STE3A01 HW Solutions Chapter 3 Part 1

STE3A01: Part 1 Solutions Chapter 3 Semester 1, 2015

Statistics for Engineers 3A (STE3A01)Part 1 Solutions to Selected Homework Exercises from Chapter 3 of

Probability and Statistics for Engineering and the Sciencesby Jay L. Devore

Section 3.1, p.95

Question 1, p.95

Refer to p.33 of the Appendix at the back of the textbook.

Question 6, p.96

X = {1, 2, 3, . . .}

Possible Outcomes

Associated X Value

L 1

RL or AL 2

RRL, AAL, ARL or RAL 3

RRRL, RRAL, RARL, ARRL,

RAAL, ARAL, AARL or RRRL 4

RRRRL, RRRAL, RRARL, RARRL, ARRRL,

RRAAL, RARAL, ARARL, AARRL, RAARL,

ARRAL, AAARL, AARAL, ARAAL, RAAAL or AAAAL 5

You only need to list any five outcomes in this table. There are more possibilities as well.

Question 7, p.96


Question 8, p.96

See next page.

Question 10, p.96

a. T = {0, 1, 2, . . . ,10}

b. Assume that station 1 is the six-pump station and station 2 is the four-pump station.X = {4, 3, 2, . . .6}

c. U = {0, 1, 2, 3, 4, 5, 6}

d. Z = {0, 1, 2}

University of Johannesburg Page 1 of 14


Question 8, p.96

Five Smallest Possible

Possible Outcomes Values of Y

SSS 3

FSSS, SFSS or SSFS 4

FFSSS, FSFSS, FSSFS, SFFSS, SSFFS or SFSFS 5

FFFSSS, FFSFSS, FSFFSS, SFFFSS, FFSSFS,

FSSFFS, SSFFFS, SFFFSS, SFSFFS or FSFFSS 6

FFFFSSS, FFFSFSS, FFSFFSS, FSFFFSS, FFFSSFS,

FFSSFFS, FSSFFFS, FSFFFSS, FSFSFFS, FFSFFSS,

SSFFFFS, SFSFFFS, SFFSFFS, SFFFSFS or SFFFFSS 7

Section 3.2, p.104

Question 12, p.104

a. The flight can accommodate a maximum of 50 passengers:

P (Y 50) = P (Y = 45) + P (Y = 46) + P (Y = 47) + P (Y = 48) + P (Y = 49) + P (Y = 50)

= 0.05 + 0.1 + 0.12 + 0.14 + 0.25 + 0.17

= 0.83

b. Not all ticketed passengers who show up can be accommodated when more than 50 people showup:

P (Y > 50) = 1 P (Y 50) first proposition (complement rule), p.59

= 1 0.83 from question (a) above

= 0.17

c. The first person on the standby list will be accommodated when 49 or less ticketed passangers showup:

P (Y 49) = P (Y 50) P (Y = 50)

= 0.83l 0.17 lfrom (a) above

= 0.66

The third person on the standby list will be accommodated when 47 or less ticketed passangersshow up:

P (Y 47) = P (Y = 45) + P (Y = 46) + P (Y = 47)

= 0.05 + 0.1 + 0.12

= 0.27



Question 13, p.104


Question 14, p.105

a. We know that the total probability must be 1:

5

i=1

ky = 1

k5

i=1

y = 1

k =1

5i=1 y

k =1

15

b. Calculate the probability by using the probability mass function p(y) = 115y. It will be shorter to

calculate the probability that at most three forms will be required by using the complement rule:

P (Y 3) = 1 P (Y 4) first proposition (complement rule), p.59

= 1 [P (Y = 4) + P (Y = 5)]= 1 [ 1

15 4 +

1

15 5]

= 0.4

c. Calculate the probability by using the probability mass function p(y) = 115y. It will be shorter to

calculate the probability that at most three forms will be required by using the complement rule:

P (Y 3) = 1 P (Y 4) first proposition (complement rule), p.59= 1 [P (Y = 4) + P (Y = 5)]= 1 [ 1

15 4 +

1

15 5]

= 0.4

d. The probability that between two and four forms (inclusive) will be required has been calculatedbelow:

P (2 Y 4) = P (Y = 2) + P (Y = 3) + P (Y = 4)=

1

15 2 +

1

15 3 +

1

15 4

= 0.6

e. No, p(y) = y2/50 could not be the probability mass function of Y , because it does not assign thesame probabilities to Y as p(y) = y/15:



Y p(y) = y/15 p(y) = y2/50 y/15 = y2/50?1 1/15 = 0.067 1/50 = 0.020 No2 2/15 = 0.133 4/50 = 2/25 = 0.080 No3 3/15 = 1/5 = 0.200 9/50 = 0.180 No4 4/15 = 0.267 16/50 = 8/25 = 0.32 No5 5/15 = 1/3 = 0.333 25/50 = 1/2 = 0.5 No

Question 15, p.105


Question 17, p.105


Question 23, p.106


Question 27, p.106


Section 3.3, p.113

Question 29, p.113


Question 32, p.113

a. E(X) = xp(x) definition, p.107E(X) = 13.5 0.2 + 15.9 0.5 + 19.1 0.3 = 16.38 cubic feet of storage space per freezer.

E(X2) = x2p(x)E(X2) = 13.52 0.2 + 15.92 0.5 + 19.12 0.3 = 272.298 cubic feet2.

V (X) = E(X2] [E(X)]2 first proposition, p.112V (X) = 272.298 [16.38]2 = 3.9936 cubic feet2.



b. E(25X 8.5) = 25E(X) 8.5 = 25(16.38) 8.5 = 401 units of currency. proposition, p.110c. V (25X 8.5) = 252V (X) 0 = 252 3.9936 = 2 496 (units of currency)2.d. E(h(X)) = E(X) E(0.01X2) = E(X) 0.01E(X2) = 16.38 0.01 272.298 = 13.65702 cubic feet.

proposition, p.109

Question 39, p.114


Question 42, p.114

a. We have been given that E(X) = 5 and E[X(X 1)] = 27.5. We need to find E(X2):E[X(X 1)] = 27.5E[X2 X] = 27.5

E(X2) E(X) = 27.5E(X2) = E(X) + 27.5E(X2) = 5 + 27.5E(X2) = 32.5

b. V (X) = E(X2) [E(X)]2 = 32.5 52 = 7.5.c. Start with V (X):

V (X) = E(X2) [E(X)]2Subtract E(X) on both sides:

V (X) E(X) = E(X2) [E(X)]2 E(X)so that

V (X) E(X) = E[X(X 1)] [E(X)]2

Section 3.4, p.120

Question 46, p.120

b(x; n, p) = (nx)px (1 p)nx theorem, p.117

a. b(3; 8, 0.35) = (83)0.353 0.655 = 0.279, which is the probability of observing 3 successes when an

experiment is repeated 8 times and the probability of a single successful trial is 0.35.

b. b(5; 8, 0.6) = (85)0.65 0.43 = 0.279, which is the probability of observing 5 successes when an exper-

iment is repeated 8 times and the probability of a single successful trial is 0.6.

c. P (3 X 5) when n = 7 and p = 0.6:P (3 X 5) = P (X = 3) + P (X = 4) + P (X = 5)

= (73)0.63 0.44 + (7

4)0.64 0.43 + (7

5)0.65 0.42

= 0.588



This is the probability of observing between 3 and 5 successes (inclusive) when an experiment isrepeated 7 times and the probability of a single successful trial is 0.6.

d. Use the complement rule to solve P (1 X) when n = 9 and p = 0.1:P (1 X) = 1 P (X < 1)

= 1 P (X = 0)= 1 (9

0)0.10 0.99

= 0.613

This is the probability of observing at least one success when an experiment is repeated 9 timesand the probability of a single successful trial is 0.1.

Question 48, p.120

X Bin(25, 0.05) so that P (X = x) = (25x)0.05x 0.9525x.

a. At most two defective circuit boards in a sample of 25:

P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)= (25

0)0.050 0.9525 + (25

1)0.051 0.9524 + (25

2)0.052 0.9523

= 0.872894

b. At least five defective circuit boards in a sample of 25:

P (X 5) = P (X = 5) + P (X = 6) + . . . + P (X = 25)= 1 [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]

first proposition (complement rule), p.59

= 1 [0.872894 + (253)0.053 0.9522 + (25

4)0.054 0.9521]

from (a) above

= 0.00716

c. Between one and four (inclusive) defective circuit boards in a sample of 25:

P (1 X 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)= (25

1)0.051 0.9524 + (25

2)0.052 0.9523 + (25

3)0.053 0.9522 + (25

4)0.054 0.9521

= 0.715

d. No defective circuit boards in a sample of 25:

P (X = 0) = (250)0.050 0.9525

= 0.277

e. The expected number of defective circuit boards in a sample of 25:

E(X) = np proposition, p120= 25 0.05

= 1.25 defective circuit boards do not round to a whole number!



The standard deviation of defective circuit boards in a sample of 25:

X =npq proposition, p120

=25 0.05 0.95

= 1.0897 defective circuit boards do not round to a whole number!

Question 55, p.121


Question 57, p.121


Question 58, p.121

Let X represent the number of defective components in the sample of ten. Then X will be a discreterandom variable with the following assumptions that can be made 3.4, p.114 :

1. The experiment consists of testing 10 components (n = 10 trials).

2. Each component can either be defective or not (only two possible outcomes per trial).

3. The ten components will be assumed to be independent of each other.

4. The probability of a defective component (p) will be assumed to be constant for all ten components.

Because all the afore-mentioned assumptions can be assumed to hold, X Bin(10, p), so that theprobability of observing x defective components in the batch of ten can be calculated as

P (X = x) = (10x)px (1 p)10x.

a. Batch will be accepted when number of defective components in sample of 10 items is at most two:

p=0.01:

P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)= (10

0)0.010 0.9910 + (10

1)0.011 0.999 + (10

2)0.012 0.998

= 0.99989

p=0.05:

P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)= (10

0)0.050 0.9510 + (10

1)0.051 0.959 + (10

2)0.052 0.958

= 0.98850

p=0.1:

P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)= (10

0)0.10 0.910 + (10

1)0.11 0.99 + (10

2)0.12 0.98

= 0.92981



p=0.2:

P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)= (10

0)0.20 0.810 + (10

1)0.21 0.89 + (10

2)0.22 0.88

= 0.67780

p=0.25:

P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)= (10

0)0.250 0.7510 + (10

1)0.251 0.759 + (10

2)0.252 0.758

= 0.52559

b. The operating characteristic curve is shown in Figure 1:

Actual proportion of defectives (p)

P (Batch is accepted)

0 0.025 0.05 0.075 0.1 0.125 0.15 0.175 0.2 0.225 0.25 0.275

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

(0.01, 0.99989)(0.05, 0.986)(0.1, 0.930)

(0.2, 0.678)

(0.25, 0.526)

b

b

b

b

b

Figure 1: Operating characteristic curve for at most two defective items in a batch of 10.

c. (a) Batch will be accepted when number of defective components in sample of 10 is at most one:

p=0.01:

P (X 1) = P (X = 0) + P (X = 1)= (10

0)0.010 0.9910 + (10

1)0.011 0.999

= 0.99573



p=0.05:

P (X 1) = P (X = 0) + P (X = 1)= (10

0)0.050 0.9510 + (10

1)0.051 0.959

= 0.91386

p=0.1:

P (X 1) = P (X = 0) + P (X = 1)= (10

0)0.10 0.910 + (10

1)0.11 0.99

= 0.73610

p=0.2:

P (X 1) = P (X = 0) + P (X = 1)= 0.37581

p=0.25:

P (X 1) = P (X = 0) + P (X = 1)= 0.24403

The operating characteristic curve is shown in Figure 2:



0 0.025 0.05 0.075 0.1 0.125 0.15 0.175 0.2 0.225 0.25 0.275

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0(0.01, 0.996)

(0.05, 0.914)

(0.1, 0.736)

(0.2, 0.376)

(0.25, 0.224)

b

b

b

b

b

Figure 2: Operating characteristic curve for at most one defective item in a batch of 10.



d. (a) Batch will be accepted when number of defective components in sample of 15 items is at mosttwo:

p=0.01:

P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)= (15

0)0.010 0.9915 + (15

1)0.011 0.9914 + (14

2)0.012 0.9913

= 0.99958

p=0.05:

P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)= (15

0)0.050 0.9515 + (15

1)0.051 0.9514 + (15

2)0.052 0.9513

= 0.96380

p=0.1:

P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)= (15

0)0.10 0.915 + (15

1)0.11 0.914 + (15

2)0.12 0.913

= 0.81594

p=0.2:

P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)= (15

0)0.20 0.815 + (15

1)0.21 0.814 + (15

2)0.22 0.813

= 0.39802

p=0.25:

P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)= (15

0)0.250 0.7515 + (15

1)0.251 0.7514 + (15

2)0.252 0.7513

= 0.23609

The operating characteristic curve is shown in Figure 3:





0 0.025 0.05 0.075 0.1 0.125 0.15 0.175 0.2 0.225 0.25 0.275

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

(0.01, 0.99958)(0.05, 0.964)

(0.1, 0.816)

(0.2, 0.398)

(0.25, 0.236)

b

b

b

b

b

Figure 3: Operating characteristic curve for at most two defective items in a batch of 15.

e. Sampling plan (b) appears most satisfactory, because the more stringent the acceptance criteriaare, the lower the probability of accepting a batch will become.

Question 67, p.122


Section 3.5, p.127

You can do this section for self-study. However, you will not be assessed on it in any quizzes, semestertests or examinations.

Section 3.6, p.132

Question 79, p.132




Question 80, p.132

An anomoly is an abnormality or an irregularity; something that is against the norm.

We have been given that = 4. Now, X Poi(4) so that the probability of observing x anomolies ina particular region of an aircraft gas-turbine disk can be calculated by using the formula

P (X = x) = e4 4xx!

definition, p.128

a. The probability that there are at most four anomolies:

P (X 4) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)=e4 40

0!+e4 41

1!+e4 42

2!+e4 43

3!+e4 44

4!= 0.0183 + 0.0733 + 0.147 + 0.195 + 0.195

= 0.629

The probability that there are less than four anomolies:

P (X < 4) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)=e4 40

0!+e4 41

1!+e4 42

2!+e4 43

3!= 0.0183 + 0.0733 + 0.147 + 0.195

= 0.433

When working with all the decimals, answer is 0.433. When worrking with rounded answers,answer is 0.434. More accurate to only round final answer: i.e. P (X < 4) = 0.433

b. The probability that there are between four and eight (inclusive) anomolies:

P (4 X 8) = P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8)=e4 44

4!+e4 45

5!+e4 46

6!+e4 47

7!+e4 48

8!= 0.545

c. The probability that there are at least eight anomolies:

P (X 8) = 1 P (X 7) first proposition (complement rule), p.59= 1 P (X 4) P (4 X 8) + P (X = 8) from (a) above; from (b) above= 1 0.433 0.545 +

e4 48

8!= 0.0518

d. First determine the mean and standard deviation number of anomolies:

E(X) = = 4 proposition, p.130X =

V (X) = =4 = 2 proposition, p.130



P (Number of anomolies exceeds its mean value by no more than one standard deviation)= P (


Question 100, p.134

Let X represent the number of defective chips in the sample of 25. Then X will be a discrete randomvariable with the following assumptions that can be made 3.4, p.114 :

1. The experiment consists of testing 25 chips (n = 25 trials).

2. Each chip can either be defective or not (only two possible outcomes per trial).

3. The 25 chips will be assumed to be independent of each other.

4. The probability of a defective chip (p) will be assumed to be constant for all 25 components.

Because all the afore-mentioned assumptions can be assumed to hold, X Bin(25, p), so that theprobability of observing x defective chips from the selection of 25 can be calculated as

P (X = x) = (25x)px (1 p)25x.

a. p = 0.05. The batch will be rejected when five or more of the chips are defective:

P (X 5)= 1 P (X 4)= 1 [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]= 1 [(25

0)0.050 0.9525 + (25

1)0.051 0.9524 + . . . + (25

4)0.054 0.9521]

= 0.00716

b. p = 0.1. The batch will be rejected when five or more of the chips are defective:

P (X 5)= 1 P (X 4)= 1 [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]= 1 [(25

0)0.10 0.925 + (25

1)0.11 0.924 + . . . + (25

4)0.14 0.921]

= 0.09799

c. p = 0.2. The batch will be rejected when five or more of the chips are defective:

P (X 5)= 1 P (X 4)= 1 [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]= 1 [(25

0)0.20 0.825 + (25

1)0.21 0.824 + . . . + (25

4)0.24 0.821]

= 0.579

d. The probabilities of rejection will all decrease.

Question 117, p.135


Question 119, p.136



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STE3A01 HW Solutions Chapter 3 Part 1