0 = y− β1x = 1.578083.

y = 1.578083−0.00246354x

(b) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.765621, s=

√(1− r2)∑n

i=1(yi −y)2

n−2= 0.0992651.

β1 = −0.00246354, sβ1=

s√∑n

i=1(xi −x)2= 0.00036429.


Therefore a 95% confidence interval for the slope is−0.00246354±2.145(0.00036429), or(−0.003245,−0.001682).



(c) x = 100, y = β0 + β1(100) = 1.33173.

sy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 = 0.030124.



(d) x = 100, y = β0 + β1(100) = 1.33173.

spred = s

√

1+1n

+(x−x)2

∑ni=1(xi −x)2 = 0.103735.



(e) The confidence interval is likely to be more useful. It is likely to be of greater importance to estimate the meandiffusivity, rather than the diffusivity of a specific specimen.

9. (a) lny = β0 + β1 lnx, whereβ0 = lnk andβ1 = r.

(b) Letui = lnxi and letvi = lnyi .

u = 1.755803,v = −0.563989,∑ni=1(ui −u)2 = 0.376685, ∑n

i=1(vi −v)2 = 0.160487,

∑ni=1(ui −u)(vi −v) = 0.244969, n = 5.

β1 =∑n

i=1(ui −u)(vi −v)

∑ni=1(ui −u)2 = 0.650328 andβ0 = v− β1u = −1.705838.

The least-squares line is lny = −1.705838+0.650328lnx.

Thereforer = 0.650328 andk = e−1.705838= 0.18162.

(c) The null and alternate hypotheses areH0 : r = 0.5 versusH1 : r 6= 0.5.

r = 0.650328,s= 0.0198005,sr =s√

∑ni=1(ui −u)2

= 0.0322616.

There are 5−2= 3 degrees of freedom.t = (0.650328−0.5)/0.0322616= 4.660.

Since the alternate hypothesis is of the formr 6= r0, theP-value is the sum of the areas to the right oft = 4.660and to the left oft = −4.660.


We can conclude thatr 6= 0.5.


374 CHAPTER 7

10. (a)x = 28.078261,y = 29.221739,∑ni=1(xi −x)2 = 287.999130, ∑n

i=1(yi −y)2 = 1692.179130,

∑ni=1(xi −x)(yi −y) = −347.219130, n = 23.

β1 =∑n


∑ni=1(xi −x)2 = −1.205626 andβ0 = y− β1x = 63.073610,

The least-squares line isy = 63.073610−1.205626x.

(b) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.247383, s=

√(1− r2)∑n

i=1(yi −y)2

n−2= 7.787545.

β0 = 63.073610, sβ0= s

√1n

+x2

∑ni=1(xi −x)2 = 12.986644.



β1 = −1.205626, sβ1=

s√∑n

i=1(xi −x)2= 0.458886.



(c) By 1.205626(2) = 2.411 days.

(d) The null and alternate hypotheses areH0 : β1 ≥ 0 versusH1 : β1 < 0.

β1 = −1.205626, sβ1=

s√∑n

i=1(xi −x)2= 0.458886.

There are 23−2= 21 degrees of freedom.t = (−1.205626−0)/0.458886= −2.627.

Since the alternate hypothesis is of the formβ1 < b, theP-value is the area to the left oft = −2.627.


We can conclude that butterflies with larger wingspans have shorter flight periods on the average.

(e) Lety30 = β0 + β1(30) denote the mean flight period for butterflies whose wingspan is 30 mm.

The null and alternate hypotheses areH0 : y30 ≥ 28 versusH1 : y30 < 28.

y = β0 + β1(30) = 26.904841,sy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 = 1.847824.

t = (26.904841−28)/1.847824= −0.593. There are 23−2= 21 degrees of freedom.

Since the alternate hypothesis isy30 < 0.93, theP-value is the area to the left oft = −0.593.


We cannot conclude that the mean flight period for butterflieswith wingspan 30 mm is less than 28 days.

(f) x = 28.5, y = β0 + β1(28.5) = 28.713280.



spred = s

√

1+1n

+(x−x)2

∑ni=1(xi −x)2 = 7.957392.



11. (a)x = 2812.130435,y = 612.739130,∑ni=1(xi −x)2 = 335069724.6, ∑n

i=1(yi −y)2 = 18441216.4,

∑ni=1(xi −x)(yi −y) = 32838847.8.

β1 =∑n


∑ni=1(xi −x)2 = 0.098006 andβ0 = y− β1x = 337.133439.


(b)

0 500 1000 1500 2000−2000

−1000

0

1000

2000

3000

Fitted Value

Re

sid

ua

l

(c) Letui = lnxi and letvi = lnyi .

u = 6.776864,v = 5.089504,∑ni=1(ui −u)2 = 76.576108, ∑n

i=1(vi −v)2 = 78.642836,

∑ni=1(ui −u)(vi −v) = 62.773323.

β1 =∑n

i=1(ui −u)(vi −v)

∑ni=1(ui −u)2 = 0.819751 andβ0 = v− β1u = −0.465835.

The least-squares line is lny = −0.466+0.820lnx.

(d)

0 2 4 6 8−3

−2

−1

0

1

2

3

Fitted Value

Res

idua

l

(e) Letui = lnxi and letvi = lnyi .

x = 3000, sou = lnx = ln3000, andv = β0 + β1(ln3000) = 6.09739080.


376 CHAPTER 7

spred = s

√

1+1n

+(u−u)2

∑ni=1(ui −u)2 = 1.17317089.


Therefore a 95% prediction interval forv= lny is 6.09739080±2.080(1.17317089), or (3.657195, 8.537586).

A 95% prediction interval fory is (e3.657195,e8.537586), or (38.75,5103.01).

12. (iv). Since Newton’s law of cooling is a physical law, theplot of ln temperature versus time should be linear.The non-linearity of the plot suggests that something went wrong with the experiment.

13. (a)x = 225,y = 86.48333,∑ni=1(xi −x)2 = 37500, ∑n

i=1(yi −y)2 = 513.116667,

∑ni=1(xi −x)(yi −y) = 4370, n = 12.

β1 =∑n


∑ni=1(xi −x)2 = 0.116533 andβ0 = y− β1x = 60.263333.

The estimate ofσ2 is s2 =(1− r2)∑n

i=1(yi −y)2

n−2.

r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.992466, s2 = 0.386600.

(b) The null and alternate hypotheses areH0 : β0 = 0 versusH1 : β0 6= 0.

s=√

s2 = 0.621772. β0 = 60.263333, sβ0= s

√1n

+x2

∑ni=1(xi −x)2 = 0.744397.






s=√

s2 = 0.621772. β1 = 0.116533, sβ1=

s√∑n

i=1(xi −x)2= 0.0032108.


Since the alternate hypothesis is of the formβ1 6= b, theP-value is the sum of the areas to the right oft = 36.294



and to the left oft = −36.294.



(d)

80 85 90 95

−1

−0.5

0

0.5

1

Fitted Value

Re

sid

ua

l

The linear model appears to be appropriate.

(e) β1 = 0.116533, sβ1= 0.0032108.

There aren−2 = 10 degrees of freedom.t10,.025= 2.228.

Therefore a 95% confidence interval for the slope is 0.116533±2.228(0.0032108), or (0.10938, 0.12369).

(f) x = 225, y = β0 + β1(225) = 86.483333.

sy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 = 0.179490.



(g) x = 225, y = β0 + β1(225) = 86.483333.

spred = s

√

1+1n

+(x−x)2

∑ni=1(xi −x)2 = 0.647160.



14. (a)y = −0.23429+0.72868(2.5)= 1.5874.

(b) The null and alternate hypotheses areH0 : β1 = 1 versusH1 : β1 6= 1.


378 CHAPTER 7

β1 = 0.72868, sβ1= 0.06353




We can conclude thatβ1 6= 1. The claim is not plausible.


From the output,P = 0.350.

The data are consistent with the fact that there is no runoff when there is no rainfall.

15. (ii). The standard deviationsy is not given in the output. To computesy, the quantity∑ni=1(xi − x)2 must be

known.

16. The width of the confidence interval is proportional tosy = s

√1n

+(x−x)2

∑ni=1(xi −x)2 .

Sinces, n, and∑ni=1(xi −x)2 are the same for each confidence interval, the width of the confidence interval is

an increasing function of(x−x).

x = (11+37+54+70+93)/5= 53.

Therefore the narrowest is (i) and the widest is (ii).

17. (a) If f = 1/2 then 1/ f = 2. The estimate ist = 145.736−0.05180(2)= 145.63.

(b) Yes.r = −√R-Sq= −0.988. Note thatr is negative because the slope of the least-squares line is negative.

(c) If f = 1 then 1/ f = 1. The estimate ist = 145.736−0.05180(1)= 145.68.

18. (a) Ify is unbiased, theny = true value+ ε. If the calculated valuex is equal to the true value, theny = x+ ε.

(b) x = 0.775714, y = 0.767857, ∑ni=1(xi −x)2 = 0.955743, ∑n

i=1(yi −y)2 = 0.968036,

∑ni=1(xi −x)(yi −y) = 0.935871, n = 14.



β1 =∑n


∑ni=1(xi −x)2 = 0.979208 andβ0 = y− β1x = 0.00827120.

y = 0.00827120+0.979208x.

(c) If the calculated value is accurate, theny = 0+1x+ ε.

(d) r2 =[∑n

i=1(xi −x)(yi −y)]2

∑ni=1(xi −x)2 ∑n

i=1(yi −y)2 = 0.946673, s=

√(1− r2)∑n

i=1(yi −y)2

n−2= 0.0655887.

β0 = 0.00827120, sβ0= s

√1n

+x2

∑ni=1(xi −x)2 = 0.0549156.




From thet table,P > 0.80. A computer package givesP = 0.88.



β1 = 0.979208, sβ1=

s√∑n

i=1(xi −x)2= 0.0670901.


Since the alternate hypothesis is of the formβ0 6= b, theP-value is the sum of the areas to the right oft =−0.310and to the left oft = −0.310.



(e) It is plausible that the calculated value is accurate. Neither of the hypothesesβ0 = 0 norβ1 = 1 can be rejected.

19. (a) We need to minimize the sum of squaresS= ∑(yi − βxi)2.

We take the derivative with respect toβ and set it equal to 0, obtaining−2∑xi(yi − βxi) = 0.

Then∑xiyi − β∑x2i = 0, soβ = ∑xiyi/∑x2

i .

(b) Letci = xi/∑x2i . Thenβ = ∑ciyi , soσ2

β= ∑c2

i σ2 = σ2 ∑x2i /(∑x2

i )2 = σ2/∑x2

i .


380 CHAPTER 7

20. First note that∑ni=1(xi −x) = ∑n

i=1xi −nx= 0.

Sincex is constant,∑ni=1x(xi −x) = x∑n

i=1(xi −x) = 0 as well.

We now show that∑ni=1xi(xi −x) = ∑n

i=1(xi −x)2. This follows since∑n

i=1xi(xi −x) = ∑ni=1xi(xi −x)−∑n

i=1x(xi −x) = ∑ni=1[xi(xi −x)−x(xi −x)] =

∑ni=1(xi −x)(xi −x) = ∑n

i=1(xi −x)2.

Now

µβ1=

n

∑i=1

(xi −x)

∑ni=1(xi −x)2 µyi

=n

∑i=1

(xi −x)

∑ni=1(xi −x)2 (β0−β1xi)

=1

∑ni=1(xi −x)2

[n

∑i=1

(xi −x)(β0 + β1xi)

]

=1

∑ni=1(xi −x)2

[β0

n

∑i=1

(xi −x)+ β1

n

∑i=1

(xi −x)xi

]

=1

∑ni=1(xi −x)2

[β0(0)+ β1

n

∑i=1

(xi −x)2

]

=1

∑ni=1(xi −x)2

[β1

n

∑i=1

(xi −x)2

]

= β1

21. From the answer to Exercise 20, we know that∑ni=1(xi −x) = 0, ∑n

i=1x(xi −x) = 0, and∑n

i=1xi(xi −x) = ∑ni=1(xi −x)2. Now

µβ0=

n

∑i=1

[1n− x(xi −x)

∑ni=1(xi −x)2

]µyi

=n

∑i=1

[1n− x(xi −x)

∑ni=1(xi −x)2

](β0 + β1xi)

= β0

n

∑i=1

1n

+ β1

n

∑i=1

xi

n−β0

∑ni=1x(xi −x)

∑ni=1(xi −x)2 −β1

∑ni=1xix(xi −x)

∑ni=1(xi −x)2

= β0 + β1x−0−β1x∑n

i=1xi(xi −x)

∑ni=1(xi −x)2

= β0 + β1x−0−β1x

= β0



22. σ2β1

=n

∑i=1

[(xi −x)

∑ni=1(xi −x)2

]2

σ2

=∑n

i=1(xi −x)2

[∑ni=1(xi −x)2]2

σ2

=σ2

∑ni=1(xi −x)2

23.

σ2β0

=n

∑i=1

[1n− x(xi −x)

∑ni=1(xi −x)2

]2

σ2

=n

∑i=1

[1n2 −

2xn

(xi −x)

∑ni=1(xi −x)2 +x2 ∑n

i=1(xi −x)2

[∑ni=1(xi −x)2]2

]σ2

=

[n

∑i=1

1n2 −2

xn

∑ni=1(xi −x)

∑ni=1(xi −x)2 +x2 ∑n

i=1(xi −x)2

[∑ni=1(xi −x)2]2

]σ2

=

[1n−2

xn(0)+

x2

∑ni=1(xi −x)2

]σ2

=

[1n

+x2

∑ni=1(xi −x)2

]σ2


382 CHAPTER 8

Chapter 8

Section 8.1

1. (a) y = 8.2407−0.10826(10)−0.0039249(50)= 6.9619 miles per gallon.

(b) By 0.0039249(10)= 0.03925 miles per gallon.

(c) By 0.10826(5) = 0.5413 miles per gallon.

2. (a) β1 = −0.10826,sβ1= 0.01194,n = 15.

There aren−3= 12 degrees of freedom.t12,.025= 2.179.

A 95% confidence interval is−0.10826±2.179(0.01194), or (−0.1343,−0.0822).

(b) β2 = −0.0039249,sβ2= 0.001406,n = 15.

There aren−3= 12 degrees of freedom.t12,.005= 3.055.

A 95% confidence interval is−0.0039249± (3.055)0.001406, or(−0.00822,0.000370).

(c) β1 = −0.10826,sβ1= 0.01194,n = 15.

The null and alternate hypotheses areH0 : β1 ≥−0.05 versusH1 : β1 < −0.05.

There aren−3= 12 degrees of freedom.t = [−0.10826− (−0.05)]/0.01194= −4.879.

Since the alternate hypothesis is of the formβ1 < b, theP-value is the area to the left oft = −4.879. From thet table,P < 0.0005. A computer package givesP = 0.00019. We can conclude thatβ1 < −0.05.

(d) β2 = −0.0039249,sβ2= 0.001406,n = 15.

The null and alternate hypotheses areH0 : β2 ≤−0.005 versusH1 : β2 > −0.005.

There aren−3= 12 degrees of freedom.t = [−0.0039249− (−0.005)]/0.001406= 0.765.

Since the alternate hypothesis is of the formβ2 > b, theP-value is the area to the right oft = 0.765. From thet table, 0.10 −0.005.


SECTION 8.1 383

3.

4 5 6 7 8−1

−0.5

0

0.5

Fitted Value

Re

sid

ua

l

There is no obvious pattern to the residualplot, so the linear model appears to fit well.

4. (a) y = −0.83+0.017(20)+0.0895(17)+42.771(0.006)+0.027(10)−0.0043(17)(10)= 0.8271 seconds.

(b) Yes, the predicted change is 0.017(2) = 0.034 seconds.

(c) No, the predicted change depends on the value ofx4, because of the interaction termx2x4.

(d) R2 = SSR/SST= SSR/(SSR+SSE) = 1.08613/(1.08613+0.036310)= 0.967651.

(e) There are 5 degrees of freedom for regression and 44 degrees of freedom for error.

F5,44 =SSR/5SSE/44

= 263.23. From theF table,P < 0.001. A computer package givesP≈ 0.

The null hypothesis can be rejected.

5. (a) y= 56.145−9.046(3)−33.421(1.5)+0.243(20)−0.5963(3)(1.5)−0.0394(3)(20)+0.6022(1.5)(20)+0.6901(32)+11.7244(1.52)−0.0097(202) = 25.465.

(b) No, the predicted change depends on the values of the other independent variables, because of the interactionterms.

(c) R2 = SSR/SST= (SST−SSE)/SST= (6777.5−209.55)/6777.5= 0.9691.

(d) There are 9 degrees of freedom for regression and 27−9−1= 17 degrees of freedom for error.

F9,17 =SSR/9SSE/17

=(SST−SSE)/9

SSE/17= 59.204.

From theF table,P < 0.001. A computer package givesP = 4.6×10−11.


384 CHAPTER 8

The null hypothesis can be rejected.

6. (a) Yes, the predicted productivity would increase by 151.680 m3.

(b) No, the number of trucks, the match factor, and the truck travel time must also be known, because thesevariables appear in interactions with haul length.

7. (a) y = −0.21947+0.779(2.113)−0.10827(0)+1.3536(1.4)−0.0013431(730)= 2.3411 liters

(b) By 1.3536(0.05) = 0.06768 liters

(c) Nothing is wrong. In theory, the constant estimates FEV1 for an individual whose values for the other variablesare all equal to zero. Since these values are outside the range of the data (e.g., no one has zero height), theconstant need not represent a realistic value for an actual person.

8. (a) β1 = 0.779, sβ1= 0.04909. There are 160 degrees of freedom for error.

Since the number of degrees of freedom is large, use thez table to compute a confidence interval.


(b) β3 = 1.3536, sβ3= 0.2880. There are 160 degrees of freedom for error.

Since the number of degrees of freedom is large, use thez table to compute a confidence interval.


(c) β2 = −0.10827, sβ2= 0.0352.

The null and alternate hypotheses areH0 : β2 ≥−0.08 versusH1 : β2 < −0.08.

There are 160 degrees of freedom for error.

Since the number of degrees of freedom is large, use thez table to find theP-value.

z= [−0.10827− (−0.08)]/0.0352= −0.80.

Since the alternate hypothesis is of the formβ2 < b, theP-value is the area to the left ofz= −0.80. From theztable,P = 0.2119. We cannot conclude thatβ2 < −0.08.

(d) β3 = 1.3536, sβ3= 0.2880.

The null and alternate hypotheses areH0 : β3 ≤ 0.5 versusH1 : β3 > 0.5.


SECTION 8.1 385

There are 160 degrees of freedom for error.

Since the number of degrees of freedom is large, use thez table to find theP-value.

z= (1.3536−0.5)/0.2880= 2.96.

Since the alternate hypothesis is of the formβ3 > b, theP-value is the area to the right ofz= 2.96. From theztable,P = 0.0015. We can conclude thatβ3 > 0.5.

9. (a) y = −1.7914+0.00026626(1500)+9.8184(1.04)−0.29982(17.5)= 3.572.

(b) By 9.8184(0.01) = 0.098184.

(c) Nothing is wrong. The constant estimates the pH for a pulpwhose values for the other variables are all equalto zero. Since these values are outside the range of the data (e.g., no pulp has zero density), the constant neednot represent a realistic value for an actual pulp.

(d) From the output, the confidence interval is (3.4207, 4.0496).

(e) From the output, the prediction interval is (2.2333, 3.9416).

(f) Pulp B. The standard deviation of its predicted pH (SE Fit) is smaller than that of Pulp A (0.1351 vs. 0.2510).

10. lny = β0 + β1x1 + β2x2 + ε, whereβ0 = lnα andε = lnδ.

11. (a)t = −0.58762/0.2873= −2.05.

(b) sβ1satisfies the equation 4.30= 1.5102/sβ1

, sosβ1= 0.3512.

(c) β2 satisfies the equation−0.62= β2/0.3944, soβ2 = −0.2445.

(d) t = 1.8233/0.3867= 4.72.

(e) MSR= 41.76/3= 13.92.

(f) F = MSR/MSE= 13.92/0.76= 18.316.


386 CHAPTER 8

(g) SSE= 46.30−41.76= 4.54.

(h) 3+6= 9.

12. (a)β0 satisfies the equation 5.91= β0/1.4553, soβ0 = 8.6008.


, sosβ1= 0.7092.

(c) t = 7.8369/3.2109= 2.44.

(d) β3 satisfies the equation−3.56= β3/0.8943, soβ3 = −3.1837

(e) 13−10= 3.

(f) SSR= 8.1291(3) = 24.3876.

(g) MSE= 6.8784/10= 0.68784.

(h) SST= SSR+SSE= 24.3876+6.8784= 31.2660.

13. (a)y = 267.53−1.5926(30)−1.3897(35)−1.0934(30)−0.002658(30)(30)= 135.92F

(b) No. The change in the predicted flash point due to a change in acetic acid concentration depends on the butyricacid concentration as well, because of the interaction between these two variables.

(c) Yes. The predicted flash point will change by−1.3897(10) = −13.897F.

14. lnp = β0 + β1t + β2(1/t)+ β3 ln t + ε, whereε = lnδ.

15. (a) The residuals are the valuesei = yi − yi for eachi. They are shown in the following table.


SECTION 8.1 387

Fitted Value Residualx y y e= y− y

138 5390 5399.17819 −9.17819140 5610 5611.42732 −1.42732146 5670 5546.73182 123.26818148 5140 5291.35236 −151.35236152 4480 4429.87200 50.12800153 4130 4141.43494 −11.43494

(b) SSE= ∑6i=1e2

i = 40832.432, SST= ∑ni=1(yi −y)2 = 1990600.

(c) s2 = SSE/(n−3) = 13610.811

(d) R2 = 1−SSE/SST= 0.979487

(e) F =SSR/2SSE/3

=(SST−SSE)/2

SSE/3= 71.6257. There are 2 and 3 degrees of freedom.

(f) Yes. From theF table, 0.001< P < 0.01. A computer package givesP = 0.0029. SinceP ≤ 0.05, thehypothesisH0 : β1 = β2 = 0 can be rejected at the 5% level.

16. (a) The residuals are the valuesei = yi − yi for eachi. They are shown in the following table.

Fitted Value Residualx y y e= y− y5 16 15.485712 0.514288

10 21 22.457139 −1.45713915 27 25.714280 1.28572020 25 25.257136 −0.25713625 21 21.085706 −0.085706

(b) SSE= ∑6i=1e2

i = 4.11429, SST= ∑ni=1(yi −y)2 = 72.0000.

(c) s2 = SSE/(n−3) = 2.05714

(d) R2 = 1−SSE/SST= 0.942857

(e) F =SSR/2SSE/2

= 16.5000. There are 2 and 2 degrees of freedom.


388 CHAPTER 8

(f) No, the upper 5% point of theF statistic with 2 and 2 degrees of freedom is 19.00, which is greater than 16.5.ThereforeP > 0.05, soH0 : β1 = β2 = 0 cannot be rejected at the 5% level.

17. (a)y = 1.18957+0.17326(0.5)+0.17918(5.7)+0.17591(3.0)−0.18393(4.1)= 2.0711.

(b) 0.17918

(c) PP is more useful, because itsP-value is small, while theP-value of CP is fairly large.

(d) The percent change in GDP would be expected to be larger inSweden, because the coefficient of PP is negative.

18. (a) Predictor Coef StDev T PConstant −372.98 242.37 −1.5389 0.158Days Ice Cover 3.5368 1.1948 2.9602 0.016Days Below 8 3.7345 1.1069 3.3738 0.008Avg Snow Depth −2.1661 0.85177 −2.5430 0.032

(b) By 3.5368(2) = 7.0736.

(c) Less thickness, because the coefficient is negative.

19. (a) Predictor Coef StDev T PConstant −0.012167 0.01034 −1.1766 0.278Time 0.043258 0.043186 1.0017 0.350Time2 2.9205 0.038261 76.33 0.000

y = −0.012167+0.043258t+2.9205t2

(b) β2 = 2.9205, sβ2= 0.038261. There aren−3= 7 degrees of freedom.

t7,.025 = 2.365. A 95% confidence interval is therefore 2.9205±2.365(0.038261), or (2.830, 3.011).

(c) Sincea = 2β2, the confidence limits for a 95% confidence interval fora are twice the limits of the confidenceinterval forβ2. Therefore a 95% confidence interval fora is (5.660, 6.022).

(d) β0: t7 = −1.1766,P = 0.278,β1: t7 = 1.0017,P = 0.350,β2: t7 = 76.33,P = 0.000.


SECTION 8.2 389

(e) No, theP-value of 0.278 is not small enough to reject the null hypothesis thatβ0 = 0.

(f) No, theP-value of 0.350 is not small enough to reject the null hypothesis thatβ1 = 0.

Section 8.2

1. (a) Predictor Coef StDev T PConstant 6.3347 2.1740 2.9138 0.009x1 1.2915 0.1392 9.2776 0.000

β0 differs from 0 (P = 0.009),β1 differs from 0 (P = 0.000).

(b) Predictor Coef StDev T PConstant 53.964 8.7737 6.1506 0.000x2 −0.9192 0.2821 −3.2580 0.004


(c) Predictor Coef StDev T PConstant 12.844 7.5139 1.7094 0.104x1 1.2029 0.1707 7.0479 0.000x2 −0.1682 0.1858 −0.90537 0.377

β0 may not differ from 0 (P = 0.104),β1 differs from 0 (P= 0.000),β2 may not differ from 0 (P = 0.377).

(d) The model in part (a) is the best. When bothx1 andx2 are in the model, only the coefficient ofx1 is significantlydifferent from 0. In addition, the value ofR2 is only slightly greater (0.819 vs. 0.811) for the model containingbothx1 andx2 than for the model containingx1 alone.

2. Torque and HP are most nearly collinear. We can tell because their coefficients are significantly different from0 when either one is in the model without the other, but not significantly different from 0 when both are in themodel together.

3. (a) Plot (i) came from Engineer B, and plot (ii) came from Engineer A. We know this because the variablesx1 andx2 are both significantly different from 0 for Engineer A but notfor Engineer B. Therefore Engineer B is theone who designed the experiment to have the dependent variables nearly collinear.


390 CHAPTER 8

(b) Engineer A’s experiment produced the more reliable results. In Engineer B’s experiment, the two dependentvariables are nearly collinear.

4. (a) Predictor Coef StDev T PConstant −1.6415 0.54133 −3.0323 0.013x1 3.1415 0.61125 5.1394 0.000


(b) Predictor Coef StDev T PConstant −1.3701 0.57649 −2.3765 0.039x2 2.8892 0.6634 4.3552 0.001


(c) Predictor Coef StDev T PConstant −1.8339 0.53319 −3.4395 0.007x1 2.0828 0.94564 2.2026 0.055x2 1.3006 0.91539 1.4209 0.189

β0 differs from 0 (P = 0.007), the evidence regardingβ1 is borderline (P = 0.055),β2 may not differ from 0(P = 0.189).

(d) The modely = β0 + β1x1 + ε. When bothx1 andx2 are in the model, the coefficient ofx2 is not significantlydifferent from 0. In addition, the value ofR2 is only slightly greater (0.776 vs. 0.725) for the model containingbothx1 andx2 than for the model containingx1 alone.

5. (a) ForR1 < 4, the least squares line isR2 = 1.233+0.264R1. ForR1 ≥ 4, the least squares line isR2 = −0.190+0.710R1.

(b) The relationship is clearly non-linear whenR1 < 4.

Quadratic ModelPredictor Coef StDev T PConstant 1.2840 0.26454 4.8536 0.000R1 0.21661 0.23558 0.91947 0.368R2

1 0.0090189 0.044984 0.20049 0.843

Cubic ModelPredictor Coef StDev T PConstant −1.8396 0.56292 −3.2680 0.004R1 4.4987 0.75218 5.9809 0.000R2

1 −1.7709 0.30789 −5.7518 0.000R3

1 0.22904 0.039454 5.8053 0.000


SECTION 8.2 391

Quartic ModelPredictor Coef StDev T PConstant −2.6714 2.0117 −1.3279 0.200R1 6.0208 3.6106 1.6675 0.112R2

1 −2.7520 2.2957 −1.1988 0.245R3

1 0.49423 0.61599 0.80234 0.432R4

1 −0.02558 0.05930 −0.43143 0.671

(c)

1.5 2 2.5−0.5

0

0.5

Fitted Value

Re

sid

ua

l

Quadratic Model

1 1.5 2 2.5−0.5

0

0.5

Fitted Value

Re

sid

ua

l

Cubic Model

1 1.5 2 2.5−0.5

0

0.5

Fitted Value

Re

sid

ua

l

Quartic Model

(d)

0 20 40 600

50

100

150

200

250

R13

R14

The correlation coefficient betweenR31 andR4

1is 0.997.

(e) R31 andR4

1 are nearly collinear.

(f) The cubic model is best. The quadratic is inappropriate because the residual plot exhibits a pattern. The residualplots for both the cubic and quartic models look good, however, there is no reason to includeR4

1 in the modelsince it merely confounds the effect ofR3

1.

6. (a) Predictor Coef StDev T PConstant 95.9120 5.0911 18.839 0.000x1 −7.3597 1.4259 −5.1614 0.000


392 CHAPTER 8

Yes, there is a linear relationship.

(b) Predictor Coef StDev T PConstant 119.14 6.8619 17.363 0.000x2 −0.68333 0.095153 −7.1814 0.000

Yes, there is a linear relationship.

(c) Predictor Coef StDev T PConstant 120.15 7.4121 16.210 0.000x1 0.89597 2.3684 0.37831 0.707x2 −0.74088 0.17982 −4.1202 0.000

(d) x2. If x2 were known, it would not be much additional help to knowx1. The value of the coefficient ofx1

is not significantly different from 0 whenx2 is in the model. Equivalently, the value ofR2 does not increasesignificantly whenx1 is added to the model containingx2. The value ofR2 for the model in part (b) is 0.47067,and the value ofR2 for the model in part (c) is 0.47199.

Section 8.3

1. (a) False. There are usually several models that are aboutequally good.

(b) True.

(c) False. Model selection methods can suggest models that fit the data well.

(d) True.

2. (iii). x1x2, x1x3, andx2x3 all have largeP-values and thus may not contribute significantly to the fit.

3. (v). x22, x1x2, andx1x3 all have largeP-values and thus may not contribute significantly to the fit.

4. (iv). x1 andx3 both have largeP-values and thus may not contribute significantly to the fit.


SECTION 8.3 393

5. (i). Rest, Light*Rest, and Sit*Rest all have largeP-values and thus may not contribute significantly to the fit.

6. (a)X1, X3, andX4

(b) X1, X2, X3, andX4

(c) Yes, for example the model containingX1, X3, X4, andX5 is reasonable.

7. The four-variable model with the highest value ofR2 has a lowerR2 than the three-variable model with thehighest value ofR2. This is impossible.

8. SSEfull = 4.6409, SSEreduced= 11.5820, n = 12, p = 6, k = 3. The value of theF statistic for testing theplausibility of the reduced model is

F =(SSEreduced−SSEfull)/(p−k)

SSEfull/(n− p−1)= 2.4927.

There arep−k = 3 andn− p−1= 5 degrees of freedom.


The reduced model is plausible.

9. (a) SSEfull = 7.7302, SSEreduced= 7.7716, n = 165, p = 7, k = 4.

F =(SSEreduced−SSEfull)/(p−k)

SSEfull/(n− p−1)= 0.2803.

(b) 3 degrees of freedom in the numerator and 157 in the denominator.

(c) P > 0.10 (a computer package givesP = 0.840). The reduced model is plausible.

(d) This is not correct. It is possible for a group of variables to be fairly strongly related to an independent variable,even though none of the variables individually is strongly related.

(e) No mistake. Ify is the dependent variable, then the total sum of squares is∑(yi −y)2. This quantity does notinvolve the independent variables.


394 CHAPTER 8

10. (a) Predictor Coef StDev T PConstant −3.4641 1.2468 −2.7785 0.032pH 0.47196 0.20246 2.3311 0.059

Analysis of VarianceSource DF SS MS F PRegression 1 2.3478 2.3478 5.4341 0.059Residual Error 6 2.5923 0.43205Total 7 4.9401

(b) Predictor Coef StDev T PConstant −20.485 1.2132 −16.885 0.000pH 5.8329 0.37782 15.438 0.000pH2 −0.40645 0.028526 −14.248 0.000


(c) Predictor Coef StDev T PConstant −32.027 8.544 −3.7484 0.020pH 11.383 4.0878 2.7845 0.050pH2 −1.276 0.63871 −1.9978 0.116pH3 0.044329 0.032532 1.3626 0.245


(d) The quadratic model (b) is best. TheF statistic for the plausibility of the quadratic model against the cubicmodel is(0.062311− 0.042556)/0.010639= 1.857. The degrees of freedom are 1 and 4, soP > 0.10 (acomputer package givesP = 0.245). Therefore adding the cubic term is not justified. TheF statistic for theplausibility of the linear model against the quadratic model is (2.5923−0.062311)/0.012462= 203.02. Thedegrees of freedom are 1 and 5, soP < 0.001 (a computer package givesP = 3.1× 10−5). Therefore thequadratic term is necessary; the linear model is not sufficient.

11. (a) Predictor Coef StDev T PConstant 0.087324 0.16403 0.53237 0.597x 0.52203 0.055003 9.4909 0.000x2 −0.02333 0.0043235 −5.396 0.000


SECTION 8.3 395

(b) Predictor Coef StDev T PConstant 0.19878 0.11677 1.7023 0.096lnx 1.2066 0.068272 17.673 0.000

(c)

0 1 2 3 4−1

−0.5

0

0.5

1

Fitted Value

Re

sid

ua

l

Quadratic Model

0 1 2 3 4−1

−0.5

0

0.5

1

Fitted ValueR

esid

ua

l

Log Model

Neither residual plot reveals gross violations of assumptions.

(d) Quadratic Model Log Modelx Prediction Prediction

3.0 1.44 1.525.0 2.11 2.147.0 2.60 2.559.0 2.90 2.85

(e) TheF-test cannot be used. TheF-test can only be used when one model is formed by dropping oneor moreindependent variables from another model.

(f) The predictions of the two models do not differ much. The log model has only one independent variable insteadof two, which is an advantage.

12. (a) Predictor Coef StDev T PConstant 6.6320 1.6806 3.9462 0.002ln Sc. Dist. −1.6778 0.36395 −4.6101 0.001ln Charge −0.19583 0.20846 −0.9394 0.366

(b) Predictor Coef StDev T PConstant 5.2782 0.86076 6.1320 0.000ln Sc. Dist. −1.4652 0.28366 −5.1653 0.000


396 CHAPTER 8

(c) The model in part (b) is preferred. In model (a), the coefficient of the variable ln Charge may not be differentfrom 0. It is better to leave this variable out of the model.

13. (a) Predictor Coef StDev T PConstant 37.989 53.502 0.71004 0.487x 1.0774 0.041608 25.894 0.000

(b) Predictor Coef StDev T PConstant −253.45 132.93 −1.9067 0.074x 1.592 0.22215 7.1665 0.000x2 −0.00020052 0.000085328 −2.3499 0.031

(c)

500 1000 1500 2000−150

−100

−50

0

50

100

150

200

Fitted Value

Re

sid

ua

l

Linear Model

(d)

500 1000 1500 2000 2500−150

−100

−50

0

50

100

150

Fitted Value

Re

sid

ua

l

Quadratic Model

(e) The quadratic model seems more appropriate. TheP-value for the quadratic term is fairly small (0.031), andthe residual plot for the quadratic model exhibits less pattern. (There are a couple of points somewhat detachedfrom the rest of the plot, however.)


SECTION 8.3 397

(f) 1683.5

(g) (1634.7, 1732.2)

14. (a) Predictor Coef StDev T PConstant 0.046975 0.021292 2.2062 0.041x1 −0.00084581 0.00028086 −3.0115 0.008x2 0.00041568 0.00017662 2.3536 0.031

y = 0.046975−0.00084581x1+0.00041568x2.

−0.1 0 0.1−0.2

0

0.2

Fitted Value

Re

sid

ua

l

The residual plot shows a definite pattern. The model is notappropriate.

(b) Predictor Coef StDev T PConstant −1.8898 1.761 −1.0732 0.298lnx1 −3.0315 0.45905 −6.6039 0.000lnx2 2.06 0.46648 4.416 0.000

lny = −1.8898−3.0315x1+2.0600x2.

−10 −8 −6 −4 −2−2

−1

0

1

2

Fitted Value

Re

sid

ua

l

The residual plot shows no definite pattern. The model appearsto be appropriate.

(c) lny = −1.8898−3.0315ln50+2.0600ln100= −4.2624, soy = e−4.2624= 0.01409.


398 CHAPTER 8

(d) Predictor Coef StDev T PConstant −5.6416 10.491 −0.53777 0.598lnx1 −2.0919 2.6307 −0.79518 0.438lnx2 2.9432 2.4793 1.1871 0.253lnx1 lnx2 −0.2174 0.5988 −0.36306 0.721

No. TheP-value for testing the null hypothesis that the coefficient of the interaction term lnx1 lnx2 is zero is0.721. Therefore the interaction term should not be included.

15. (a) Predictor Coef StDev T PConstant 25.613 10.424 2.4572 0.044x1 0.18387 0.12353 1.4885 0.180x2 −0.015878 0.0040542 −3.9164 0.006

(b) Predictor Coef StDev T PConstant 14.444 16.754 0.86215 0.414x1 0.17334 0.20637 0.83993 0.425

(c) Predictor Coef StDev T PConstant 40.370 3.4545 11.686 0.000x2 −0.015747 0.0043503 −3.6197 0.007

(d) The model containingx2 as the only independent variable is best. There is no evidence that the coefficient ofx1 differs from 0.

16. Answers will vary. One good model isy = 73.228+0.17511x2.

Predictor Coef StDev T PConstant 73.228 2.0671 35.425 0.000x2 0.17511 0.063783 2.7454 0.025

17. The modely = β0 + β1x2 + ε is a good one. One way to see this is to compare the fit of this model to the fullquadratic model. The ANOVA table for the full model is

Source DF SS MS F PRegression 5 4.1007 0.82013 1.881 0.193Residual Error 9 3.9241 0.43601Total 14 8.0248

The ANOVA table for the modely = β0 + β1x2 + ε is



Source DF SS MS F PRegression 1 2.7636 2.7636 6.8285 0.021Residual Error 13 5.2612 0.40471Total 14 8.0248

From these two tables, theF statistic for testing the plausibility of the reduced modelis

(5.2612−3.9241)/(5−1)

3.9241/9= 0.7667.

The null distribution isF4,9, soP > 0.10 (a computer package givesP = 0.573). The largeP-value indicatesthat the reduced model is plausible.

18. (a) False

(b) True

(c) False

(d) True


1. (a) y = 46.802−130.11(0.15)−807.10(0.01)+3580.5(0.15)(0.01)= 24.6%.

(b) By 130.11(0.05)−3580.5(0.006)(0.05)= 5.43%.

(c) No, we need to know the oxygen content, because of the interaction term.

2. (a) β1 = −130.11, sβ1= 20.467.

There are 35 degrees of freedom for error.t35,.025= 2.030.


(b) β2 = −807.10, sβ2= 158.03.

There are 35 degrees of freedom for error.t35,.005= 2.724.



400 CHAPTER 8

(c) β3 = 3580.5, sβ3= 958.05.

There are 35 degrees of freedom for error.t35,.01 = 2.438.

A 98% confidence interval is 3580.5±2.438(958.05), or (1244.8, 5916.2).

(d) β1 = −130.11, sβ1= 20.467.

The null and alternate hypotheses areH0 : β1 ≥−75 versusH1 : β1 < −75.

There are 35 degrees of freedom for error.t = [−130.11− (−75)]/20.467= −2.693.

Since the alternate hypothesis is of the formβ1 < b, theP-value is the area to the left oft = −2.693. From thet table, 0.005 −1000.

There are 35 degrees of freedom for error.t = [−807.10− (−1000)]/158.03= 1.221.

Since the alternate hypothesis is of the formβ2 > b, theP-value is the area to the right oft = 1.221. From thet table, 0.10 −1000.

3. (a) β0 satisfies the equation 0.59= β0/0.3501, soβ0 = 0.207.


, sosβ1= 0.8015.

(c) t = 2.7241/0.7124= 3.82.

(d) s=√

MSE=√

1.44= 1.200.

(e) There are 2 independent variables in the model, so there are 2 degrees of freedom for regression.

(f) SSR= SST−SSE= 104.09−17.28= 86.81.

(g) MSR= 86.81/2= 43.405.

(h) F = MSR/MSE= 43.405/1.44= 30.14.

(i) 2+12= 14.



4. The colleague is right. The models with the second or thirdbestR2 or Mallows’ Cp are often just as good inpractice as the model with the best values.

5. (a) Predictor Coef StDev T PConstant 10.84 0.2749 39.432 0.000Speed −0.073851 0.023379 −3.1589 0.004Pause −0.12743 0.013934 −9.1456 0.000Speed2 0.0011098 0.00048887 2.2702 0.032Pause2 0.0016736 0.00024304 6.8861 0.000Speed·Pause −0.00024272 0.00027719 −0.87563 0.390

S = 0.33205 R-sq = 92.2% R-sq(adj) = 90.6%


(b) We drop the interaction term Speed·Pause.

Predictor Coef StDev T PConstant 10.967 0.23213 47.246 0.000Speed −0.079919 0.022223 −3.5961 0.001Pause −0.13253 0.01260 −10.518 0.000Speed2 0.0011098 0.00048658 2.2809 0.031Pause2 0.0016736 0.0002419 6.9185 0.000

S = 0.33050 R-sq = 92.0% R-sq(adj) = 90.7%


Comparing this model with the one in part (a),F1,24 =(2.7307−2.6462)/(5−4)

2.6462/24= 0.77,P > 0.10. A com-

puter package givesP = 0.390 (the same as theP-value for the dropped variable).


402 CHAPTER 8

(c)

6 7 8 9 10 11−1

−0.5

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There is a some suggestion of heteroscedas-ticity, but it is hard to be sure without moredata.

(d)

Predictor Coef StDev T PConstant 9.9601 0.21842 45.601 0.000Pause -0.13253 0.020545 -6.4507 0.000Pause2 0.0016736 0.00039442 4.2431 0.000

S = 0.53888 R-sq = 76.9% R-sq(adj) = 75.2%


Comparing this model with the one in part (a),F3,24 =(7.8405−2.6462)/(5−2)

2.6462/24= 15.70, P < 0.001. A

computer package givesP = 7.3×10−6.



(e)

Speed

S P *S P p a Pp a e u ae u e s ue s d e s

Vars R-Sq R-Sq(adj) C-p S d e 2 2 e1 61.5 60.1 92.5 0.68318 X1 60.0 58.6 97.0 0.69600 X2 76.9 75.2 47.1 0.53888 X X2 74.9 73.0 53.3 0.56198 X X3 90.3 89.2 7.9 0.35621 X X X3 87.8 86.4 15.5 0.39903 X X X4 92.0 90.7 4.8 0.33050 X X X X4 90.5 89.0 9.2 0.35858 X X X X5 92.2 90.6 6.0 0.33205 X X X X X

(f) The model containing the dependent variables Speed, Pause, Speed2 and Pause2 has both the lowest value ofCp and the largest value of adjustedR2.

6. (a) Predictor Coef StDev T PConstant 13.135 4.8104 2.7305 0.008x1 −0.14429 0.13189 −1.094 0.277x2 0.75476 0.27414 2.7533 0.007x3 1.6266 0.33675 4.8304 0.000

(b) We dropx1:

Predictor Coef StDev T PConstant 10.517 4.1776 2.5174 0.014x2 0.49962 0.14425 3.4637 0.001x3 1.6370 0.33699 4.8577 0.000

(c) Predictor Coef StDev T PConstant 17.306 12.490 1.3856 0.169x2 0.22459 0.49812 0.45087 0.653x3 0.54875 1.9160 0.2864 0.775x2x3 0.043892 0.07606 0.57706 0.565

(d) y = 10.517+0.49962x2+1.6370x3


404 CHAPTER 8

(e) Yes

7.

100 200 300 400−80

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The residual plot shows an obvious curvedpattern, so the linear model is not appropri-ate.

0 100 200 300 400−20

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20Quadratic Model

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There is no obvious pattern to the residualplot, so the quadratic model appears to fitwell.

0 100 200 300 400−15

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15Cubic Model

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There is no obvious pattern to the residualplot, so the cubic model appears to fit well.

8. (i) The linear model is best. The plot shows that all three models make nearly identical predictions, so the bestmodel is the one that contains the fewest independent variables.



9. (a) Under model 1, the prediction is−320.59+0.37820(1500)−0.16047(400)= 182.52.

Under model 2, the prediction is−380.1+0.41641(1500)−0.5198(150)= 166.55.

Under model 3, the prediction is−301.8+0.3660(1500)−0.2106(400)+0.164(150)= 187.56.

(b) Under model 1, the prediction is−320.59+0.37820(1600)−0.16047(300)= 236.39.



(c) Under model 1, the prediction is−320.59+0.37820(1400)−0.16047(200)= 176.80.



(d) (iv). The output does not provide much to choose from between the two-variable models. In the three-variablemodel, none of the coefficients are significantly different from 0, even though they were significant in thetwo-variable models. This suggest collinearity.

10. (a) Yes, the new model isy = β∗0 + β∗

1F + ε, whereβ∗0 = −57.6β1 andβ∗

1 = 1.8β1.

(b) Yes, the new model isy = β∗0−β∗

1F +β∗2F2 + ε, whereβ∗

0 = β0 +57.62β2, β∗1 = −207.36β2, andβ∗

2 = 3.24β2.

(c) If y = βz theny = β(a+bx) 6= βx.

(d) If y = β0 + β2z2 theny = β0 + β2(a+bx)2 6= β0 + β2x2.

11. (a) Linear ModelPredictor Coef StDev T PConstant 40.751 5.4533 7.4728 0.000Hardwood 0.54013 0.61141 0.88341 0.389

S = 12.308 R-sq = 4.2% R-sq(adj) =−1.2%



406 CHAPTER 8

Quadratic ModelPredictor Coef StDev T PConstant 12.683 3.9388 3.2199 0.005Hardwood 10.067 1.1028 9.1287 0.000Hardwood2 −0.56928 0.063974 −8.8986 0.000

S = 5.3242 R-sq = 83.1% R-sq(adj) = 81.1%


Cubic ModelPredictor Coef StDev T PConstant 27.937 2.9175 9.5755 0.000Hardwood 0.48749 1.453 0.3355 0.742Hardwood2 0.85104 0.20165 4.2204 0.001Hardwood3 −0.057254 0.0080239 −7.1354 0.000

S = 2.6836 R-sq = 95.9% R-sq(adj) = 95.2%


Quartic ModelPredictor Coef StDev T PConstant 30.368 4.6469 6.5351 0.000Hardwood −1.7962 3.6697 −0.48946 0.632Hardwood2 1.4211 0.8632 1.6463 0.120Hardwood3 −0.10878 0.076229 −1.4271 0.174Hardwood4 0.0015256 0.0022438 0.67989 0.507

S = 2.7299 R-sq = 96.1% R-sq(adj) = 95.0%

Analysis of VarianceSource DF SS MS F PRegression 4 2733 683.24 91.683 0.00Residual Error 15 111.78 7.4522Total 19 2844.8

The values of SSE and their degrees of freedom for models of degrees 1, 2, 3, and 4 are:



Linear 18 2726.55Quadratic 17 481.90Cubic 16 115.23Quartic 15 111.78

To compare quadratic vs. linear,F1,17 =(2726.55−481.90)/(18−17)

481.90/17= 79.185.

P≈ 0. A computer package givesP = 8.3×10−8.

To compare cubic vs. quadratic,F1,16 =(481.90−115.23)/(17−16)

115.23/16= 50.913.

P≈ 0. A computer package givesP = 2.4×10−6.

To compare quartic vs. cubic,F1,15 =(115.23−111.78)/(16−15)

111.78/15= 0.463.

P > 0.10. A computer package givesP = 0.507.

The cubic model is selected by this procedure.

(b) The cubic model isy = 27.937+ 0.48749x+ 0.85104x2− 0.057254x3. The estimatey is maximized whendy/dx = 0. dy/dx = 0.48749+ 1.70208x− 0.171762x2. Thereforex = 10.188 (x = −0.2786 is a spuriousroot).

12. (a) Linear ModelPredictor Coef StDev T PConstant 16.887 4.1077 4.111 0.003x −1.2482 0.069411 −17.983 0.000

Quadratic ModelPredictor Coef StDev T PConstant 3.814 1.234 3.0908 0.015x −0.44982 0.057106 −7.8769 0.000x2 −0.0077691 0.00053834 −14.432 0.000

Cubic ModelPredictor Coef StDev T PConstant 5.4977 1.6043 3.4268 0.011x −0.64732 0.14185 −4.5635 0.003x2 −0.003024 0.0031997 −0.94511 0.376x3 −0.000030317 0.000020192 −1.5015 0.177

(b) The quadratic model is most appropriate. This can be seenby noticing that the coefficient ofx3 in the cubicmodel is not significantly different from 0. Another way to see this is to perform successiveF tests, comparingeach model with the model of degree one less.


408 CHAPTER 8

(c) There are 11 observations, and 2 independent variables,so there are 11−2−1= 8 degrees of freedom for errorin the quadratic model.

t8,.005 = 3.355.

A 99% confidence interval forβ0 is 3.814±3.355(1.234), or (−0.326,7.95).

A 99% confidence interval forβ1 is−0.44982±3.355(0.057106), or (−0.641,−0.258).

A 99% confidence interval forβ2 is−0.0077691±3.355(0.00053834), or (−0.009575,−0.005963).

13. (a) Lety1 represent the lifetime of the sponsor’s paint,y2 represent the lifetime of the competitor’s paint,x1 repre-sent January temperature,x2 represent July temperature, andx3 represent Precipitation.

Then one good model fory1 is y1 = −4.2342+ 0.79037x1 + 0.20554x2− 0.082363x3− 0.0079983x1x2 −0.0018349x2

1.

A good model fory2 is y2 = 6.8544+0.58898x1+0.054759x2−0.15058x3−0.0046519x1x2+0.0019029x1x3−0.0035069x2

1.

(b) Substitutex1 = 26.1, x2 = 68.9, andx3 = 13.3 to obtainy1 = 13.83 andy2 = 13.90.

14. (a) Linear ModelPredictor Coef StDev T PConstant −171.9 1.9125 −89.885 0.000x 209.66 2.3188 90.417 0.000

(b) Quadratic ModelPredictor Coef StDev T PConstant 453.22 53.355 8.4945 0.000x −1306.7 129.42 −10.097 0.000x2 919.47 78.475 11.717 0.000

(c) Cubic ModelPredictor Coef StDev T PConstant −12099 5832.2 −2.0744 0.107x 44365 21221 2.0906 0.105x2 −54472 25737 −2.1165 0.102x3 22392 10405 2.1522 0.098

Note: The design matrix for the cubic model is highly ill-conditioned, so the values calculated for the coeffi-cients and standard deviations are unstable.

(d) The quadratic model. The coefficients are not significantin the cubic model.



(e) The log of the flow rate is estimated to bey = 453.22−1306.7(0.83)+919.47(0.832) = 2.1, so the flow rateis estimated to bee2.1 = 8.2.

15. (a) Linear ModelPredictor Coef StDev T PConstant 0.25317 0.0065217 38.819 0.000x −0.041561 0.040281 −1.0318 0.320

(b) Quadratic ModelPredictor Coef StDev T PConstant 0.21995 0.0038434 57.23 0.000x 0.58931 0.06146 9.5886 0.000x2 −2.2679 0.2155 −10.524 0.000

(c) Cubic ModelPredictor Coef StDev T PConstant 0.22514 0.0068959 32.648 0.000x 0.41105 0.20576 1.9977 0.069x2 −0.74651 1.6887 −0.44206 0.666x3 −3.6728 4.043 −0.90843 0.382

(d) Quartic ModelPredictor Coef StDev T PConstant 0.23152 0.013498 17.152 0.000x 0.10911 0.58342 0.18702 0.855x2 3.4544 7.7602 0.44515 0.665x3 −26.022 40.45 −0.64333 0.533x4 40.157 72.293 0.55548 0.590

(e) The quadratic model. The coefficient ofx3 in the cubic model is not significantly different from 0. Neither isthe coefficient ofx4 in the quartic model.

(f) y = 0.21995+0.58931(0.12)−2.2679(0.122) = 0.258

16. (a) Predictor Coef StDev T PConstant 0.094806 3.0827 0.030754 0.976x1 0.096663 0.17372 0.55643 0.585x2 −0.074354 0.071246 −1.0436 0.310x1x2 0.0034715 0.0042389 0.81896 0.424


410 CHAPTER 8


(b) Predictor Coef StDev T PConstant −1.9749 1.7499 −1.1286 0.273x1 0.21634 0.093111 2.3235 0.031x2 −0.017619 0.016492 −1.0683 0.299


(c) Predictor Coef StDev T PConstant −3.0827 1.4146 −2.1792 0.041x1 0.25439 0.086334 2.9466 0.008


(d) For comparing (b) with (a),F =(50.888−49.06)/(3−2)

49.06/18= 0.671.

There are 1 and 18 degrees of freedom.


For comparing (c) with (a),F =(53.945−49.06)/(3−1)

49.06/18= 0.896.

There are 2 and 18 degrees of freedom.


The comparison of (c) with (a) shows that it is reasonable to drop bothx2 andx1x2 from the model. Thereforemodel (c) is preferred.



17. (a) Predictor Coef StDev T PConstant −0.093765 0.092621 −1.0123 0.335x1 0.63318 2.2088 0.28666 0.780x2 2.5095 0.30151 8.3233 0.000x2

1 5.318 8.2231 0.64672 0.532x2

2 −0.3214 0.17396 −1.8475 0.094x1x2 0.15209 1.5778 0.09639 0.925


(b) The model containing the variablesx1, x2, andx22 is a good one. Here are the coefficients along with their

standard deviations, followed by the analysis of variance table.

Predictor Coef StDev T PConstant −0.088618 0.068181 −1.2997 0.218x1 2.1282 0.30057 7.0805 0.000x2 2.4079 0.13985 17.218 0.000x2

2 −0.27994 0.059211 −4.7279 0.000


TheF statistic for comparing this model to the full quadratic model is

F2,10 =(0.048329−0.045513)/(12−10)

0.045513/10= 0.309,P > 0.10,

so it is reasonable to dropx21 andx1x2 from the full quadratic model. All the remaining coefficients are signifi-

cantly different from 0, so it would not be reasonable to reduce the model further.

(c) The output from the MINITAB best subsets procedure is

Response is y

x x x1 2 1

Mallows x x ˆ ˆ xVars R-Sq R-Sq(adj) C-p S 1 2 2 2 2

1 98.4 98.2 61.6 0.15470 X1 91.8 91.2 354.1 0.34497 X2 99.3 99.2 20.4 0.10316 X X


412 CHAPTER 8

2 99.2 99.1 25.7 0.11182 X X3 99.8 99.7 2.2 0.062169 X X X3 99.8 99.7 2.6 0.063462 X X X4 99.8 99.7 4.0 0.064354 X X X X4 99.8 99.7 4.1 0.064588 X X X X5 99.8 99.7 6.0 0.067463 X X X X X

The model with the best adjustedR2 (0.99716) contains the variablesx2, x21, andx2

2. This model is also themodel with the smallest value of Mallows’ Cp (2.2). This is not the best model, since it containsx2

1 but notx1.The model containingx1, x2, andx2

2, suggested in the answer to part (b), is better. Note that theadjustedR2 forthe model in part (b) is 0.99704, which differs negligibly from that of the model with the largest adjustedR2

value.

18. (a) Linear ModelPredictor Coef StDev T PConstant 10.408 3.9496 2.6353 0.027x 6.7508 0.58519 11.536 0.000

0 20 40 60 80 100−15

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The least-squares line isy = 10.408+ 6.5708x. Theresidual plot shows a curved pattern, so the linearmodel is not appropriate.

(b) Log-linear ModelPredictor Coef StDev T PConstant 1.5037 1.107 1.3584 0.207lnx 31.397 0.64619 48.589 0.000



0 20 40 60 80−2

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The least-squares line isy = 1.5037+ 31.397ln(x).The residual plot shows a curved pattern, so the log-linear model is not appropriate.

(c) Quadratic ModelPredictor Coef StDev T PConstant −4.3677 2.7567 −1.5844 0.152x 13.733 1.0758 12.766 0.000x2 −0.575 0.086308 −6.6622 0.000

0 20 40 60 80−4

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The residual plot shows a curved pattern, so thequadratic model is not appropriate.

Cubic ModelPredictor Coef StDev T PConstant −14.542 1.3195 −11.021 0.000x 22.112 0.92605 23.878 0.000x2 −2.2255 0.17391 −12.797 0.000x3 0.089953 0.0093835 9.5863 0.000


414 CHAPTER 8

0 20 40 60 80 100−1

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The residual plot shows a curved pattern, so the cubicmodel is not appropriate.

Quartic ModelPredictor Coef StDev T PConstant −19.005 0.94669 −20.075 0.000x 27.258 0.96693 28.19 0.000x2 −3.9152 0.29976 −13.061 0.000x3 0.29679 0.035807 8.2885 0.000x4 −0.0084092 0.001447 −5.8116 0.001

0 20 40 60 80 100−0.6

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The residual plot for the quartic model is acceptable.

Quintic ModelPredictor Coef StDev T PConstant −18.657 1.8428 −10.124 0.000x 26.743 2.4948 10.719 0.000x2 −3.6711 1.1193 −3.2797 0.022x3 0.24775 0.21856 1.1335 0.308x4 −0.0040467 0.019197 −0.2108 0.841x5 −0.00014162 0.00062108 −0.22802 0.829



0 20 40 60 80 100−0.6

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The residual plot for the quintic model is acceptable.

(d) Yes. All the coefficients of the quartic model are significantly different from 0, while several of the coefficientsof the quintic model are not.

19. (a) Predictor Coef StDev T PConstant 1.1623 0.17042 6.8201 0.006t 0.059718 0.0088901 6.7174 0.007t2 −0.00027482 0.000069662 −3.9450 0.029

(b) Letx be the time at which the reaction rate will be equal to 0.05. Then 0.059718−2(0.00027482)x= 0.05, sox = 17.68 minutes.

(c) β1 = 0.059718, sβ1= 0.0088901.

There are 6 observations and 2 dependent variables, so thereare 6−2−1= 3 degrees of freedom for error.

t3,.025 = 3.182.

A 95% confidence interval is 0.059718±3.182(0.0088901), or (0.0314, 0.0880).

(d) The reaction rate is decreasing with time ifβ2 < 0. We therefore testH0 : β2 ≥ 0 versusH1 : β2 < 0.

From the output, the test statistic for testingH0 : β2 = 0 versusH1 : β2 6= 0 is ist = −3.945.

The output givesP = 0.029, but this is the value for a two-tailed test.

For the one-tailed test,P = 0.029/2= 0.0145.

It is reasonable to conclude that the reaction rate decreases with time.

20. (a) Predictor Coef StDev T PConstant −5.3862 11.184 −0.48160 0.638x1 −0.3545 1.0714 −0.33087 0.746x2 5.7697 2.8274 2.0406 0.061


416 CHAPTER 8

(b) Predictor Coef StDev T PConstant −6.7265 10.11 −0.66532 0.516x2 5.0425 1.7251 2.923 0.010

(c)

10 15 20 25 30 35−10

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The two uppermost points in the plot could be char-acterized as outliers. If these are dropped and themodel is refit, the least-squares line isy = −2.4236+4.0416x2. These coefficients are well inside the 95%confidence intervals computed from the data that in-cludes the outliers. Therefore it is reasonable to con-clude that the outliers are not particularly influential,and to accept the model fit to the entire data set.

21. y = β0 + β1x1 + β2x2 + β3x1x2 + ε.

22. There are several good models. The variabley should be transformed to√

y or perhaps lny to avoid het-eroscedasticity. Models that obey the exceptions to the principle of parsimony are better than ones that havebetter goodness-of-fit statistics such as Mallows’Cp or adjustedR2, but which do not obey these exceptions.One reasonably good model is

√y = −6.7381+ 1.7014m−0.016352d+ 3.1571s2−0.67773ms2. There are

others.

23. (a) The 17-variable model containing the independent variablesx1, x2, x3, x6, x7, x8, x9, x11, x13, x14, x16, x18, x19,x20, x21, x22, andx23 has adjustedR2 equal to 0.98446. The fitted model is

y = −1569.8−24.909x1+196.95x2+8.8669x3−2.2359x6

−0.077581x7+0.057329x8−1.3057x9−12.227x11+44.143x13

+4.1883x14+0.97071x16+74.775x18+21.656x19−18.253x20

+82.591x21−37.553x22+329.8x23

(b) The 8-variable model containing the independent variablesx1, x2, x5, x8, x10, x11, x14, andx21 has Mallows’ Cp

equal to 1.7. The fitted model is

y = −665.98−24.782x1+76.499x2+121.96x5+0.024247x8+20.4x10−7.1313x11+2.4466x14+47.85x21



(c) Using a value of 0.15 for bothα-to-enter andα-to-remove, the equation chosen by stepwise regression isy = −927.72+142.40x5+0.081701x7+21.698x10+0.41270x16+45.672x21.

(d) The 13-variable model below has adjustedR2 equal to 0.95402. (There are also two 12-variable models whoseadjustedR2 is only very slightly lower.)

z = 8663.2−313.31x3−14.46x6+0.358x7−0.078746x8

+13.998x9+230.24x10−188.16x13+5.4133x14+1928.2x15

−8.2533x16+294.94x19+129.79x22−3020.7x23

(e) The 2-variable modelz= −1660.9+0.67152x7+134.28x10 has Mallows’ Cp equal to−4.0.

(f) Using a value of 0.15 for bothα-to-enter andα-to-remove, the equation chosen by stepwise regression isz= −1660.9+0.67152x7+134.28x10

(g) The 17-variable model below has adjustedR2 equal to 0.97783.

w = 700.56−21.701x2−20.000x3+21.813x4+62.599x5+0.016156x7−0.012689x8

+1.1315x9+15.245x10+1.1103x11−20.523x13−90.189x15−0.77442x16+7.5559x19

+5.9163x20−7.5497x21+12.994x22−271.32x23

(h) The 13-variable model below has Mallows’ Cp equal to 8.0.

w = 567.06−23.582x2−16.766x3+90.482x5+0.0082274x7−0.011004x8+0.89554x9

+12.131x10−11.984x13−0.67302x16+11.097x19+4.6448x20+11.108x22−217.82x23

(i) Using a value of 0.15 for bothα-to-enter andα-to-remove, the equation chosen by stepwise regression isw = 130.92−28.085x2+113.49x5+0.16802x9−0.20216x16+11.417x19+12.068x21−78.371x23.

24. (a, b) The predicted values and prediction errors for thenew dataset are:


418 CHAPTER 8

y Pred Error1713 508.99 1204344 296.3 47.695474 369.33 104.67

1336 1280.9 55.0631916 668.93 1247.11280 557.7 722.31683 1512.5 170.52901 826.94 74.062460 −88.779 548.78826 599 227

(c) The fitted values and residuals for the 28 row dataset are:

y Fit Resid y Fit Resid1850 1900.3 −50.307 214 266.69 −52.693932 895.21 36.793 300 272.96 27.038556 518.32 37.676 771 710.37 60.632217 256.95 −39.953 189 181.04 7.9603199 237.73 −38.733 494 574.73 −80.73456 7.0631 48.937 389 395.78 −6.776764 74.546 −10.546 441 451.23 −10.225

397 404.85 −7.8539 768 746.81 21.191926 1923.3 2.7333 797 786.35 10.652724 722.13 1.8669 261 238.32 22.684711 716.06 −5.0634 524 495.5 28.501

1818 1768.1 49.891 1262 1253.3 8.6612619 674.77 −55.772 839 776.76 62.24375 374.13 0.8659 1003 1072.7 −69.667

(d) The prediction errors are on the whole larger than the residuals. The model is chosen to fit the first data set,in other words, the model is chosen to make the residuals small. The fit to another data set will therefore ingeneral be less good.


SECTION 9.1 419

Chapter 9

Section 9.1

1. (a) Source DF SS MS F PDuration 4 1.3280 0.33200 7.1143 0.002Error 15 0.7000 0.046667Total 19 2.0280

(b) Yes.F4,15 = 7.1143, 0.001< P < 0.01 (P= 0.002).

2. (a) Source DF SS MS F PCooling Rate 2 272.40 136.20 3.818 0.038Error 22 784.81 35.673Total 24 1057.2

(b) Yes.F2,22 = 3.818, 0.01 0.10 (P = 0.117).

4. (a) Source DF SS MS F PDay 3 2264.1 754.72 3.7515 0.033Error 16 3218.8 201.18Total 19 5482.9

(b) Yes.F3,16 = 3.7515, 0.01 0.10 (P = 0.111).

6. (a) Source DF SS MS F PAge 3 20.964 6.988 0.89095 0.453Error 45 352.95 7.8433Total 48 373.91

(b) No. F3,45 = 0.89095,P > 0.10 (P = 0.453).

7. (a) Source DF SS MS F PGroup 3 0.19218 0.064062 1.8795 0.142Error 62 2.1133 0.034085Total 65 2.3055

(b) No. F3,62 = 1.8795,P > 0.10 (P = 0.142).

8. (a) Source DF SS MS F PPlant 3 12712.7 4237.6 4.8179 0.003Error 116 102027.8 879.55Total 119 114740.5

(b) Yes.F3,116= 4.8179,P < 0.01 (P = 0.003).

9. (a) Source DF SS MS F PCatalyst 2 61.960 30.980 5.9926 0.037Error 6 31.019 5.1698Total 8 92.979

(b) Yes.F2,6 = 5.9926, 0.01< P < 0.05 (P = 0.037).

10. (a) MSTr= 176.482/3= 58.827

(b) 19−3= 16

(c) SSE= SST−SSTr= 235.958−176.482= 59.476


SECTION 9.1 421

(d) MSE= 59.476/16= 3.717

(e) F = MSTr/MSE= 15.83

(f) There are 3 and 16 degrees of freedom.

From theF table,P < 0.001. A computer package givesP = 4.8×10−5.

11. No,F3,16 = 15.83,P < 0.001 (P≈ 4.8×10−5).

12. (a) Source DF SS MS F PMethod 2 12.936 6.4678 7.4724 0.024Error 6 5.193 0.86556Total 8 18.129

(b) Yes.F2,6 = 7.4724, 0.01< P < 0.05 (P = 0.024).

13. (a) Source DF SS MS F PTemperature 3 58.650 19.550 8.4914 0.001Error 16 36.837 2.3023Total 19 95.487

(b) Yes,F3,16 = 8.4914, 0.001< P < 0.01 (P= 0.0013).

14. (a) From Exercise 12, MSE= 0.86556, sos=√

0.86556= 0.930.

(b) The MINITAB output for the power calculation is

Power and Sample Size

One-way ANOVA

Alpha = 0.05 Assumed standard deviation = 0.93 Number of Leve ls = 3

SS Sample Target MaximumMeans Size Power Actual Power Difference


422 CHAPTER 9

2 7 0.9 0.920718 2

The sample size is for each level.

(c) The MINITAB output for the power calculation is


One-way ANOVA

Alpha = 0.05 Assumed standard deviation = 1.395 Number of Lev els = 3


2 14 0.9 0.914584 2



2.3023= 1.517.



One-way ANOVA



2 18 0.9 0.912468 2




One-way ANOVA

Alpha = 0.05 Assumed standard deviation = 2.2755 Number of Le vels = 4


SECTION 9.1 423


2 38 0.9 0.902703 2


16. (a) Source DF SS MS F PDiameter 2 20.565 10.283 7.4959 0.015Error 8 10.974 1.3718Total 10 31.539

(b) Yes,F2,8 = 7.4959, 0.01< P < 0.05 (P= 0.015).

17. (a) Source DF SS MS F PGrade 3 1721.4 573.81 9.4431 0.000Error 96 5833.4 60.765Total 99 7554.9

(b) Yes,F3,96 = 9.4431,P < 0.001 (P≈ 0).

18. (a) Source DF SS MS F PNail Type 2 316.62 158.31 2.2887 0.121Error 27 1867.6 69.171Total 29 2184.2

(b) No,F2,27 = 2.2887,P > 0.10 (P = 0.121).

19. (a) Source DF SS MS F PSoil 2 2.1615 1.0808 5.6099 0.0104Error 23 4.4309 0.19265Total 25 6.5924

(b) Yes.F2,23 = 5.6099, 0.01< P < 0.05 (P = 0.0104).

20. (a) The target power is 0.85.


424 CHAPTER 9

(b) The are 4 levels, or groups, with a sample size of 14 in eachgroup, for a total of 56 observations in all.

(c) The maximum difference is 200.

(d) Greater than 0.864138. The larger the difference is, thegreater the probability is to detect it.

Section 9.2

1. (a) Yes,F5,6 = 46.64,P≈ 0.

(b) q6,6,.05 = 5.63. The value of MSE is 0.00508. The 5% critical value is therefore 5.63√

0.00508/2= 0.284. Anypair that differs by more than 0.284 can be concluded to be different. The following pairs meet this criterion:A and B, A and C, A and D, A and E, B and C, B and D, B and E, B and F, D and F.

(c) t6,.025/15 = 4.698 (the value obtained by interpolating is 4.958). The value of MSE is 0.00508. The 5% critical

value is therefore 4.698√

2(0.00508)/2= 0.335. Any pair that differs by more than 0.335 may be concludedto be different. The following pairs meet this criterion: A and B, A and C, A and D, A and E, B and C, B andD, B and E, B and F, D and F.

(d) The Tukey-Kramer method is more powerful, since its critical value is smaller (0.284 vs. 0.335).

(e) Either the Bonferroni or the Tukey-Kramer method can be used.

2. (a) Yes,F2,24 = 5.071,P = 0.015.


5023.0/9= 83.394. Wemay conclude that treatments B and C differ.

(c) t24,.025/3 = 2.5736 (the value obtained by interpolating is 2.594). The value of MSE is 5023.0. The 5% critical


2(5023.0)/9= 85.984. We may conclude that treatments B and C differ.

(d) The Tukey-Kramer method is more powerful, because its critical value is smaller.

3. (a) MSE= 2.9659,Ji = 12 for all i. There are 7 comparisons to be made. Nowt88,.025/7 = 2.754, so the 5% critical

value is 2.754√

2.9659(1/12+1/12) = 1.936. All the sample means of the non-control formulations differfrom the sample mean of the control formulation by more than this amount. Therefore we conclude at the 5%level that all the non-control formulations differ from thecontrol formulation.


SECTION 9.2 425

(b) There are 7 comparisons to be made. We should use the Studentized range valueq7,88,.05. This value is not in thetable, so we will useq7,60,.05 = 4.31, which is only slightly larger. The 5% critical value is 4.31

√2.9659/12=

2.14. All of the non-control formulations differ from the sample mean of the control formulation by more thanthis amount. Therefore we conclude at the 5% level that all the non-control formulations differ from the controlformulation.

(c) The Bonferroni method is more powerful, because it is based on the actual number of comparisons being made,which is 7. The Tukey-Kramer method is based on the largest number of comparisons that could be made,which is(7)(8)/2 = 28.

4. (a)t15,.025/10 = 3.2860 (the value obtained by interpolating is 3.4383). The value of MSE is 0.046667. The 5%

critical value is therefore 3.2860√

2(0.046667)/4= 0.5019. We may conclude that the mean amount absorbedin one hour differs from the mean amount absorbed in 24 hours.


0.046667/4= 0.472.We may conclude amounts absorbed in one hour and in 2 hours differs from the mean amount absorbed in 24hours.

(c) The Tukey-Kramer method is more powerful, because its critical value is smaller.

5. (a)t16,.025/6 = 3.0083 (the value obtained by interpolating is 3.080). The value of MSE is 2.3023. The 5% critical


2(2.3023)/5= 2.8869. We may conclude that the mean for 750C differs from themeans for 850C and 900C, and that the mean for 800 differs from the mean for 900C.


2.3023/5 = 2.75. Wemay conclude that the mean for 750C differs from the means for 850C and 900C, and that the mean for 800

differs from the mean for 900C.

(c) The Tukey-Kramer method is more powerful, because its critical value is smaller.

6. (a)t15,.025/4 = 2.8366 (the value obtained by interpolating is 2.8607). The value of MSE is 0.046667. The 5% crit-

ical value is therefore 2.8366√

2(0.046667)/4= 0.43330. We may conclude that the mean amount absorbedin 24 hours differs from the mean amounts absorbed in 1 hour and in 2 hours.


0.046667/4= 0.472.We may conclude that the mean amount absorbed in 24 hours differs from the mean amounts absorbed in 1hour and in 2 hours.

(c) The Bonferroni method is more powerful, because its critical value is smaller.


426 CHAPTER 9

7. (a)t16,.025/3 = 2.6730 (the value obtained by interpolating is 2.696). The value of MSE is 2.3023. The 5% critical


2(2.3023)/5= 2.5651. We may conclude that the mean for 900C differs from themeans for 750C and 800C.


2.3023/5 = 2.75. Wemay conclude that the mean for 900C differs from the means for 750C and 800C.

(c) The Bonferroni method is more powerful, because its critical value is smaller.

8. (a)t11,.025= 2.201, MSE= 2.0133, the sample sizes are 4 and 3. The sample means areXB = 5.2525,XD = 6.990.The 95% confidence interval is 6.990−5.2525±2.201

√2.0133(1/4+1/3), or (−0.6477,4.1227).

(b) The sample sizes areJ1 = 2,J2 = 4,J3 = 3,J4 = 3,J5 = 4. MSE= 2.0133. The 0.95 quantile of the StudentizedRange distribution isq5,11,.05 = 4.57. The values ofq5,11,.05

√(MSE/2)(1/Ji +1/Jj) are presented in the table

on the left, and the values of the differences|Xi. −X j .| are presented in the table on the right.

1 2 3 4 51 − 3.9709 4.1857 4.1857 3.97092 3.9709 − 3.5020 3.5020 3.24223 4.1857 3.5020 − 3.7438 3.50204 4.1857 3.5020 3.7438 − 3.50205 3.9709 3.2422 3.5020 3.5020 −

1 2 3 4 51 0 0.3225 1.2667 2.06 1.032 0.3225 0 0.94417 1.7375 1.35253 1.2667 0.94417 0 0.79333 2.29674 2.06 1.7375 0.79333 0 3.095 1.03 1.3525 2.2967 3.09 0

None of the differences exceeds its critical value, so we cannot conclude at the 5% level that any of the treatmentmeans differ.

9. (a)t47,.025 = 2.012, MSE= 0.22815, the sample sizes are 16 and 9. The sample means areX1 = 1.255625,X2 = 1.756667. The 95% confidence interval is 0.501042±2.012

√0.22815(1/16+1/9), or(0.1006,0.9015).

(b) The sample sizes areJ1 = 16, J2 = 9, J3 = 14, J4 = 12. MSE= 0.22815. We should use the Studentizedrange valueq4,47,.05. This value is not in the table, so we will useq4,40,.05 = 3.79, which is only slightly larger.The values ofq4,40,.05

√(MSE/2)(1/Ji +1/Jj) are presented in the table on the left, and the values of the

differences|Xi. −X j .| are presented in the table on the right.

1 2 3 41 − 0.53336 0.46846 0.488842 0.53336 − 0.54691 0.564463 0.46846 0.54691 − 0.503584 0.48884 0.56446 0.50358 −

1 2 3 41 0 0.50104 0.16223 0.183542 0.50104 0 0.33881 0.31753 0.16223 0.33881 0 0.021314 0.18354 0.31750 0.02131 0

None of the differences exceeds its critical value, so we cannot conclude at the 5% level that any of the treatmentmeans differ.


SECTION 9.2 427

10. (a)t6,.025 = 2.447, MSE= 5.1698, and both sample sizes are 3. The sample means areX1 = 86.733 andX3 =

92.903. The 95% confidence interval is 6.170±2.447√

2(5.1698)/3, or(1.627,10.713).

(b) The common sample sizes isJ = 3. The value of MSE is 5.1698. The Studentized range value isq3,6,.05 = 4.34.The 5% critical value is therefore 4.34

√5.1698/3= 5.70. The sample means areX1 = 86.733,X2 = 88.260,

X3 = 92.903. We may therefore conclude that the mean for catalyst 3 differs from the mean for catalyst 1.

11. (a)t8,.025= 2.306, MSE= 1.3718. The sample means areX1 = 1.998 andX3 = 5.300. The sample sizes areJ1 = 5andJ3 = 3. The 95% confidence interval is therefore 3.302±2.306

√1.3718(1/5+1/3), or (1.330,5.274).

(b) The sample means areX1 = 1.998,X2 = 3.0000,X3 = 5.300. The sample sizes areJ1 = 5,J2 = J3 = 3. The up-per 5% point of the Studentized range isq3,8,.05 = 4.04. The 5% critical value for|X1−X2| and for|X1−X3| is4.04

√(1.3718/2)(1/5+1/3)= 2.44, and the 5% critical value for|X2−X3| is 4.04

√(1.3718/2)(1/3+1/3)=

2.73. Therefore means 1 and 3 differ at the 5% level.

12. (a)t27,.025 = 2.052, MSE= 69.171. The sample means areX1 = 32.019 andX3 = 24.239. The sample sizes areJ1 = J3 = 10. The 95% confidence interval is therefore 7.780±2.052

√69.171(1/10+1/10), or(0.148,15.412).

(b) The table doesn’t containq3,27,.05, so we’ll useq3,24,.05 = 3.53. The 5% critical value is 3.53√

69.171/10=9.28. The three sample means areX1 = 32.019,X2 = 26.681,X3 = 24.239. No pair of means differs at the 5%level.

13. (a)X.. = 88.04, I = 4, J = 5, MSTr= ∑Ii=1J(Xi.−X..)

2/(I −1) = 19.554.

F = MSTr/MSE = 19.554/3.85 = 5.08. There are 3 and 16 degrees of freedom, so 0.01 < P < 0.05(a computer package givesP = 0.012). The null hypothesis of no difference is rejected at the5% level.

(b) q4,16.05 = 4.05, so catalysts whose means differ by more than 4.05√

3.85/5 = 3.55 are significantly differentat the 5% level. Catalyst 1 and Catalyst 2 both differ significantly from Catalyst 4.

14. (a)X.. = 1351.5, I = 4, J = 4, MSTr= ∑Ii=1J(Xi. −X..)

2/(I −1) = 5158.67.

F = MSTr/MSE = 5158.67/875.2 = 5.89. There are 3 and 12 degrees of freedom, soP ≈ 0.01(a computer package givesP = 0.010). The null hypothesis of no difference is rejected at the5% level.

(b) q4,12.05 = 4.20, so catalysts whose means differ by more than 4.20√

875.2/4= 62.13 are significantly differentat the 5% level. Curing times 1 and 2 both differ significantlyfrom curing time 4.


428 CHAPTER 9

15. The value of theF statistic isF = MSTr/MSE= 19.554/MSE. The upper 5% point of theF3,16 distribution is3.24. Therefore theF test will reject at the 5% level if 19.554/MSE≥ 3.24, or, equivalently, if MSE≤ 6.035.

The largest difference between the sample means is 89.88−85.79= 4.09. The upper 5% point of the Studen-tized range distribution isq4,16,.05 = 4.05. Therefore the Tukey-Kramer test will fail to find any differencessignificant at the 5% level if 4.09< 4.05

√MSE/5, or equivalently, if MSE> 5.099.

Therefore theF test will reject the null hypothesis that all the means are equal, but the Tukey-Kramer test willnot find any pair of means to differ at the 5% level, for any value of MSE satisfying 5.099< MSE< 6.035.

16. The value of theF statistic isF = MSTr/MSE= 5158.67/MSE. The upper 5% point of theF3,12 distributionis 3.49. Therefore theF test will reject at the 5% level if 5158.67/MSE≥ 3.49, or, equivalently, if MSE≤1478.13.

The largest difference between the sample means is 1389−1316= 73. The upper 5% point of the Studentizedrange distribution isq4,12,.05= 4.20. Therefore the Tukey-Kramer test will fail to find any differences significantat the 5% level if 73< 4.20

√MSE/4, or equivalently, if MSE> 1208.39.

Therefore theF test will reject the null hypothesis that all the means are equal, but the Tukey-Kramer test willnot find any pair of means to differ at the 5% level, for any value of MSE satisfying 1208.39< MSE< 1478.13.

Section 9.3

1. Let I be the number of levels of oil type, letJ be the number of levels of piston ring type, and letK be thenumber of replications. ThenI = 4, J = 3, andK = 3.

(a) The number of degrees of freedom for oil type isI −1 = 3.

(b) The number of degrees of freedom for piston ring type isJ−1 = 2.

(c) The number of degrees of freedom for interaction is(I −1)(J−1) = 6.

(d) The number of degrees of freedom for error isIJ(K−1) = 24.

(e) The mean squares are found by dividing the sums of squaresby their respective degrees of freedom.

TheF statistics are found by dividing each mean square by the meansquare for error. The number of degreesof freedom for the numerator of anF statistic is the number of degrees of freedom for its effect,and the numberof degrees of freedom for the denominator is the number of degrees of freedom for error.

P-values may be obtained from theF table, or from a computer software package.


SECTION 9.3 429

Source DF SS MS F POil 3 1.0926 0.36420 5.1314 0.007Ring 2 0.9340 0.46700 6.5798 0.005Interaction 6 0.2485 0.041417 0.58354 0.740Error 24 1.7034 0.070975Total 35 3.9785

(f) Yes. F6,24 = 0.58354,P > 0.10 (P = 0.740).

(g) No, some of the main effects of oil type are non-zero.F3,24 = 5.1314, 0.001< P < 0.01 (P = 0.007).

(h) No, some of the main effects of piston ring type are non-zero. F2,24 = 6.5798, 0.001< P < 0.01 (P = 0.005).

2. Let I be the number of levels of operator, letJ be the number of levels of machine, and letK be the number ofreplications. ThenI = 3, J = 3, andK = 4.

(a) The number of degrees of freedom for operator isI −1 = 2.

(b) The number of degrees of freedom for machine isJ−1 = 2.

(c) The number of degrees of freedom for interaction is(I −1)(J−1) = 4.

(d) The number of degrees of freedom for error isIJ(K−1) = 27.

(e) The mean squares are found by dividing the sums of squaresby their respective degrees of freedom.



Source DF SS MS F POperator 2 3147.0 1573.5 27.914 0.000Machine 2 136.50 68.250 1.2107 0.314Interaction 4 411.70 102.92 1.8259 0.153Error 27 1522.0 56.370Total 35 5217.2

(f) Yes. F4,27 = 1.8259,P > 0.10 (P = 0.153).

(g) No, some of the main effects of operator are non-zero.F2,27 = 27.914,P < 0.001 (P≈ 0).


430 CHAPTER 9

(h) Yes.F2,27 = 1.2107,P > 0.10 (P = 0.314).

3. (a) Let I be the number of levels of mold temperature, letJ be the number of levels of alloy, and letK be thenumber of replications. ThenI = 5, J = 3, andK = 4.

The number of degrees of freedom for mold temperature isI −1 = 4.

The number of degrees of freedom for alloy isJ−1 = 2.

The number of degrees of freedom for interaction is(I −1)(J−1) = 8.

The number of degrees of freedom for error isIJ(K−1) = 45.

The mean squares are found by dividing the sums of squares by their respective degrees of freedom.



Source DF SS MS F PMold Temp. 4 69738 17434.5 6.7724 0.000Alloy 2 8958 4479.0 1.7399 0.187Interaction 8 7275 909.38 0.35325 0.939Error 45 115845 2574.3Total 59 201816

(b) Yes.F8,45 = 0.35325,P > 0.10 (P = 0.939).

(c) No, some of the main effects of mold temperature are non-zero.F4,45 = 6.7724,P < 0.001 (P≈ 0).

(d) Yes.F3,45 = 1.7399,P > 0.10, (P = 0.187).

4. (a) LetI be the number of levels of adhesive, letJ be the number of levels of curing pressure, and letK be thenumber of replications. ThenI = 4, J = 3, andK = 5.

The number of degrees of freedom for adhesive isI −1 = 3.

The number of degrees of freedom for curing pressure isJ−1 = 2.




TheF statistics are found by dividing each mean square by the meansquare for error. The number of degreesof freedom for the numerator of anF statistic is the number of degrees of freedom for its effect,and the number


SECTION 9.3 431

of degrees of freedom for the denominator is the number of degrees of freedom for error.


Source DF SS MS F PAdhesive 3 155.7 51.900 6.2751 0.001Curing Press. 2 287.9 143.95 17.405 0.000Interaction 6 156.7 26.117 3.1577 0.011Error 48 397 8.2708Total 59 997.3

(b) No. F6,48 = 3.1577, 0.01< P < 0.001 (P = 0.011).

(c) No, some of the main effects of curing pressure are non-zero. F2,48 = 17.405,P < 0.001 (P≈ 0). Note thatsince the additive model does not hold, it is difficult to assign a practical interpretation to the main effects.

(d) No, some of the main effects of adhesive are non-zero.F3,48 = 6.2751,P≈ 0.001. Note that since the additivemodel does not hold, it is difficult to assign a practical interpretation to the main effects.

5. (a) Main Effects ofSolution

NaCl −9.1148Na2HPO4 9.1148

Main Effects ofTemperature

25C 1.810137C −1.8101

InteractionsTemperature

Solution 25C 37CNaCl −0.49983 0.49983

Na2HPO4 0.49983 −0.49983

(b) Source DF SS MS F PSolution 1 1993.9 1993.9 5.1983 0.034Temperature 1 78.634 78.634 0.20500 0.656Interaction 1 5.9960 5.9960 0.015632 0.902Error 20 7671.4 383.57Total 23 9750.0

(c) Yes,F1,20 = 0.015632,P > 0.10 (P = 0.902).

(d) Yes, since the additive model is plausible. The mean yield stress differs between Na2HPO4 and NaCl:F1,20 = 5.1983, 0.01 0.10 (P = 0.656).


432 CHAPTER 9

6. (a) Main Effects ofWater Type

De-ionized 14.667Tap −14.667

Main Effects ofGlycerol

Present 5.5000Absent −5.5000

InteractionsGlycerol

Present AbsentDe-ionized 1.6667 −1.6667

Tap −1.6667 1.6667

(b) Source DF SS MS F PWater Type 1 2581.3 2581.3 14.543 0.005Glycerol 1 363.00 363.00 2.0451 0.191Interaction 1 33.333 33.333 0.18779 0.676Error 8 1420.0 177.50Total 11 4397.7

(c) Yes.F1,8 = 0.18779,P > 0.10 (P = 0.676).

(d) Yes, since the additive model is plausible. The mean amount of foam differs between de-ionized water and tapwater:F1,8 = 14.543, 0.001 0.10 (P = 0.191).

7. (a) Main Effects Main Effectsof Weight of Lauric Acid

15 1.9125 10 2.837525 −1.9125 30 −2.8375

InteractionsFraction = 10 Fraction = 30

Weight = 15 −3.9125 3.9125Weight = 25 3.9125 −3.9125

(b) Source DF SS MS F PWeight 1 29.261 29.261 7.4432 0.053Lauric Acid 1 64.411 64.411 16.384 0.016Interaction 1 122.46 122.46 31.151 0.005Error 4 15.725 3.9312Total 7 231.86

(c) No. F1,4 = 31.151, 0.001< P < 0.01 (P= 0.005).

(d) No, because the additive model is rejected.


SECTION 9.3 433

(e) No, because the additive model is rejected.

8. (a) Main Effects Main Effectsof Location of Depth

A 3.3156 Upper −4.9719B −3.3156 Lower 4.9719

InteractionsUpper Lower

Location A −21.559 21.559Location B 21.559 −21.559

(b) Source DF SS MS F PLocation 1 351.79 351.79 0.28563 0.597Depth 1 791.03 791.03 0.64227 0.430Interaction 1 14873.8 14873.8 12.077 0.002Error 28 34485.1 1231.6Total 31 50501.8

(c) No. F1,28 = 12.077, 0.001< P < 0.01 (P = 0.002).

(d)

Cell MeansUpper Lower

Location A 91.800 144.86Location B 128.29 95.113

The greatest and least flux both occur at location A, while thetwo middle values occur at location B. At locationA, the greater flux occurs at the lower depth, while at location B, the greater flux occurs at the upper depth.

9. (a) Main Effects ofDepth

0.05 mm 11.3100.53 mm 5.91811.01 mm 1.93471.49 mm −2.34031.97 mm −6.45692.45 mm −10.365

Main Effects ofDistance

0 mm 8.41816 mm 1.4931

12 mm −9.9111

InteractionsDistance

Depth 0 mm 6 mm 12 mm0.05 mm 0.25694 0.081944 −0.338890.53 mm 0.62361 −0.62639 0.00277781.01 mm 1.4569 −0.29306 −1.16391.49 mm 0.081944 0.30694 −0.388891.97 mm −1.2014 1.0236 0.177782.45 mm −1.2181 −0.49306 1.7111


434 CHAPTER 9

(b) Source DF SS MS F PDepth 5 3855.4 771.08 33.874 0.000Distance 2 4111.8 2055.9 90.314 0.000Interaction 10 47.846 4.7846 0.21019 0.994Error 54 1229.2 22.764Total 71 9244.2

(c) Yes. The value of the test statistic is 0.21019, its null distribution isF10,54, andP > 0.10 (P= 0.994).

(d) Yes, the depth does affect the Knoop hardness:F5,54 = 33.874, P ≈ 0. To determine which effects differat the 5% level, we should useq6,54,.05. This value is not found in Table A.8, so we approximate it withq6,40,.05 = 4.23. We compute 4.23

√22.764/12= 5.826. We conclude that all pairs of depths differ in their

effects, except 0.05 mm and 0.53 mm, 0.53 mm and 1.01 mm, 1.01 mm and 1.49 mm, 1.49 mm and 1.97 mm,and 1.97 mm and 2.45 mm.

(e) Yes, the distance of the curing tip does affect the Knoop hardness:F2,54 = 90.314,P≈ 0. To determine whicheffects differ at the 5% level, we should useq3,54,.05. This value is not found in Table A.8, so we approximateit with q3,40,.05 = 3.44. We compute 3.44

√22.764/24= 3.350. We conclude that the effects of 0 mm, 6 mm,

and 12 mm all differ from one another.

10. (a) Main Effects ofBinder Content6% 6.6593

7.5% 15.0049% −21.663

Main Effects ofRubber Content19% 10.68122% 3.670424% −14.352

InteractionsRubber Content

Binder Content 0 mm 6 mm 12 mm0.05 mm −1.6593 0.31852 1.34070.53 mm −2.0037 −2.9926 4.99631.01 mm 3.663 2.6741 −6.337

(b) Source DF SS MS F PBinder Content 2 6648.7 3324.3 1.1906 0.327Rubber Content 2 3001.9 1500.9 0.53756 0.593Interaction 4 309.93 77.484 0.027751 0.998Error 18 50258 2792.1Total 26 60219

(c) Yes. The value of the test statistic is 0.027751, its nulldistribution isF4,18, andP > 0.10 (P = 0.998).

(d) Yes, the main effects may be interpreted. There is no evidence that binder content affects tensile strength:F2,18 = 1.1906,P > 0.10 (P = 0.327).


SECTION 9.3 435

(e) Yes, the main effects may be interpreted. There is no evidence that rubber content affects tensile strength:F2,18 = 0.53756,P > 0.10 (P= 0.593).

11. (a) Main Effects ofMaterial

CPTi-ZrO2 0.044367TiAlloy-ZrO2 −0.044367

Main Effects ofNeck Length

Short −0.018533Medium −0.024833

Long 0.043367

InteractionsNeck Length

Short Medium LongCPTi-ZrO2 0.0063333 −0.023767 0.017433

TiAlloy-ZrO2 0.0063333 0.023767 −0.017433

(b) Source DF SS MS F PTaper Material 1 0.059052 0.059052 23.630 0.000Neck Length 2 0.028408 0.014204 5.6840 0.010Interaction 2 0.0090089 0.0045444 1.8185 0.184Error 24 0.059976 0.002499Total 29 0.15652

(c) Yes, the interactions may plausibly be equal to 0. The value of the test statistic is 1.8185, its null distribution isF2,24, andP > 0.10 (P = 0.184).

(d) Yes, since the additive model is plausible. The mean coefficient of friction differs between CPTi-ZrO2 andTiAlloy-ZrO2: F1,24 = 23.630,P < 0.001.

(e) Yes, since the additive model is plausible. The mean coefficient of friction is not the same for all neck lengths:F2,24 = 5.6840,P ≈ 0.01. To determine which pairs of effects differ, we useq3,24,.05 = 3.53. We compute3.53

√0.002499/10= 0.056. We conclude that the effect of long neck length differs from both short and

medium lengths, but we cannot conclude that the effects of short and medium lengths differ from each other.

12. (a) Main Effects ofConcentration15 0.7247240 −0.32528

100 −0.39944

Main Effects ofDelivery Ratio1:0 5.22671:1 −1.00891:5 −2.0422

1:10 −2.1756

InteractionsDelivery Ratio

Concentration 1:0 1:1 1:5 1:1015 1.6742 0.10972 −0.86028 −0.9236140 −0.77250 0.056389 0.19306 0.52306

100 −0.90167 −0.16611 0.66722 0.40056


436 CHAPTER 9

(b) Source DF SS MS F PConcentration 2 9.487 4.7435 11.641 0.000Delivery Ratio 3 335.16 111.72 274.16 0.000Interaction 6 20.295 3.3825 8.3008 0.000Error 24 9.7799 0.40749Total 35 374.72

(c) No. The value of the test statistic is 8.3008, its null distribution isF6,24, andP≈ 0.

13. (a) Main Effects ofConcentration15 0.1666740 −0.067778

100 −0.098889

Main Effects ofDelivery Ratio1:1 0.733331:5 −0.30000

1:10 −0.43333

InteractionsDelivery Ratio

Concentration 1:1 1:5 1:1015 0.66778 −0.30222 −0.3655640 −0.20111 −0.064444 0.26556

100 −0.46667 0.36667 0.10000

(b) Source DF SS MS F PConcentration 2 0.37936 0.18968 3.8736 0.040Delivery Ratio 2 7.34 3.67 74.949 0.000Interaction 4 3.4447 0.86118 17.587 0.000Error 18 0.8814 0.048967Total 26 12.045

(c) No. The The value of the test statistic is 17.587, its nulldistribution isF4,18, andP≈ 0.

(d)

1:1 1:5 1:100

0.5

1

1.5

2

2.5

3

Delivery Ratio

So

rptio

n (

%)

concentration = 15

concentration = 40

concentration = 100

The slopes of the line segments are quite different from oneanother, indicating a high degree of interaction.


SECTION 9.3 437

14. (a) Main Effects ofTravel Speed

10 −61.38920 18.72230 42.667

Main Effects ofAccelerating Voltage10 66.05625 −2.944450 −63.111

InteractionsAccelerating Voltage

Travel Speed 10 25 5010 9.0556 −22.944 13.88920 −16.722 9.6111 7.111130 7.6667 13.333 −21.000

(b) Source DF SS MS F PTravel Speed 2 106912.1 53456.1 8.7401 0.001Acceleration 2 150390.3 75195.2 12.294 0.000Interaction 4 11408.8 2852.2 0.46634 0.760Error 45 275228.0 6116.2Total 53 543939.3

(c) Yes,F4,45 = 0.46634,P = 0.760.

(d) The travel speed does affect the hardness:F2,45 = 8.7401,P≈ 0.001. To determine which effects differ at the5% level, we should useq3,45,.05. This value is not found in Table A.8, so we approximate it withq3,40,.05= 3.44.We compute 3.44

√6116.2/18= 63.41. We conclude that the effect of a travel speed of 10 differsfrom those

of both 20 and 30, but we cannot conclude that the effects of 20and 30 differ from each other.

(e) The accelerating voltage does affect the hardness:F2,45 = 12.294, P = 0.000. To determine which effectsdiffer at the 5% level, we should useq3,45,.05. This value is not found in Table A.8, so we approximate it withq3,40,.05 = 3.44. We compute 3.44

√6116.2/18= 63.41. We conclude that the effect of an accelerating voltage

of 10 differs from those of both 25 and 50, but we cannot conclude that the effects of 25 and 50 differ fromeach other.

15. (a) Main Effects ofAttachmentNail −1.3832

Adhesive 1.3832

Main Effects ofLength

Quarter −7.1165Half −2.5665Full 9.683

InteractionsLength

Quarter Half FullNail 0.48317 0.33167 −0.51633

Adhesive −0.48317 −0.33167 0.51633

(b) Source DF SS MS F PAttachment 1 114.79 114.79 57.773 0.000Length 2 3019.8 1509.9 759.94 0.000Interaction 2 10.023 5.0115 2.5223 0.090Error 54 107.29 1.9869Total 59 3251.9


438 CHAPTER 9

(c) The additive model is barely plausible:F2,54 = 2.5223, 0.05< P < 0.10 (P = 0.090).

(d) Yes, the attachment method does affect the critical buckling load:F1,54 = 57.773,P≈ 0.

(e) Yes, the side member length does affect the critical buckling load: F2,54 = 759.94,P≈ 0. To determine whicheffects differ at the 5% level, we should useq3,54,.05. This value is not found in Table A.8, so we approximate itwith q3,40,.05 = 3.44. We compute 3.44

√1.9869/20= 1.08. We conclude that the effects of quarter, half and

full all differ from each other.

16. (a) Main Effects ofAttachment

Nail −0.043833Adhesive 0.043833

Main Effects ofLength

Quarter −0.11917Half 0.31933Full −0.20017

InteractionsLength

Quarter Half FullNail −0.029167 0.024333 0.0048333

Adhesive 0.029167 −0.024333 −0.0048333

(b) Source DF SS MS F PAttachment 1 0.11528 0.11528 0.14024 0.710Length 2 3.1248 1.5624 1.9006 0.159Interaction 2 0.029323 0.014662 0.017835 0.982Error 54 44.391 0.82206Total 59 47.66

(c) The additive model is quite plausible:F2,54 = 0.017835,P > 0.10 (P = 0.982).

(d) There is no evidence that Young’s modulus differs between attachment methods:F1,54 = 0.14024,P > 0.10(P = 0.710).

(e) There is no evidence that Young’s modulus differs among side member lengths:F2,54 = 1.9006,P > 0.10(P = 0.159).


SECTION 9.3 439

17. (a) Source DF SS MS F PWafer 2 114661.4 57330.7 11340.1 0.000Operator 2 136.78 68.389 13.53 0.002Interaction 4 6.5556 1.6389 0.32 0.855Error 9 45.500 5.0556Total 17 114850.3

(b) There are differences among the operators.F2,9 = 13.53, 0.01< P < 0.001 (P = 0.002).

18. (a) Source DF SS MS F PWafer 2 262627.1 131313.6 24118.8 0.000Operator 2 24.778 12.389 2.28 0.159Interaction 4 16.222 4.0556 0.74 0.585Error 9 49.000 5.4444Total 17 262717.1

(b) With the balance powered up continuously, there are no apparent differences in measurements obtained bydifferent operators. When one of the operators was using thebalance shortly after power-up, there were differ-ences. It is reasonable to conclude that the balance should be powered up continuously.

19. (a) Source DF SS MS F PPVAL 2 125.41 62.704 8.2424 0.003DCM 2 1647.9 823.94 108.31 0.000Interaction 4 159.96 39.990 5.2567 0.006Error 18 136.94 7.6075Total 26 2070.2

(b) Since the interaction terms are not equal to 0, (F4,18 = 5.2567,P= 0.006), we cannot interpret the main effects.Therefore we compute the cell means. These are

DCM (ml)PVAL 50 40 300.5 97.8 92.7 74.21.0 93.5 80.8 75.42.0 94.2 88.6 78.8

We conclude that a DCM level of 50 ml produces greater encapsulation efficiency than either of the other levels.If DCM = 50, the PVAL concentration does not have much effect.Note that for DCM = 50, encapsulation


440 CHAPTER 9

efficiency is maximized at the lowest PVAL concentration, but for DCM = 30 it is maximized at the highestPVAL concentration. This is the source of the significant interaction.

Section 9.4

1. (a) Liming is the blocking factor, soil is the treatment factor.

(b) Source DF SS MS F PSoil 3 1.178 0.39267 18.335 0.000Block 4 5.047 1.2617 58.914 0.000Error 12 0.257 0.021417Total 19 6.482

(c) Yes,F3,12 = 18.335,P≈ 0.

(d) q4,12,.05 = 4.20, MSAB= 0.021417, andJ = 5. The 5% critical value is therefore 4.20√

0.021417/5= 0.275.The sample means areXA = 6.32,XB = 6.02,XC = 6.28,XD = 6.70. We can therefore conclude that D differsfrom A, B, and C, and that A differs from B.

2. (a) Source DF SS MS F PMachine 3 257.678 85.893 4.7754 0.021Block 2 592.428 296.21 16.469 0.000Interaction 6 30.704 5.1173 0.28451 0.933Error 12 215.836 17.986Total 23 1096.646

(b) Yes, theP-value for interactions is large.

(c) Yes, theP-value for the main effect of machine is small (0.021).

3. (a) LetI be the number of levels for treatment. letJ be the number of levels for blocks, and letK be the number ofreplications. ThenI = 3, J = 4, andK = 3.


SECTION 9.4 441

The number of degrees of freedom for treatments isI −1 = 2.

The number of degrees of freedom for blocks isJ−1 = 3.






Source DF SS MS F PConcentration 2 756.7 378.35 5.0185 0.015Block 3 504.7 168.23 2.2315 0.110Interaction 6 415.3 69.217 0.91809 0.499Error 24 1809.4 75.392Total 35 3486.1

(b) Yes. TheP-value for interactions is large (0.499).

(c) Yes. TheP-value for concentration is small (0.015).

4. (a) LetI be the number of levels for treatment. letJ be the number of levels for blocks, and letK be the number ofreplications. ThenI = 3, J = 4, andK = 2.

The number of degrees of freedom for treatments isI −1 = 2.

The number of degrees of freedom for blocks isJ−1 = 3.






Source DF SS MS F PMachine 2 30.83 15.415 4.1299 0.043Block 3 48.47 16.157 4.3286 0.028Interaction 6 93.27 15.545 4.1648 0.017Error 12 44.79 3.7325Total 23 217.36


442 CHAPTER 9

(b) No. TheP-value for interactions is 0.017, so we cannot conclude thatthere is no interaction between treatmentand block.

(c) No, we cannot interpret the main effect, since we cannot conclude that there is no interaction.

5. (a) Source DF SS MS F PVariety 9 339032 37670 2.5677 0.018Block 5 1860838 372168 25.367 0.000Error 45 660198 14671Total 59 2860069

(b) Yes,F9,45 = 2.5677,P = 0.018.

6. (a) Day is the blocking factor, Sprinkler is the treatmentfactor.

(b) Source DF SS MS F PSprinkler 3 328.69 109.56 55.308 0.000Day 3 30.693 10.231 5.1648 0.011Interaction 9 31.553 3.5059 1.7698 0.153Error 16 31.695 1.9809Total 31 422.63

(c) Yes, theP-value for interactions is reasonably large (0.153), so it is plausible that the interactions are equal to0.

(d) Yes, theP-value for the main effect of Sprinkler is≈ 0.

(e) q4,16,.05 = 4.05, MSE= 1.9809, so the critical value is 4.05√

1.9809/(4 ·2) = 2.015. The mean runoffs forthe sprinklers are A: 7.975, B: 6.4875, C: 1.4625, D: 0.4375.We can therefore conclude at the 5% level that Adiffers from both C and D, and B differs from both C and D.


SECTION 9.4 443

7. (a) Source DF SS MS F PWaterworks 3 1253.5 417.84 4.8953 0.005Block 14 1006.1 71.864 0.84193 0.622Error 42 3585.0 85.356Total 59 5844.6

(b) Yes,F3,42 = 4.8953,P = 0.005.

(c) To determine which effects differ at the 5% level, we should useq4,42,.05. This value is not found in Table A.8,so we approximate it withq4,40,.05 = 3.79. The 5% critical value is 3.79

√85.356/15= 9.04. The sample

means areXA = 34.000,XB = 22.933,XC = 24.800,XD = 31.467. We can conclude that A differs from bothB and C.

(d) TheP-value for the blocking factor is large (0.622), suggestingthat the blocking factor (time) has only a smalleffect on the outcome. It might therefore be reasonable to ignore the blocking factor and perform a one-wayANOVA.

8. (a) Source DF SS MS F PLocation 5 0.95333 0.19067 1.0451 0.428Block 3 7.1883 2.3961 13.133 0.000Error 15 2.7367 0.18244Total 23 10.878

(b) No. F5,15 = 1.0451,P = 0.428.

(c) q6,15,.05 = 4.60, MSE= 0.18244, so the 5% critical value is 4.60√

0.18244/4= 0.982. The sample means are:XA = 3.85, XB = 4.15, XC = 3.85, XD = 3.725,XE = 3.475,XF = 3.8. We cannot conclude that any of thelocations differ in their mean concentrations.

9. (a) One motor of each type should be tested on each day. The order in which the motors are tested on any givenday should be chosen at random. This is a randomized block design, in which the days are the blocks. It is nota completely randomized design, since randomization occurs only within blocks.

(b) The test statistic is∑5

i=1(Xi. −X..)2

∑4j=1 ∑5

i=1(Xi j −Xi.−X. j −X..)2/12.


444 CHAPTER 9

10. (a) Use of the randomized block design assumes that thereare no interactions between treatments and blocks.Since different students often do best in different courses, there are likely to be interactions between studentsand courses.

(b) There are clearly interactions present. For example, student 1 did best in chemistry, while student 2 did best instatistics, and student 4 did best in physics.

Section 9.5

1. A B C D1 − − − −ad + − − +bd − + − +ab + + − −cd − − + +ac + − + −bc − + + −abcd + + + +

The alias pairs areA,BCD, B,ACD, C,ABD, D,ABC,AB,CD, AC,BD, andAD,BC

2. An experiment in which the effect of the presence or absence of several catalysts on yield is being studied,and in which the ambient temperature also affects yield. If all the runs with catalyst A present are run in themorning, when the ambient temperature is low, and all the runs with catalyst A absent are run in the afternoon,when the ambient temperature is high, then the effect of catalyst A will be incorrectly estimated.

3. (a) Sum of MeanVariable Effect DF Squares Square F PA 23.0000 1 2116.0000 2116.0000 264.5000 0.000B −5.0000 1 100.0000 100.0000 12.5000 0.008C 1.5000 1 9.0000 9.0000 1.1250 0.320AB 7.5000 1 225.0000 225.0000 28.1250 0.001AC 3.0000 1 36.0000 36.0000 4.5000 0.067BC 0.0000 1 0.0000 0.0000 0.0000 1.000ABC 0.5000 1 1.0000 1.0000 0.1250 0.733Error 8 64.0000 8.0000Total 15 2551.0000


SECTION 9.5 445

(b) Main effectsA andB, and interactionABseem most important. TheAC interaction is borderline.

(c) The mean yield is higher when the temperature is 180C.

4. (a) Sum of MeanVariable Effect DF Squares Square F PA −0.04625 1 0.0085562 0.0085562 1.5718 0.245B −0.19875 1 0.15801 0.15801 29.025 0.001C 0.01625 1 0.0010563 0.0010563 0.19403 0.671AB −0.02375 1 0.0022563 0.0022563 0.41447 0.538AC −0.02375 1 0.0022563 0.0022563 0.41447 0.538BC 0.00875 1 0.00030625 0.00030625 0.056257 0.818ABC −0.01125 1 0.00050625 0.00050625 0.092997 0.768Error 8 0.043550 0.0054437Total 15 0.21649

(b) The additive model is appropriate, since none of the interactions are significant.

(c) FactorB has an effect on yield; there is no evidence that eitherA or C does.

5. (a) Variable EffectA 3.3750B 23.625C 1.1250AB −2.8750AC −1.3750BC −1.6250ABC 1.8750

(b) No, since the design is unreplicated, there is no error sum of squares.

(c) No, none of the interaction terms are nearly as large as the main effect of factorB.

(d) If the additive model is known to hold, then the ANOVA table below shows that the main effect ofB is notequal to 0, while the main effects ofA andC may be equal to 0.


446 CHAPTER 9

Sum of MeanVariable Effect DF Squares Square F PA 3.3750 1 22.781 22.781 2.7931 0.170B 23.625 1 1116.3 1116.3 136.86 0.000C 1.1250 1 2.5312 2.5312 0.31034 0.607Error 4 32.625 8.1562Total 7 1174.2

6. (a) Variable EffectA −1.5B 5.5C 11.0AB −3.0AC −1.5BC 2.5ABC 8.0


(c) The three-way interactionABC is the second largest effect.

(d) TheABC interaction is large. This suggests that the additive modeldoes not hold.

7. (a) Variable EffectA 2.445B 0.140C −0.250AB 1.450AC 0.610BC 0.645ABC −0.935



SECTION 9.5 447

(c) The estimates lie nearly on a straight line, so none ofthe factors can clearly be said to influence the resis-tance.

−1 0 1 2

0.001

0.010.050.1

0.250.5

0.750.9

0.950.99

0.999

Effect

8. (a) Letx be the required value.

Thenx satisfies the equation−x+84.49+82.15−86.37+82.60−85.14−82.87+86.44= 0.

Thereforex = 81.30.

(b) Variable EffectA 3.38B 1.075C 0.685AB 0.515AC −0.325BC −0.290ABC 0


448 CHAPTER 9

9. (a) Variable EffectA 1.2B 3.25C −16.05D −2.55AB 2AC 2.9AD −1.2BC 1.05BD −1.45CD −1.6ABC −0.8ABD −1.9ACD −0.15BCD 0.8ABCD 0.65

(b) FactorC is the only one that really stands out.

10. (a) Term EffectConstant 1.783 A*B*C 0.07225A −0.040375 A*B*D 0.016875B 0.026785 A*B*E 0.014875C 0.78400 A*C*D 0.025D 0.43287 A*C*E 0.00525E −0.025625 A*D*E −0.001875A*B 0.14550 B*C*D 0.032A*C 0.089125 B*C*E 0.03525A*D −0.01175 B*D*E −0.049125A*E 0.05500 C*D*E 0.19075B*C −0.064375 A*B*C*D −0.011125B*D −0.05900 A*B*C*E −0.12763B*E −0.01175 A*B*D*E 0.05225C*D 0.048125 A*C*D*E 0.025625C*E 0.10963 B*C*D*E −0.011875D*E −0.03575 A*B*C*D*E −0.019

(b) FactorsC andD. These have by far the largest effects.


SECTION 9.5 449

11. (a) Sum of MeanVariable Effect DF Squares Square F PA 14.245 1 811.68 811.68 691.2 0.000B 8.0275 1 257.76 257.76 219.5 0.000C −6.385 1 163.07 163.07 138.87 0.000AB −1.68 1 11.29 11.29 9.6139 0.015AC −1.1175 1 4.9952 4.9952 4.2538 0.073BC −0.535 1 1.1449 1.1449 0.97496 0.352ABC −1.2175 1 5.9292 5.9292 5.0492 0.055Error 8 9.3944 1.1743Total 15 1265.3

(b) All main effects are significant, as is theAB interaction. Only theBC interaction has aP value that is reasonablylarge. All three factors appear to be important, and they seem to interact considerably with each other.

12. (a) Variable Effect Variable EffectA 49.337 AE 16.112B 51.788 BC 15.588C 17.812 BD −9.9375D −8.7125 BE 12.862E 18.537 CD 13.637AB 40.112 CE −8.0625AC 17.188 DE 12.263AD −3.2375

(b) The probability plot shows that factorsA, B and Dappear to affect the solubility. TheCE interaction issomewhat large as well; this is more difficult to inter-pret.

0 20 40

0.001

0.01 0.05 0.1 0.25 0.5 0.75 0.9 0.95 0.99

0.999

Effect

B

A AB

BD

D

CE AD


450 CHAPTER 9

(c) Variable Effect Variable EffectA 7.50 AE 7.00B 19.75 BC 3.00C 1.25 BD −1.75D 0.00 BE 0.25E 44.50 CD 2.25AB 5.25 CE −6.25AC 1.25 DE 3.50AD −4.00

(d) The probability plot shows that factorsB andE appearto affect the yield. None of the interactions appear tobe noticeably large.

−10 0 10 20 30 40

0.001

0.010.05

0.10.25

0.50.75

0.90.950.99

0.999

Effect

E

B

13. (ii) The sum of the main effect ofA and theBCDE interaction.


1. Source DF SS MS F PGypsum 3 0.013092 0.0043639 0.28916 0.832Error 8 0.12073 0.015092Total 11 0.13383

The value of the test statistic isF3,8 = 0.28916;P > 0.10 (P = 0.832). There is no evidence that the pH differswith the amount of gypsum added.



2. Source DF SS MS F PGypsum 2 0.096717 0.048358 0.94836 0.439Soil Type 1 0.10268 0.10268 2.0136 0.206Interaction 2 0.03515 0.017575 0.34466 0.722Error 6 0.30595 0.050992Total 11 0.54049

(a) No.F2,6 = 0.34466,P > 0.10 (P = 0.722).

(b) No. F1,6 = 2.0136,P > 0.10 (P = 0.206).

(c) No. F2,6 = 0.94836,P > 0.10 (P = 0.439).

3. Source DF SS MS F PDay 2 1.0908 0.54538 22.35 0.000Error 36 0.87846 0.024402Total 38 1.9692

We conclude that the mean sugar content differs among the three days (F2,36 = 22.35,P≈ 0).

4. (a) 23−3−2−6= 12

(b) SSAB= 4.200(6) = 25.200

(c) SSE= SST−SSA−SSB−SSAB= 32.253

(d) MSA = 145.375/3= 48.4583

(e) MSB= 15.042/2= 7.521

(f) MSE = 32.253/12= 2.68775

(g) MSA/MSE= 48.4583/2.68775= 18.029


452 CHAPTER 9

(h) MSB/MSE= 7.521/2.68775= 2.798

(i) MSAB/MSE= 4.200/2.68775= 1.563

(j) P < 0.001 (F3,12 = 18.029,P≈ 9.7×10−5)

(k) P≈ 0.10 (F2,12 = 2.798,P= 0.101)

(l) P > 0.10 (F6,12 = 1.563,P≈ 0.240)

5. (a) No. The variances are not constant across groups. In particular, there is an outlier in group 1.

(b) No, for the same reasons as in part (a).

(c) Source DF SS MS F PGroup 4 5.2029 1.3007 8.9126 0.000Error 35 5.1080 0.14594Total 39 10.311

We conclude that the mean dissolve time differs among the groups (F4,35 = 8.9126,P≈ 0).

6. (a) Main Effects Main Effectsof Al2O3 of C4S4

1.0 0.3244 Low 1.00035.0 −0.01222 Medium −0.11722

10.0 −0.3122 High −0.88306

InteractionsC4S3 Low C4S3 medium C4S3 High

Al2O3 = 1.0 0.021389 0.033889 −0.055278Al2O3 = 5.0 0.040556 −0.074444 0.033889

Al2O3 = 10.0 −0.061944 0.040556 0.021389

(b) Source DF SS MS F PAl2O3 2 2.4348 1.2174 11.507 0.000C4S3 2 21.529 10.765 101.75 0.000Interaction 4 0.075744 0.018936 0.17898 0.947Error 27 2.8566 0.1058Total 35 26.896

(c) No. F4,27 = 0.17898,P > 0.01 (P= 0.947).

(d) Yes.F2,27 = 11.507,P≈ 0.



(e) Yes.F2,27 = 101.75,P≈ 0.

7. The recommendation is not a good one. The engineer is trying to interpret the main effects without looking atthe interactions. The smallP-value for the interactions indicates that they must be taken into account. Lookingat the cell means, it is clear that if design 2 is used, then theless expensive material performs just as well asthe more expensive material. The best recommendation, therefore, is to use design 2 with the less expensivematerial.

8. (a) The main effects are:A: (1/4)(−79.8+69.0−72.3+71.2−91.3+95.4−92.7+91.5)= −2.25.B: (1/4)(−79.8−69.0+72.3+71.2−91.3−95.4+92.7+91.5)= −1.95.C: (1/4)(−79.8−69.0−72.3−71.2+91.3+95.4+92.7+91.5)= 19.65.D: (1/4)(−79.8+69.0+72.3−71.2+91.3−95.4−92.7+91.5)= −3.75.

(b) The interaction sums of squares areAB: 2.42,AC: 27.38,BC: 0.98. The pooled sum of squares is 2.42 + 27.38+ 0.98 = 30.78. This is used as an error sum of squares.

(c) Sum of MeanVariable Effect DF Squares Square F PA −2.25 1 10.125 10.125 0.98684 0.394B −1.95 1 7.605 7.605 0.74123 0.453C 19.65 1 772.25 772.25 75.268 0.003D −3.75 1 28.125 28.125 2.7412 0.196

Interaction 3 30.78 10.26

Total 7 848.88

The interaction mean square is 30.78/3 = 10.26. The test statistics for the main effects are found by dividingtheir respective mean squares by the interaction mean square. Each test statistic has null distributionF1,3. Wecan conclude that factor C affects the yield (F1,3 = 75.268,P = 0.003).


454 CHAPTER 9

9. (a) Source DF SS MS F PBase 3 13495 4498.3 7.5307 0.000Instrument 2 90990 45495 76.164 0.000Interaction 6 12050 2008.3 3.3622 0.003Error 708 422912 597.33Total 719 539447

(b) No, it is not appropriate because there are interactionsbetween the row and column effects (F6,708 = 3.3622,P = 0.003).

10. (a) Main Effects Main Effectsof Fertilizer of Zone

Low −5.2630 Southern −2.3852Medium 5.8926 Central 3.8370

High −0.62963 Northern −1.4519

InteractionsSouthern Zone Central Zone Northern Zone

Low Fertilizer −2.3259 −0.74815 3.0741Medium Fertilizer 1.2852 −0.27037 −1.0148

High Fertilizer 1.0407 1.0185 −2.0593

(b) Source DF SS MS F PFertilizer 2 565.36 282.68 33.896 0.000Zone 2 202.68 101.34 12.152 0.000Interaction 4 73.606 18.401 2.2065 0.109Error 18 150.11 8.3396Total 26 991.76

(c) It is plausible that there is no interaction. (F4,18 = 2.2065,P = 0.109).

(d) There are differences in sap production between the three fertilizer levels (F2,18 = 33.896,P≈ 0).

11. (a) Source DF SS MS F PChannel Type 4 1011.7 252.93 8.7139 0.001Error 15 435.39 29.026Total 19 1447.1

Yes.F4,15 = 8.7139,P = 0.001.



(b) q5,20,.05 = 4.23, MSE= 29.026,J = 4. The 5% critical value is therefore 4.23√

29.026/4= 11.39. The samplemeans for the five channels areX1 = 44.000,X2 = 44.100,X3 = 30.900,X4 = 28.575,X5 = 44.425. We cantherefore conclude that channels 3 and 4 differ from channels 1, 2, and 5.

12. (a) The estimates are: A: 1.6875, B: 0.8125, C: 1.1875, AB: 0.6875, AC: 1.8125, BC:−0.3125,

ABC: −0.1875.

(b) Sum of MeanVariable Effect DF Squares Square F PA 1.6875 1 22.781 22.781 9.3863 0.005B 0.8125 1 5.2812 5.2812 2.176 0.153C 1.1875 1 11.281 11.281 4.6481 0.041AB 0.6875 1 3.7812 3.7812 1.5579 0.224AC 1.8125 1 26.281 26.281 10.828 0.003BC −0.3125 1 0.78125 0.78125 0.32189 0.576ABC −0.1875 1 0.28125 0.28125 0.11588 0.737Error 24 58.25 2.4271Total 31 128.72

(c) The sum of squares for interactions is 3.7812 + 26.281 + 0.78125 + 0.28125 = 31.1247. There are four in-teractions (AB, AC, BC, andABC), so there are four degrees of freedom for interaction. The mean square forinteraction is therefore 31.1247/4= 7.781175. The mean square for error is 2.4271, so theF statistic for test-ing interactions is 7.781175/2.4271= 3.206. The null distribution isF4,24, and 0.01< P < 0.05 (P = 0.030).The additive model does not seem appropriate.

(d) The effects with smallP-values areA, C, and theAC interaction. It is reasonable to conclude that factorsA andC affect the outcome, whileB may have no effect.

13. Source DF SS MS F PWell Type 4 5.7523 1.4381 1.5974 0.175Error 289 260.18 0.90028Total 293 265.93

No. F4,289= 1.5974,P > 0.10 (P = 0.175).


456 CHAPTER 9

14. (a) Source DF SS MS F PConcentration 3 8.5768 2.8589 345.58 0.000Error 12 0.099275 0.0082729Total 15 8.6761

No. F3,12 = 345.58,P≈ 0.

(b) q4,12,.05 = 4.20, MSE= 0.0082729, so the 5% critical point is 4.20√

0.0082729/4= 0.191. The sample meansareX1 = 1.625,X2 = 2.7, X3 = 3.545,X4 = 3.2475. We can conclude that all pairs of concentrations havediffering enthalpies.


29.026= 5.388.



One-way ANOVA



50 10 0.9 0.901970 10




One-way ANOVA



50 22 0.9 0.913650 10





0.0082729= 0.0910.



One-way ANOVA


SS Sample Target MaximumMeans Size Power Actual Power Difference0.005 20 0.8 0.821699 0.1




One-way ANOVA

Alpha = 0.05 Assumed standard deviation = 0.1365 Number of Le vels = 4

SS Sample Target MaximumMeans Size Power Actual Power Difference0.005 42 0.8 0.804081 0.1


17. (a) Variable Effect Variable Effect Variable Effect Variable EffectA 3.9875 AB −0.1125 BD −0.0875 ACD 0.4875B 2.0375 AC 0.0125 CD 0.6375 BCD −0.3125C 1.7125 AD −0.9375 ABC −0.2375 ABCD −0.7125D 3.7125 BC 0.7125 ABD 0.5125

(b) The main effects are noticeably larger than the interactions, and the main effects forA andD are noticeablylarger than those forB andC.


458 CHAPTER 9

(c) Sum of MeanVariable Effect DF Squares Square F PA 3.9875 1 63.601 63.601 68.415 0.000B 2.0375 1 16.606 16.606 17.863 0.008C 1.7125 1 11.731 11.731 12.619 0.016D 3.7125 1 55.131 55.131 59.304 0.001AB −0.1125 1 0.050625 0.050625 0.054457 0.825AC 0.0125 1 0.000625 0.000625 0.00067231 0.980AD −0.9375 1 3.5156 3.5156 3.7818 0.109BC 0.7125 1 2.0306 2.0306 2.1843 0.199BD −0.0875 1 0.030625 0.030625 0.032943 0.863CD 0.6375 1 1.6256 1.6256 1.7487 0.243Interaction 5 4.6481 0.92963Total 15 158.97

We can conclude that each of the factorsA, B, C, andD has an effect on the outcome.

(d) TheF statistics are computed by dividing the mean square for eacheffect (equal to its sum of squares) by theerror mean square 1.04. The degrees of freedom for eachF statistic are 1 and 4. The results are summarized inthe following table.

Sum of MeanVariable Effect DF Squares Square F PA 3.9875 1 63.601 63.601 61.154 0.001B 2.0375 1 16.606 16.606 15.967 0.016C 1.7125 1 11.731 11.731 11.279 0.028D 3.7125 1 55.131 55.131 53.01 0.002AB −0.1125 1 0.050625 0.050625 0.048678 0.836AC 0.0125 1 0.000625 0.000625 0.00060096 0.982AD −0.9375 1 3.5156 3.5156 3.3804 0.140BC 0.7125 1 2.0306 2.0306 1.9525 0.235BD −0.0875 1 0.030625 0.030625 0.029447 0.872CD 0.6375 1 1.6256 1.6256 1.5631 0.279ABC −0.2375 1 0.22563 0.22563 0.21695 0.666ABD 0.5125 1 1.0506 1.0506 1.0102 0.372ACD 0.4875 1 0.95063 0.95063 0.91406 0.393BCD −0.3125 1 0.39062 0.39062 0.3756 0.573ABCD −0.7125 1 2.0306 2.0306 1.9525 0.235

(e) Yes. None of theP-values for the third- or higher-order interactions are small.

(f) We can conclude that each of the factorsA, B, C, andD has an effect on the outcome.



18. (a) Source DF SS MS F PCircumference 7 6589643 941378 5.3437 0.000Depth 3 2877565 959188 5.4448 0.002Interaction 21 3455204 164534 0.93397 0.550Error 96 16911875 176165.4Total 127 29834287

(b) No. F21,96 = 0.93397,P > 0.10 (P = 0.550).

(c) Yes.F7,96 = 5.3437,P≈ 0.

(d) Yes.F3,96 = 5.4448, 0.001< P < 0.01 (P = 0.002).

19. Source DF SS MS F PLocation 2 3.0287 1.5144 9.4427 0.000Error 107 17.16 0.16037Total 109 20.189

Yes,F2,107= 9.4427,P≈ 0.

20. The sample standard deviations vary by a factor of about 9. Since the sample sizes are reasonably large, thesample standard deviations are close to the population standard deviations, so we conclude that the assumptionof equal population variances is seriously violated.

21. (a) Source DF SS MS F PH2SO4 2 457.65 228.83 8.8447 0.008CaCl2 2 38783 19391 749.53 0.000Interaction 4 279.78 69.946 2.7036 0.099Error 9 232.85 25.872Total 17 39753

(b) TheP-value for interactions is 0.099. One cannot rule out the additive model.


460 CHAPTER 9

(c) Yes,F2,9 = 8.8447, 0.001< P < 0.01 (P = 0.008).

(d) Yes,F2,9 = 749.53,P≈ 0.000.

22. Source DF SS MS F PArea 4 245.13 61.283 13.224 0.000Error 16 74.147 4.6342Total 20 319.28

Yes,F4,16 = 13.224,P≈ 0.

23. Source DF SS MS F PAngle 6 936.4 156.07 20.302 0.000Error 39 299.81 7.6874Total 45 1236.2

Yes,F6,39 = 20.302,P≈ 0.

24. (a) Source DF SS MS F PDistance 5 3.2226 0.64452 3.4668 0.012Time 2 153.33 76.665 412.37 0.000Interaction 10 3.5999 0.35999 1.9363 0.072Error 36 6.6929 0.18591Total 53 166.84

(b) F10,36 = 1.9363, 0.05< P < 0.10 (P = 0.072). The additive model is barely plausible.


SECTION 10.1 461

Chapter 10

Section 10.1

1. (a) Continuous

(b) Count

(c) Binary

(d) Continuous

2. (a) True

(b) False. If no special causes are operating, then the variability is constant. It is still possible for much of theoutput to fail to conform to specifications, for example through incorrect calibration, or because of a largevariation in common causes.

(c) True. This is part of the definition of a common cause.

(d) False. Rational subgroups should be chosen in such way that the variation within them is due to commoncauses.

(e) False. There will be variation from common causes.

3. (a) is in control

(b) has high capability

4. (i) It must still be monitored continually.

5. (a) False. Being in a state of statistical control means only that no special causes are operating. It is still possiblefor the process to be calibrated incorrectly, or for the variation due to common causes to be so great that muchof the output fails to conform to specifications.


462 CHAPTER 10

(b) False. Being out of control means that some special causes are operating. It is still possible for much of theoutput to meet specifications.

(c) True. This is the definition of statistical control.

(d) True. This is the definition of statistical control.

6. (ii) it is more important to sample frequently than to choose large samples, so that special causes can be detectedmore quickly.

Section 10.2

1. (a) The sample size isn = 4. The upper and lower limits for theR-chart areD3RandD4R, respectively.

From the control chart table,D3 = 0 andD4 = 2.282.

R= 143.7/30= 4.79. Therefore LCL = 0, and UCL = 10.931.

(b) The sample size isn = 4. The upper and lower limits for theS-chart areB3sandB4s, respectively.

From the control chart table,B3 = 0 andB4 = 2.266.

s= 62.5/30= 2.08333. Therefore LCL = 0 and UCL = 4.721.

(c) The upper and lower limits for theX-chart areX−A2RandX +A2R, respectively.

From the control chart table,A2 = 0.729.R= 143.7/30= 4.79 andX = 712.5/30= 23.75.

Therefore LCL = 20.258 and UCL = 27.242.

(d) The upper and lower limits for theX-chart areX−A3sandX +A3s, respectively.

From the control chart table,A3 = 1.628.s= 62.5/30= 2.08333 andX = 712.5/30= 23.75.


2. The process is never out of control.


SECTION 10.2 463



R= 0.131. Therefore LCL = 0 and UCL = 0.277. The variance is in control.

(b) The upper and lower limits for theX-chart areX−A2RandX +A2R, respectively.

From the control chart table,A2 = 0.577.R= 0.131 andX = 1.110.

Therefore LCL = 1.034 and UCL = 1.186. The process is out of control for the first time on sample 17.

(c) The 1σ limits areX−A2R/3 = 1.085 andX +A2R/3 = 1.135, respectively.

The 2σ limits areX−2A2R/3 = 1.0596 andX +2A2R/3 = 1.1604, respectively.

The process is out of control for the first time on sample 8, where 2 out of the last three samples are below thelower 2σ control limit.

4. (a) The sample size isn = 5. The upper and lower limits for theS-chart areB3sandB4s, respectively.


s= 0.057. Therefore LCL = 0 and UCL = 0.119. The variance is in control.

(b) The upper and lower limits for theX-chart areX−A3sandX +A3s, respectively.

From the control chart table,A3 = 1.427.s= 0.057 andX = 1.110.

Therefore LCL = 1.029 and UCL = 1.191. The process is out of control for the first time on sample 18.

(c) The 1σ limits areX−A3s/3 = 1.083 andX +A3s/3 = 1.137, respectively.

The 2σ limits areX−2A3s/3 = 1.056 andX +2A3s/3 = 1.164, respectively.

The process is out of control for the first time on sample 8, where 2 out of the last three samples are below thelower 2σ control limit.

5. (a)X has a normal distribution withµ= 11 andσX = 2/√

4 = 1.

The 3σ limits are 10±3(1), or 7 and 13.

The probability that a point plots outside the 3σ limits is p = P(X < 7)+P(X > 13).

Thez-score for 7 is(7−11)/1= −4. Thez-score for 13 is(13−11)/1= 2.

The probability that a point plots outside the 3σ limits is the sum of the area to the left ofz= −4 and the areato the right ofz= 2.

Thereforep = 0.0000+0.0228= 0.0228.

The ARL is 1/p = 1/0.0228= 43.86.

(b) Letm be the required value. Since the shift is upward,m> 10.

The probability that a point plots outside the 3σ limits is p = P(X < 7)+P(X > 13).


464 CHAPTER 10

Since ARL = 6,p = 1/6. Sincem> 10,P(X > 13) > P(X < 7).

Findm so thatP(X > 13) = 1/6, and check thatP(X < 7) ≈ 0.

Thez-score for 13 is(13−m)/1. Thez-score with an area of 1/6 = 0.1667 to the right is approximately z= 0.97.

Therefore 0.97= (13−m)/1, som= 12.03.

Now check thatP(X < 7) ≈ 0.

Thez-score for 7 is(7−12.03)/1= −5.03. SoP(X < 7) ≈ 0.

Thereforem= 12.03.

(c) We will find the required value forσX .

The probability that a point plots outside the 3σ limits is p = P(X < 10−3σX)+P(X > 10+3σX).

Since ARL = 6,p = 1/6. Since the process mean is 11,P(X > 10+3σX) > P(X > 10−3σX).

Find σ so thatP(X > 10+3σX) = 1/6, and check thatP(X < 10−3σX) ≈ 0.

The z-score for 10+ 3σX is (10+ 3σX − 11)/σX. The z-score with an area of 1/6 = 0.1667 to the right isapproximatelyz= 0.97.

Therefore(10+3σX −11)/σX = 0.97, soσX = 0.4926.

Now check thatP(X < 10−3σX) ≈ 0.

10−3σX = 8.522. Thez-score for 8.522 is(8.522−11)/0.4926= −5.03, soP(X < 10−3σX) ≈ 0.

ThereforeσX = 0.4926.

Sincen = 4, σX = σ/√

4 = σ/2.

Thereforeσ = 0.985.

(d) Letn be the required sample size. ThenσX = 2/√

n.

From part (c),σX = 0.4926. Therefore 2/√

n = 0.4926, son = 16.48.

Round up to obtainn = 17.

6. (a)X has a normal distribution withµ= 6.7 andσX = 1.3/√

6 = 0.530723.

The 3σ limits are 7.2±3(0.530723), or 5.6078 and 8.7922.

The probability that a point plots outside the 3σ limits is p = P(X < 5.6078)+P(X > 8.7922).

Thez-score for 5.6078 is(5.6078−6.7)/0.530723= −2.06.

Thez-score for 8.7922 is(8.7922−6.7)/0.530723= 3.94.

The probability that a point plots outside the 3σ limits is the sum of the area to the left ofz= −2.06 and thearea to the right ofz= 3.94.

Thereforep = 0.0197+0.0000= 0.0197.

The ARL is 1/p = 1/0.0197= 50.76.

(b) Letm be the required value. Since the shift is upward,m> 7.2.

The probability that a point plots outside the 3σ limits is p = P(X < 5.6078)+P(X > 8.7922).


SECTION 10.2 465

Since ARL = 15,p = 1/15. Sincem> 7.2, P(X > 8.7922) > P(X < 5.6078).

Findm so thatP(X > 8.7922) = 1/15, and check thatP(X < 5.6078)≈ 0.

The z-score for 8.7922 is(8.7922−m)/0.530723. Thez-score with an area of 1/15 = 0.0667 to the right isapproximatelyz= 1.50.

Therefore 1.50= (8.7922−m)/0.530723, som= 8.00.

Now check thatP(X < 5.6078)≈ 0.

Thez-score for 5.6078 is(5.6078−8.00)/0.530723= −4.51. SoP(X < 5.6078)≈ 0.

Thereforem= 8.00.

(c) We will find the required value forσX .

The probability that a point plots outside the 3σ limits is p = P(X < 7.2−3σX)+P(X > 7.2+3σX).

Since ARL = 15,p = 1/15. Since the process mean is 6.7,P(X < 7.2−3σX) > P(X > 7.2+3σX).

Find σ so thatP(X < 7.2−3σX) = 1/15, and check thatP(X > 7.2+3σX) ≈ 0.

The z-score for 7.2− 3σX is (7.2− 3σX − 6.7)/σX. Thez-score with an area of 1/15 = 0.0667 to the left isapproximatelyz= −1.50.

Therefore(7.2−3σX −6.7)/σX = −1.50, soσX = 0.3333.

Now check thatP(X > 7.2+3σX) ≈ 0.

7.2+3σX = 8.2. Thez-score for 8.2 is(8.2−6.7)/0.3333= 4.50, soP(X > 7.2+3σX) ≈ 0.

Sincen = 6, σX = σ/√

6.


(d) Letn be the required sample size. ThenσX = 1.3/√

n.

From part (c),σX = 0.3333. Therefore 1.3/√

n = 0.3333, son = 15.21.

Round up to obtainn = 16.

7. The probability of a false alarm on any given sample is 0.0027, and the probability that there will not be a falsealarm on any given sample is 0.9973.

(a) The probability that there will be no false alarm in the next 50 samples is 0.997350 = 0.874. Therefore theprobability that there will be a false alarm within the next 50 samples is 1−0.874= 0.126.

(b) The probability that there will be no false alarm in the next 100 samples is 0.9973100= 0.763. Therefore theprobability that there will be a false alarm within the next 50 samples is 1−0.763= 0.237.

(c) The probability that there will be no false alarm in the next 200 samples is 0.9973200= 0.582.

(d) Letnbe the required number. Then 0.9973n = 0.5, sonln0.9973= ln0.5. Solving fornyieldsn = 256.37≈ 257.


466 CHAPTER 10






Therefore LCL = 197.322, UCL = 202.310. The process is in control.



The process is in control.


From the control chart table,B3 = 0.030 andB4 = 1.970.

s= 1.961. Therefore LCL = 0.0588 and UCL = 3.863. The variance is incontrol.



Therefore LCL = 197.292 and UCL = 202.340. The process is in control.



The process is in control.









The process is out of control for the first time on sample 9, where 2 of the last three sample means are belowthe lower 2σ control limit.


SECTION 10.2 467



s= 0.4647. Therefore LCL = 0 and UCL = 0.971. The variance is in control.






The process is out of control for the first time on sample 9, where 2 of the last three sample means are belowthe lower 2σ control limit.



R= 6.97. Therefore LCL = 0 and UCL = 15.906.

The variance is out of control on sample 8. After deleting this sample,X = 150.166 andR= 6.538. The newlimits for theR-chart are 0 and 14.920. The variance is now in control.






The process is in control (recall that sample 8 has been deleted).



s= 3.082. Therefore LCL = 0 and UCL = 6.984.

The variance is out of control on sample 8. After deleting this sample,X = 150.166 ands= 2.911. The newlimits for theS-chart are 0 and 6.596. The variance is now in control.






468 CHAPTER 10


The process is in control (recall that sample 8 has been deleted).

Section 10.3

1. The sample size isn = 200. p = 1.64/30= 0.054667.

The centerline isp = 0.054667.

The LCL is p−3√

p(1− p)/200= 0.00644.

The UCL isp+3√

p(1− p)/200= 0.1029.

2. (a) The sample size isn = 400.

The mean number of defectives over the last 20 days is 24.05.

Thereforep = 24.05/400= 0.060125.

The control limits arep±3√

p(1− p)/n.


(b) The process is out of control for the first time on sample #17.

3. Yes, the only information needed to compute the control limits isp and the sample sizen. In this case,n= 100,andp = (622/50)/100= 0.1244.


p(1− p)/n, so

LCL = 0.0254 and UCL = 0.2234.

4. Yes the process is out of control on the sample with 24 defectives, since the sample proportion of 0.24 exceedsthe upper control limit of 0.2234.

5. (iv). The sample size must be large enough so the mean number of defectives per sample is at least 10.


SECTION 10.4 469

6. c = 1476/50= 29.52.

The centerline isc = 29.52.

The LCL isc−3√

c = 13.22.

The UCL isc+3√

c = 45.82.

7. It was out of control. The UCL is 45.82.

8. (a) The average number of defectives per sample isc = 45.3.

The control limits arec±3√

c


(b) The process is in control.

Section 10.4

1. (a) No samples need be deleted.

(b) The estimate ofσX is A2R/3. The sample size isn = 5.

R= 0.131. From the control chart table,A2 = 0.577.

ThereforeσX = (0.577)(0.131)/3= 0.0252.

(c)

0 5 10 15 20−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

Sample Number

Cum

ulat

ive S

um

CUSUM Chart

UCL = 0.101

LCL = −0.101


470 CHAPTER 10

(d) The process is out of control on sample 8.

(e) The Western Electric rules specify that the process is out of control on sample 8.




ThereforeσX = (0.483)(5.164)/3= 0.831.

(c)

0 5 10 15 20 25−4

−3

−2

−1

0

1

2

3

4

Sample Number

Cum

ulat

ive S

um

CUSUM Chart

UCL = 3.326

LCL = −3.326

(d) The process is in control.

(e) The Western Electric rules specify that the process is incontrol.




ThereforeσX = (0.577)(1.14)/3= 0.219.


SECTION 10.4 471

(c)

0 5 10 15 20 25 30−3

−2

−1

0

1

2

3

Sample Number

Cum

ulativ

e Su

m

CUSUM Chart

UCL = 0.877

LCL = −0.877


(e) The Western Electric rules specify that the process is out of control on sample 9.

4. (a) LCL = 0, UCL = 15.906. The variance is out of control on sample 8. After deleting this sample,X = 150.166andR= 6.538. The new limits for theR-chart are 0 and 14.920. The variance is now in control.



ThereforeσX = (0.729)(6.538)/3= 1.589.

(c)

0 5 10 15 20 25 30−8

−6

−4

−2

0

2

4

6

8

Sample Number

Cum

ulat

ive S

um

CUSUM Chart

UCL = 6.355

LCL = −6.355

(d) The process is in control.

(e) The Western Electric rules specify that the process is incontrol.


472 CHAPTER 10

5. (a)

0 10 20 30 40−80

−60

−40

−20

0

20

40

60

80

Sample Number

Cum

ulativ

e Su

m

CUSUM Chart

UCL =60

LCL =−60

(b) The process is in control.

6. The value ofSL can never be positive, and the value ofSH can never be negative. Therefore the values of0.2371 and 0.7104 forSL, and the value of−0.7345 forSH, are incorrect.

Section 10.5

1. (a) µ= X = 0.205,s= 0.002, LSL= 0.18, USL= 0.22. The sample size isn = 4.

σ = s/c4. From the control chart table,c4 = 0.9213.


Sinceµ is closer toUSLthan toLSL, Cpk = (USL− µ)/(3σ) = 2.303.

(b) Yes. SinceCpk > 1, the process capability is acceptable.

2. (a) µ= X = 12.01,R= 0.124, LSL= 11.95, USL= 12.10. The sample size isn = 5.

σ = R/d2. From the control chart table,d2 = 2.326.


Sinceµ is closer toLSL than toUSL, Cpk = (µ−LSL)/(3σ) = 0.375.

(b) No. SinceCpk < 1, the process capability is not acceptable.



3. (a) The capability is maximized when the process mean is equidistant from the specification limits.Therefore the process mean should be set to 0.20.

(b) LSL= 0.18, USL= 0.22, σ = 0.002171.

If µ= 0.20, thenCpk = (0.22−0.20)/[3(0.002171)]= 3.071.

4. (a) The capability is maximized when the process mean is equidistant from the specification limits.Therefore the process mean should be set to 12.025.

(b) No. R= 0.124, LSL= 11.95, USL= 12.10, σ = 0.05331.

If µ= 12.025, thenCpk = (12.10−12.025)/[3(0.05331)]= 0.469.

SinceCpk < 1, the process capability is not acceptable.

(c) Let σ be the required value. ThenCpk = (12.10−12.025)/(3σ)= 1, soσ = 0.025.

(d) Let σ be the required value. ThenCpk = (12.10−12.025)/(3σ)= 2, soσ = 0.0125.

5. (a) Letµ be the optimal setting for the process mean.

ThenCp = (USL−µ)/(3σ) = (µ−LSL)/(3σ), so 1.2 = (USL−µ)/(3σ) = (µ−LSL)/(3σ).

Solving forLSLandUSLyieldsLSL= µ−3.6σ andUSL= µ+3.6σ.

(b) Thez-scores for the upper and lower specification limits arez= ±3.60.

Therefore, using the normal curve, the proportion of units that are non-conforming is the sum of the areas underthe normal curve to the right ofz= 3.60 and to the left ofz= −3.60.

The proportion is 0.0002+0.0002= 0.0004.

(c) Likely. The normal approximation is likely to be inaccurate in the tails.


1. The sample size isn = 250. p = 2.98/50= 0.0596.

The centerline isp = 0.0596


474 CHAPTER 10

The LCL is p−3√

p(1− p)/250= 0.0147.

The UCL isp+3√

p(1− p)/250= 0.1045.

2. (a) The probability,p, that a point plots outside the limits±2.5σX is the sum of the areas under the normal curveto the right ofz= 2.50 and to the left ofz= −2.50.

Thereforep = 0.0062+0.0062= 0.0124.

The ARL is 1/p = 1/0.0124= 80.65.

(b) If µ= σX , then thez-scores for the upper and lower control limits are(2.5σX −σX)/σX = 1.5 and(−2.5σX −σX)/σX = −3.5 respectively.

The probability,p, that a point plots outside the limits±2.5σX is the sum of the areas under the normal curveto the right ofz= 1.50 and to the left ofz= −3.50.

Thereforep = 0.0668+0.0002= 0.0670.

The ARL is 1/p = 1/0.0670= 14.93.

(c) Letm be the process mean after the shift.

The probability that a point plots outside the±2.5σX limits is p = P(X < −2.5σX)+P(X > 2.5σX).

Since this is an upward shift,P(X > 2.5σX) > P(X < −2.5σX).

Since ARL = 10,p = 0.1. Findm so thatP(X > 2.5σX) = 0.1, then check thatP(X < −2.5σX) ≈ 0.

Thez-score for 2.5σX is (2.5σX−m)/σX. Thez-score with an area of 0.1 to the right is approximatelyz= 1.28.

Therefore 1.28= (2.5σX −m)/σX, som= 1.22σX.

Now check thatP(X < −2.5σX) ≈ 0.

Thez-score for 2.5σX is (−2.5σX −m)/σX = −3.50. SoP(X < −2.5σX) ≈ 0.

Thereforem= 1.22σX.






Therefore LCL = 4.982 and UCL = 5.208. The process is out of control on sample 3.





The process is out of control for the first time on sample 3, where a sample mean is above the upper 3σ controllimit.



s= 0.058. Therefore LCL = 0, UCL = 0.149. The variance is in control.



Therefore LCL = 4.982 and UCL = 5.208. The process is out of control on sample 3.



The process is out of control for the first time on sample 3, where a sample mean is above the upper 3σ controllimit.




ThereforeσX = (1.023)(0.110)/3= 0.0375.

(c)

0 5 10 15 20 25 30−1

−0.5

0

0.5

1

Sample Number

Cum

ulat

ive S

um

CUSUM Chart

UCL = 0.15

LCL = −0.15



476 CHAPTER 10

(e) The Western Electric rules specify that the process is out of control on sample 3. The CUSUM chart firstsignaled an out-of-control condition on sample 4.

6. (a) The average number of defectives per sample isc = 658/30= 21.933.

The centerline isc = 21.933.

The control limits arec±3√

c


(b) Yes, the process was out of control, because the number offlaws was below the lower control limit. However,the number of flaws was unusuallylow, so when the special cause is determined, steps should betaken topreserve it rather than to eliminate it.

7. (a) The sample size isn = 500.

The mean number of defectives over the last 25 days is 22.4.

Thereforep = 22.4/500= 0.0448.


p(1− p)/n.

Therefore LCL = 0.0170 and UCL = 0.0726

(b) Sample 12. The proportion of defective chips is then 7/500= 0.014, which is below the lower control limit.

(c) No, this special cause improves the process. It should bepreserved rather than eliminated.


APPENDIX B 477

Appendix B

1.∂v∂x

= 3+2y4,∂v∂y

= 8xy3

2.∂w∂x

=3x2

x2 +y2 −2x(x3 +y3)

(x2 +y2)2 ,∂w∂y

=3y2

x2 +y2 −2y(x3 +y3)

(x2 +y2)2

3.∂z∂x

= −sinxsiny2,∂z∂y

= 2ycosxcosy2

4.∂v∂x

= yexy,∂v∂y

= xexy

5.∂v∂x

= ex(cosy+sinz),∂v∂y

= −exsiny,∂v∂z

= excosz

6.∂w∂x

=x√

x2 +4y2+3z2,

∂w∂y

=4y√

x2 +4y2+3z2,

∂w∂z

=3z√

x2 +4y2+3z2

7.∂z∂x

=2x

x2 +y2 ,∂z∂y

=2y

x2 +y2

8.∂v∂x

=2xy

x2y+z−zey2

sin(xz),∂v∂y

=x2

x2y+z+2yey2

cos(xz),∂v∂z

=1

x2y+z−xey2

sin(xz)

9.∂v∂x

=

√y5

x− 3

2

√y3

x,

∂v∂y

= 5√

xy3− 92√

xy

10.∂z∂x

=xycos(x2y)√

sin(x2y),

∂z∂y

=x2cos(x2y)

2√

sin(x2y)

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