298 Chapter 5 Discrete Probability Distributions · 298 Chapter 5 Discrete Probability Distributions 5–48 Section 5–2 Expected Value 1. The expected value is the mean in a discrete

298 Chapter 5 Discrete Probability Distributions

5–48

Section 5–2 Expected Value

1. The expected value is the mean in a discrete probability

distribution.

2. We would expect variation from the expected

value of 3.

3. Answers will vary. One possible answer is that pregnant

mothers in that area might be overly concerned upon

hearing that the number of cases of kidney problems

in newborns was nearly 4 times what was usually

expected. Other mothers (particularly those who had

taken a statistics course!) might ask for more

information about the claim.

4. Answers will vary. One possible answer is that it does

seem unlikely to have 11 newborns with kidney

problems when we expect only 3 newborns to have

kidney problems.

5. The public might better be informed by percentages or

rates (e.g., rate per 1000 newborns).

6. The increase of 8 babies born with kidney problems

represents a 0.32% increase (less than %).

7. Answers will vary. One possible answer is that the

percentage increase does not seem to be something to

be overly concerned about.

Section 5–3 Unsanitary Restaurants

1. The probability of eating at 3 restaurants with

unsanitary conditions out of the 10 restaurants is

0.18793.

2. The probability of eating at 4 or 5 restaurants with

unsanitary conditions out of the 10 restaurants is

(0.24665) 1 (0.22199) 5 0.46864.

12

3. To find this probability, you could add the probabilities

for eating at 1, 2, . . . , 10 unsanitary restaurants. An

easier way to compute the probability is to subtract the

probability of eating at no unsanitary restaurants from 1

(using the complement rule).

4. The highest probability for this distribution is 4, but the

expected number of unsanitary restaurants that you

would eat at is

5. The standard deviation for this distribution is

6. We have two possible outcomes: “success” is eating

in an unsanitary restaurant; “failure” is eating in a

sanitary restaurant. The probability that one restaurant

is unsanitary is independent of the probability that any

other restaurant is unsanitary. The probability that a

restaurant is unsanitary remains constant at . And we

are looking at the number of unsanitary restaurants that

we eat at out of 10 “trials.”

7. The likelihood of success will vary from situation to

situation. Just because we have two possible outcomes,

this does not mean that each outcome occurs with

probability 0.50.

Section 5–4 Rockets and Targets

1. The theoretical values for the number of hits are as

follows:

Hits 0 1 2 3 4 5

Regions 227.5 211.3 98.2 30.4 7.1 1.3

2. The actual values are very close to the theoretical values.

3. Since the actual values are close to the theoretical values,

it does appear that the rockets were fired at random.

37

2_10+_37+_4

7+ 5 1.56.

10 •37 5 4.29.

6–1

6C H A P T E R

Outline

Introduction

6–1 Normal Distributions

6–2 Applications of the Normal Distribution

6–3 The Central Limit Theorem

6–4 The Normal Approximation to the BinomialDistribution

Summary

Objectives

After completing this chapter, you should be able to

1 Identify distributions as symmetric or skewed.

2 Identify the properties of a normal distribution.

3 Find the area under the standard normaldistribution, given various z values.

4 Find probabilities for a normally distributedvariable by transforming it into a standardnormal variable.

5 Find specific data values for givenpercentages, using the standard normaldistribution.

6 Use the central limit theorem to solveproblems involving sample means for largesamples.

7 Use the normal approximation to computeprobabilities for a binomial variable.

6The Normal

Distribution

300 Chapter 6 The Normal Distribution

6–2

StatisticsToday

What Is Normal?Medical researchers have determined so-called normal intervals for a person’s bloodpressure, cholesterol, triglycerides, and the like. For example, the normal range of sys-tolic blood pressure is 110 to 140. The normal interval for a person’s triglycerides is from30 to 200 milligrams per deciliter (mg/dl). By measuring these variables, a physician candetermine if a patient’s vital statistics are within the normal interval or if some type oftreatment is needed to correct a condition and avoid future illnesses. The question then is,How does one determine the so-called normal intervals? See Statistics Today—Revisitedat the end of the chapter.

In this chapter, you will learn how researchers determine normal intervals for specificmedical tests by using a normal distribution. You will see how the same methods are usedto determine the lifetimes of batteries, the strength of ropes, and many other traits.

Introduction

Random variables can be either discrete or continuous. Discrete variables and their dis-tributions were explained in Chapter 5. Recall that a discrete variable cannot assume allvalues between any two given values of the variables. On the other hand, a continuousvariable can assume all values between any two given values of the variables. Examplesof continuous variables are the heights of adult men, body temperatures of rats, and cho-lesterol levels of adults. Many continuous variables, such as the examples just mentioned,have distributions that are bell-shaped, and these are called approximately normally dis-

tributed variables. For example, if a researcher selects a random sample of 100 adultwomen, measures their heights, and constructs a histogram, the researcher gets a graphsimilar to the one shown in Figure 6–1(a). Now, if the researcher increases the sample sizeand decreases the width of the classes, the histograms will look like the ones shown inFigure 6–1(b) and (c). Finally, if it were possible to measure exactly the heights of alladult females in the United States and plot them, the histogram would approach what iscalled a normal distribution, shown in Figure 6–1(d). This distribution is also known as

Historical Note

The name normal

curve was used by

several statisticians,

namely, Francis

Galton, Charles

Sanders, Wilhelm

Lexis, and Karl

Pearson near the end

of the 19th century.

a bell curve or a Gaussian distribution, named for the German mathematician CarlFriedrich Gauss (1777–1855), who derived its equation.

No variable fits a normal distribution perfectly, since a normal distribution is atheoretical distribution. However, a normal distribution can be used to describe manyvariables, because the deviations from a normal distribution are very small. This conceptwill be explained further in Section 6–1.

When the data values are evenly distributed about the mean, a distribution is said tobe a symmetric distribution. (A normal distribution is symmetric.) Figure 6–2(a) showsa symmetric distribution. When the majority of the data values fall to the left or right ofthe mean, the distribution is said to be skewed. When the majority of the data values fallto the right of the mean, the distribution is said to be a negatively or left-skewed distri-

bution. The mean is to the left of the median, and the mean and the median are to the leftof the mode. See Figure 6–2(b). When the majority of the data values fall to the left ofthe mean, a distribution is said to be a positively or right-skewed distribution. Themean falls to the right of the median, and both the mean and the median fall to the rightof the mode. See Figure 6–2(c).

Chapter 6 The Normal Distribution 301

6–3

(a) Random sample of 100 women (b) Sample size increased and class width decreased

(c) Sample size increased and class width decreased further

(d) Normal distribution for the population

Figure 6–1

Histograms for the

Distribution of Heights

of Adult Women

MeanMedianMode

(a) Normal

Mode

(b) Negatively skewed

MedianMean Mean

(c) Positively skewed

MedianMode

Figure 6–2

Normal and Skewed

Distributions

Objective

Identify distributionsas symmetric orskewed.

1

The “tail” of the curve indicates the direction of skewness (right is positive, left isnegative). These distributions can be compared with the ones shown in Figure 3–1 inChapter 3. Both types follow the same principles.

This chapter will present the properties of a normal distribution and discuss itsapplications. Then a very important fact about a normal distribution called the central

limit theorem will be explained. Finally, the chapter will explain how a normaldistribution curve can be used as an approximation to other distributions, such as thebinomial distribution. Since a binomial distribution is a discrete distribution, a cor-rection for continuity may be employed when a normal distribution is used for itsapproximation.


6–4

6–1 Normal Distributions

Objective

Identify the propertiesof a normaldistribution.

2

y

Circle

Wheel

x

x 2 + y 2 = r 2

Figure 6–3

Graph of a Circle and

an Application

In mathematics, curves can be represented by equations. For example, the equation of thecircle shown in Figure 6–3 is x2 1 y2 5 r2, where r is the radius. A circle can be used torepresent many physical objects, such as a wheel or a gear. Even though it is not possi-ble to manufacture a wheel that is perfectly round, the equation and the properties of acircle can be used to study many aspects of the wheel, such as area, velocity, and accel-eration. In a similar manner, the theoretical curve, called a normal distribution curve,

can be used to study many variables that are not perfectly normally distributed but arenevertheless approximately normal.

The mathematical equation for a normal distribution is

where

e < 2.718 (< means “is approximately equal to”)

p < 3.14

m 5 population mean

s 5 population standard deviation

This equation may look formidable, but in applied statistics, tables or technology is usedfor specific problems instead of the equation.

Another important consideration in applied statistics is that the area under a normaldistribution curve is used more often than the values on the y axis. Therefore, when anormal distribution is pictured, the y axis is sometimes omitted.

Circles can be different sizes, depending on their diameters (or radii), and can beused to represent wheels of different sizes. Likewise, normal curves have different shapesand can be used to represent different variables.

The shape and position of a normal distribution curve depend on two parameters, themean and the standard deviation. Each normally distributed variable has its own normaldistribution curve, which depends on the values of the variable’s mean and standarddeviation. Figure 6–4(a) shows two normal distributions with the same mean values butdifferent standard deviations. The larger the standard deviation, the more dispersed, orspread out, the distribution is. Figure 6–4(b) shows two normal distributions with thesame standard deviation but with different means. These curves have the same shapes butare located at different positions on the x axis. Figure 6–4(c) shows two normal distribu-tions with different means and different standard deviations.

y 5e2_X2m+2/_2s2+

sÏ2p

Section 6–1 Normal Distributions 303

6–5

Curve 2

(b) Different means but same standard deviations

Curve 1

m1 m2

(a) Same means but different standard deviations

m1 = m2

Curve 2

(c) Different means and different standard deviations

Curve 1

Curve 1

m1 m2

Curve 2

1 > 2

1 > 2

1 = 2

Figure 6–4

Shapes of Normal

Distributions

Historical Notes

The discovery of theequation for a normaldistribution can betraced to threemathematicians. In1733, the FrenchmathematicianAbraham DeMoivrederived an equation fora normal distributionbased on the randomvariation of the numberof heads appearingwhen a large numberof coins were tossed.Not realizing anyconnection with thenaturally occurringvariables, he showedthis formula to onlya few friends. About100 years later, twomathematicians, PierreLaplace in France andCarl Gauss inGermany, derived theequation of the normalcurve independentlyand without anyknowledge ofDeMoivre’s work. In1924, Karl Pearsonfound that DeMoivrehad discovered theformula before Laplaceor Gauss.

The values given in item 8 of the summary follow the empirical rule for data givenin Section 3–2.

You must know these properties in order to solve problems involving distributionsthat are approximately normal.

A normal distribution is a continuous, symmetric, bell-shaped distribution of a variable.

The properties of a normal distribution, including those mentioned in the definition,are explained next.

Summary of the Properties of the Theoretical Normal Distribution

1. A normal distribution curve is bell-shaped.

2. The mean, median, and mode are equal and are located at the center of the distribution.

3. A normal distribution curve is unimodal (i.e., it has only one mode).

4. The curve is symmetric about the mean, which is equivalent to saying that its shape is thesame on both sides of a vertical line passing through the center.

5. The curve is continuous; that is, there are no gaps or holes. For each value of X, there is acorresponding value of Y.

6. The curve never touches the x axis. Theoretically, no matter how far in either directionthe curve extends, it never meets the x axis—but it gets increasingly closer.

7. The total area under a normal distribution curve is equal to 1.00, or 100%. This factmay seem unusual, since the curve never touches the x axis, but one can prove itmathematically by using calculus. (The proof is beyond the scope of this textbook.)

8. The area under the part of a normal curve that lies within 1 standard deviation of themean is approximately 0.68, or 68%; within 2 standard deviations, about 0.95, or 95%;and within 3 standard deviations, about 0.997, or 99.7%. See Figure 6–5, which alsoshows the area in each region.


6–6

2.28% 13.59%

34.13%

About 68%

m – 3 m – 2 mm – 1 m + 1 m + 2 m + 3

About 95%

About 99.7%

34.13%

13.59% 2.28%

Figure 6–5

Areas Under a Normal

Distribution Curve

Objective

Find the area underthe standard normaldistribution, givenvarious z values.

3

The Standard Normal Distribution

Since each normally distributed variable has its own mean and standard deviation, asstated earlier, the shape and location of these curves will vary. In practical applications,then, you would have to have a table of areas under the curve for each variable. To sim-plify this situation, statisticians use what is called the standard normal distribution.

The standard normal distribution is a normal distribution with a mean of 0 and astandard deviation of 1.

The standard normal distribution is shown in Figure 6–6.The values under the curve indicate the proportion of area in each section. For exam-

ple, the area between the mean and 1 standard deviation above or below the mean isabout 0.3413, or 34.13%.

The formula for the standard normal distribution is

All normally distributed variables can be transformed into the standard normally dis-tributed variable by using the formula for the standard score:

This is the same formula used in Section 3–3. The use of this formula will be explainedin Section 6–3.

As stated earlier, the area under a normal distribution curve is used to solve practi-cal application problems, such as finding the percentage of adult women whose height isbetween 5 feet 4 inches and 5 feet 7 inches, or finding the probability that a new batterywill last longer than 4 years. Hence, the major emphasis of this section will be to showthe procedure for finding the area under the standard normal distribution curve for anyz value. The applications will be shown in Section 6–2. Once the X values are trans-formed by using the preceding formula, they are called z values. The z value or z score

is actually the number of standard deviations that a particular X value is away from themean. Table E in Appendix C gives the area (to four decimal places) under the standardnormal curve for any z value from 23.49 to 3.49.

z 5value 2 mean

standard deviation or z 5

X 2 m

s

y 5e2z 2/2

Ï2p

Finding Areas Under the Standard Normal Distribution Curve

For the solution of problems using the standard normal distribution, a two-step processis recommended with the use of the Procedure Table shown.

The two steps are

Step 1 Draw the normal distribution curve and shade the area.

Step 2 Find the appropriate figure in the Procedure Table and follow the directionsgiven.

There are three basic types of problems, and all three are summarized in theProcedure Table. Note that this table is presented as an aid in understanding how to usethe standard normal distribution table and in visualizing the problems. After learningthe procedures, you should not find it necessary to refer to the Procedure Table for everyproblem.


6–7

2.28% 13.59%

34.13%

– 3 – 2 0– 1 + 1 + 2 + 3

34.13%

13.59% 2.28%

Figure 6–6

Standard Normal

Distribution

Interesting Fact

Bell-shaped

distributions occurred

quite often in early

coin-tossing and

die-rolling experiments.

Procedure Table

Finding the Area Under the Standard Normal Distribution Curve

1. To the left of any z value:Look up the z value in the table and use the area given.

0–z0

or

+z 0 +z0

or

–z

0 +z–z 00

oror

z2z1 –z2–z1

2. To the right of any z value:Look up the z value and subtract the area from 1.

3. Between any two z values:Look up both z values and subtract thecorresponding areas.


6–8

Table E in Appendix C gives the area under the normal distribution curve to the leftof any z value given in two decimal places. For example, the area to the left of a z valueof 1.39 is found by looking up 1.3 in the left column and 0.09 in the top row. Where thetwo lines meet gives an area of 0.9177. See Figure 6–7.

Example 6–1 Find the area to the left of z 5 2.06.

Solution

Step 1 Draw the figure. The desired area is shown in Figure 6–8.

0 2.06

Figure 6–8

Area Under the

Standard Normal

Distribution Curve for

Example 6–1

0–1.19

Figure 6–9

Area Under the

Standard Normal


Example 6–2

Example 6–2 Find the area to the right of z 5 21.19.

Solution

Step 1 Draw the figure. The desired area is shown in Figure 6–9.

Figure 6–7

Table E Area Value for

z 5 1.39

z 0.00 …

0.0

1.3 0.9177

......

0.09

Step 2 We are looking for the area under the standard normal distribution to the leftof z 5 2.06. Since this is an example of the first case, look up the area in thetable. It is 0.9803. Hence, 98.03% of the area is less than z 5 2.06.


6–9

Step 2 We are looking for the area to the right of z 5 21.19. This is an exampleof the second case. Look up the area for z 5 21.19. It is 0.1170. Subtract itfrom 1.0000. 1.0000 2 0.1170 5 0.8830. Hence, 88.30% of the area under thestandard normal distribution curve is to the left of z 5 21.19.

Example 6–3 Find the area between z 5 11.68 and z 5 21.37.

Solution

Step 1 Draw the figure as shown. The desired area is shown in Figure 6–10.

0 1.68–1.37

Figure 6–10

Area Under the

Standard Normal

Distribution Curve

for Example 6–3

Step 2 Since the area desired is between two given z values, look up the areascorresponding to the two z values and subtract the smaller area from thelarger area. (Do not subtract the z values.) The area for z 5 11.68 is 0.9535,and the area for z 5 21.37 is 0.0853. The area between the two z values is0.9535 2 0.0853 5 0.8682 or 86.82%.

A Normal Distribution Curve as a Probability Distribution Curve

A normal distribution curve can be used as a probability distribution curve for normallydistributed variables. Recall that a normal distribution is a continuous distribution, asopposed to a discrete probability distribution, as explained in Chapter 5. The fact that itis continuous means that there are no gaps in the curve. In other words, for every z valueon the x axis, there is a corresponding height, or frequency, value.

The area under the standard normal distribution curve can also be thought of as aprobability. That is, if it were possible to select any z value at random, the probability ofchoosing one, say, between 0 and 2.00 would be the same as the area under the curvebetween 0 and 2.00. In this case, the area is 0.4772. Therefore, the probability ofrandomly selecting any z value between 0 and 2.00 is 0.4772. The problems involvingprobability are solved in the same manner as the previous examples involving areasin this section. For example, if the problem is to find the probability of selecting az value between 2.25 and 2.94, solve it by using the method shown in case 3 of theProcedure Table.

For probabilities, a special notation is used. For example, if the problem is tofind the probability of any z value between 0 and 2.32, this probability is written asP(0 , z , 2.32).


6–10

Example 6–4 Find the probability for each.

a. P(0 , z , 2.32)

b. P(z , 1.65)

c. P(z . 1.91)

Solution

a. P(0 , z , 2.32) means to find the area under the standard normal distribution

curve between 0 and 2.32. First look up the area corresponding to 2.32. It is

0.9898. Then look up the area corresponding to z 5 0. It is 0.500. Subtract the

two areas: 0.9898 2 0.5000 5 0.4898. Hence the probability is 0.4898, or

48.98%. This is shown in Figure 6–11.

0 2.32

Figure 6–11

Area Under the

Standard Normal


Part a of Example 6–4

0 1.65

Figure 6–12

Area Under the

Standard Normal

Distribution Curve

for Part b of

Example 6–4

b. P(z , 1.65) is represented in Figure 6–12. Look up the area corresponding

to z 5 1.65 in Table E. It is 0.9505. Hence, P(z , 1.65) 5 0.9505,

or 95.05%.

c. P(z . 1.91) is shown in Figure 6–13. Look up the area that corresponds to

z 5 1.91. It is 0.9719. Then subtract this area from 1.0000. P(z . 1.91) 5

1.0000 2 0.9719 5 0.0281, or 2.81%.

Note: In a continuous distribution, the probability of any exact z value is 0 since the

area would be represented by a vertical line above the value. But vertical lines in theory

have no area. So .P_a # z # b+ 5 P_a , z , b+


6–11

Example 6–5 Find the z value such that the area under the standard normal distribution curve between0 and the z value is 0.2123.

Solution

Draw the figure. The area is shown in Figure 6–14.

0 1.91

Figure 6–13

Area Under the

Standard Normal

Distribution Curve

for Part c of

Example 6–4

In this case it is necessary to add 0.5000 to the given area of 0.2123 to get thecumulative area of 0.7123. Look up the area in Table E. The value in the left column is0.5, and the top value is 0.06. Add these two values to get z 5 0.56. See Figure 6–15.

Sometimes, one must find a specific z value for a given area under the standardnormal distribution curve. The procedure is to work backward, using Table E.

Since Table E is cumulative, it is necessary to locate the cumulative area up to agiven z value. Example 6–5 shows this.

0 z

0.2123Figure 6–14

Area Under the

Standard Normal


Example 6–5

z .00 .01 .02 .03 .04 .05 .07 .09

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.7123

...

.06 .08

Start here

Figure 6–15

Finding the z Value

from Table E for

Example 6–5


6–12

If the exact area cannot be found, use the closest value. For example, if you wantedto find the z value for an area 0.9241, the closest area is 0.9236, which gives a z value of1.43. See Table E in Appendix C.

The rationale for using an area under a continuous curve to determine a probabilitycan be understood by considering the example of a watch that is powered by a battery.When the battery goes dead, what is the probability that the minute hand will stop some-where between the numbers 2 and 5 on the face of the watch? In this case, the values ofthe variable constitute a continuous variable since the hour hand can stop anywhere onthe dial’s face between 0 and 12 (one revolution of the minute hand). Hence, the samplespace can be considered to be 12 units long, and the distance between the numbers 2 and5 is 5 2 2, or 3 units. Hence, the probability that the minute hand stops on a numberbetween 2 and 5 is . See Figure 6–16(a).

The problem could also be solved by using a graph of a continuous variable. Let usassume that since the watch can stop anytime at random, the values where the minutehand would land are spread evenly over the range of 0 through 12. The graph would thenconsist of a continuous uniform distribution with a range of 12 units. Now if we requirethe area under the curve to be 1 (like the area under the standard normal distribution), theheight of the rectangle formed by the curve and the x axis would need to be . The reasonis that the area of a rectangle is equal to the base times the height. If the base is 12 unitslong, then the height has to be since .

The area of the rectangle with a base from 2 through 5 would be or . SeeFigure 6–16(b). Notice that the area of the small rectangle is the same as the probabilityfound previously. Hence the area of this rectangle corresponds to the probability of thisevent. The same reasoning can be applied to the standard normal distribution curveshown in Example 6–5.

Finding the area under the standard normal distribution curve is the first step in solvinga wide variety of practical applications in which the variables are normally distributed.Some of these applications will be presented in Section 6–2.

143 ?

112,

12 ?1

12 5 1112

112

312 5

14

x

y

1 2 3 4 5 6 7 8 9 10 11 120

(b) Rectangle

1

12

1

12

1

12

3

12

1

4

3 units

Area 5 3 • 5 5

1

5

2

4

11

7

10

8

(a) Clock

3 units

P 53

12

1

45

Figure 6–16

The Relationship

Between Area and

Probability

Applying the Concepts 6–1

Assessing Normality

Many times in statistics it is necessary to see if a set of data values is approximately normally

distributed. There are special techniques that can be used. One technique is to draw a

histogram for the data and see if it is approximately bell-shaped. (Note: It does not have to

be exactly symmetric to be bell-shaped.)

The numbers of branches of the 50 top libraries are shown.

67 84 80 77 97 59 62 37 33 42

36 54 18 12 19 33 49 24 25 22

24 29 9 21 21 24 31 17 15 21

13 19 19 22 22 30 41 22 18 20

26 33 14 14 16 22 26 10 16 24

Source: The World Almanac and Book of Facts.

1. Construct a frequency distribution for the data.

2. Construct a histogram for the data.

3. Describe the shape of the histogram.

4. Based on your answer to question 3, do you feel that the distribution is approximately normal?

In addition to the histogram, distributions that are approximately normal have about 68%

of the values fall within 1 standard deviation of the mean, about 95% of the data values fall

within 2 standard deviations of the mean, and almost 100% of the data values fall within

3 standard deviations of the mean. (See Figure 6–5.)

5. Find the mean and standard deviation for the data.

6. What percent of the data values fall within 1 standard deviation of the mean?

7. What percent of the data values fall within 2 standard deviations of the mean?

8. What percent of the data values fall within 3 standard deviations of the mean?

9. How do your answers to questions 6, 7, and 8 compare to 68, 95, and 100%, respectively?

10. Does your answer help support the conclusion you reached in question 4? Explain.

(More techniques for assessing normality are explained in Section 6–2.)

See pages 353 and 354 for the answers.


6–13

1. What are the characteristics of a normal distribution?

2. Why is the standard normal distribution important in

statistical analysis? Many variables are normally distributed,and the distribution can be used to describe these variables.

3. What is the total area under the standard normal

distribution curve? 1 or 100%

4. What percentage of the area falls below the mean?

Above the mean? 50% of the area lies below the mean, and 50% of the area lies above the mean.

5. About what percentage of the area under the normal

distribution curve falls within 1 standard deviation

above and below the mean? 2 standard deviations?

3 standard deviations? 68%; 95%; 99.7%

For Exercises 6 through 25, find the area under the

standard normal distribution curve.

6. Between z 5 0 and z 5 1.77 0.4616

7. Between z 5 0 and z 5 0.75 0.2734

8. Between z 5 0 and z 5 20.32 0.1255

9. Between z 5 0 and z 5 22.07 0.4808

10. To the right of z 5 2.01 0.0222

11. To the right of z 5 0.29 0.3859

12. To the left of z 5 20.75 0.2266

13. To the left of z 5 21.39 0.0823

Exercises 6–1

14. Between z 5 1.23 and z 5 1.90 0.0806

15. Between z 5 1.05 and z 5 1.78 0.1094

16. Between z 5 20.96 and z 5 20.36 0.1909

17. Between z 5 21.56 and z 5 21.83 0.0258

18. Between z 5 0.24 and z 5 21.12 0.4634

19. Between z 5 21.46 and z 5 21.98 0.0482

20. To the left of z 5 1.31 0.9049

21. To the left of z 5 2.11 0.9826

22. To the right of z 5 21.92 0.9726

23. To the right of z 5 20.17 0.5675

24. To the left of z 5 22.15 and to the right of z 5 1.620.0684

25. To the right of z 5 1.92 and to the left of z 5 20.440.3574

In Exercises 26 through 39, find the probabilities for

each, using the standard normal distribution.

26. P(0 , z , 1.96) 0.4750

27. P(0 , z , 0.67) 0.2486

28. P(21.23 , z , 0) 0.3907

29. P(21.43 , z , 0) 0.4236

30. P(z . 0.82) 0.2061

31. P(z . 2.83) 0.0023

32. P(z , 21.77) 0.0384

33. P(z , 21.32) 0.0934

34. P(20.20 , z , 1.56) 0.5199

35. P(22.46 , z , 1.74) 0.9522 (TI: 0.9521)

36. P(1.12 , z , 1.43) 0.0550

37. P(1.46 , z , 2.97) 0.0706 (TI: 0.0707)

38. P(z . 21.43) 0.9236

39. P(z , 1.42) 0.9222

For Exercises 40 through 45, find the z value that

corresponds to the given area.

40. 1.32

0 z

0.4066

41. z 5 21.39

(TI: 21.3885)

42. 1.98

43. z 5 22.08

(TI: 22.0792)

44. 1.84

45. 21.26

(TI: 21.2602)

46. Find the z value to the right of the mean so that

a. 54.78% of the area under the distribution curve liesto the left of it. 0.12

b. 69.85% of the area under the distribution curve liesto the left of it. 0.52

c. 88.10% of the area under the distribution curve liesto the left of it. 1.18

47. Find the z value to the left of the mean so that

a. 98.87% of the area under the distribution curve liesto the right of it. 22.28 (TI: 22.2801)

b. 82.12% of the area under the distribution curve liesto the right of it. 20.92 (TI: 20.91995)

c. 60.64% of the area under the distribution curve liesto the right of it. 20.27 (TI: 20.26995)

0z

0.8962

0 z

0.9671

0z

0.0188

0 z

0.0239

0.4175

0z


6–14


6–15

50. In the standard normal distribution, find the values of z forthe 75th, 80th, and 92nd percentiles. 0.6745; 0.8416; 1.41

51. Find P(21 , z , 1), P(22 , z , 2), and P(23 , z , 3).How do these values compare with the empirical rule?0.6827; 0.9545; 0.9973; they are very close.

52. Find z0 such that P(z . z0) 5 0.1234. 1.16

53. Find z0 such that P(21.2 , z , z0) 5 0.8671. 2.10

54. Find z0 such that P(z0 , z , 2.5) 5 0.7672. 20.75

55. Find z0 such that the area between z0 and z 5 20.5 is0.2345 (two answers). 21.45 and 0.11

Extending the Concepts

56. Find z0 such that P(2z0 , z , z0) 5 0.76. 1.175

57. Find the equation for the standard normal distributionby substituting 0 for m and 1 for s in the equation

y 5

58. Graph by hand the standard normal distribution byusing the formula derived in Exercise 57. Let p < 3.14and e < 2.718. Use X values of 22, 21.5, 21, 20.5, 0,0.5, 1, 1.5, and 2. (Use a calculator to compute the yvalues.)

e2X 2/2

Ï2py 5

e2_X2m+2 / _2s2 +

sÏ2p

48. Find two z values so that 48% of the middle area isbounded by them. z 560.64

49. Find two z values, one positive and one negative, thatare equidistant from the mean so that the areas in thetwo tails total the following values.

a. 5% z 5 11.96 and z 5 21.96 (TI: 61.95996)

b. 10% z 5 11.65 and z 5 21.65, approximately

(TI: 61.64485)

c. 1% z 5 12.58 and z 5 22.58, approximately (TI: 62.57583)


It is possible to determine the height of the density curve given a value of z, the cumulativearea given a value of z, or a z value given a cumulative area. Examples are from Table E inAppendix C.

Find the Area to the Left of z 5 1.39

1. Select Calc>Probability Distributions>Normal. There are three options.

2. Click the button for Cumulative probability. In the center section, the mean and standarddeviation for the standard normal distribution are the defaults. The mean should be 0, andthe standard deviation should be 1.

3. Click the button for Input Constant, then click inside the text box and type in 1.39. Leavethe storage box empty.

4. Click [OK].

Technology Step by Step

MINITABStep by Step


6–16

The session window displays the result, 0.917736. If you choose the optional storage, typein a variable name such as K1. The result will be stored in the constant and will not be in thesession window.

Find the Area to the Right of 22.06

1. Select Calc>Probability Distributions>Normal.

2. Click the button for Cumulative probability.

3. Click the button for Input Constant, then enter 22.06 in the text box. Do not forget theminus sign.

4. Click in the text box for Optional storage and type K1.

5. Click [OK]. The area to the left of 22.06 is stored in K1 but not displayed in the sessionwindow.

To determine the area to the right of the z value, subtract this constant from 1, then displaythe result.

6. Select Calc>Calculator.

a) Type K2 in the text box for Store result in:.

b) Type in the expression 1 2 K1, then click [OK].

7. Select Data>Display Data. Drag the mouse over K1 and K2, then click [Select]and [OK].

The results will be in the session window and stored in the constants.

Data Display

K1 0.0196993

K2 0.980301

8. To see the constants and other information about the worksheet, click the Project Managericon. In the left pane click on the green worksheet icon, and then click the constants folder.You should see all constants and their values in the right pane of the Project Manager.

9. For the third example calculate the two probabilities and store them in K1 and K2.

10. Use the calculator to subtract K1 from K2 and store in K3.

The calculator and project manager windows are shown.

Cumulative Distribution Function

Normal with mean = 0 and standard deviation = 1

x P( X <= x )

1.39 0.917736

The graph is not shown in the output.


6–17

Calculate a z Value Given the Cumulative Probability

Find the z value for a cumulative probability of 0.025.

1. Select Calc>Probability Distributions>Normal.

2. Click the option for Inverse cumulative probability, then the option for Input constant.

3. In the text box type .025, the cumulative area, then click [OK].

4. In the dialog box, the z value will be returned, 21.960.

Inverse Cumulative Distribution Function

Normal with mean = 0 and standard deviation = 1

P ( X <= x ) x

0.025 21.95996

In the session window z is 21.95996.

TI-83 Plus orTI–84 PlusStep by Step

Standard Normal Random Variables

To find the probability for a standard normal random variable:Press 2nd [DISTR], then 2 for normalcdf(The form is normalcdf(lower z score, upper z score).Use E99 for ` (infinity) and 2E99 for 2` (negative infinity). Press 2nd [EE] to get E.

Example: Area to the right of z 5 1.11 normalcdf(1.11,E99)

Example: Area to the left of z 5 21.93 normalcdf(2E99,21.93)

Example: Area between z 5 2.00 and z 5 2.47 normalcdf(2.00,2.47)

To find the percentile for a standard normal random variable:Press 2nd [DISTR], then 3 for the invNorm(The form is invNorm(area to the left of z score)

Example: Find the z score such that the area under the standard normal curve to the left of it is0.7123 invNorm(.7123)

Excel Step by Step


Finding areas under the standard normal distribution curve

Example XL6–1

Find the area to the left of z 5 1.99.In a blank cell type: 5NORMSDIST(1.99)Answer: 0.976705

Example XL6–2

Find the area to the right of z 5 22.04.In a blank cell type: 5 1-NORMSDIST(22.04)Answer: 0.979325

Example XL6–3

Find the area between z 5 22.04 and z 5 1.99.In a blank cell type: 5NORMSDIST(1.99) 2 NORMSDIST(22.04)Answer: 0.956029

Finding a z value given an area under the standard normal distribution curve

Example XL6–4

Find a z score given the cumulative area (area to the left of z) is 0.0250.In a blank cell type: 5NORMSINV(.025)Answer: 21.95996


6–18

Objective

Find probabilities for a normallydistributed variableby transforming itinto a standardnormal variable.

4

6–2 Applications of the Normal DistributionThe standard normal distribution curve can be used to solve a wide variety of practicalproblems. The only requirement is that the variable be normally or approximately nor-mally distributed. There are several mathematical tests to determine whether a variableis normally distributed. See the Critical Thinking Challenges on page 352. For all theproblems presented in this chapter, you can assume that the variable is normally orapproximately normally distributed.

To solve problems by using the standard normal distribution, transform the originalvariable to a standard normal distribution variable by using the formula

This is the same formula presented in Section 3–3. This formula transforms the values ofthe variable into standard units or z values. Once the variable is transformed, then theProcedure Table and Table E in Appendix C can be used to solve problems.

For example, suppose that the scores for a standardized test are normally distributed,have a mean of 100, and have a standard deviation of 15. When the scores are trans-formed to z values, the two distributions coincide, as shown in Figure 6–17. (Recall thatthe z distribution has a mean of 0 and a standard deviation of 1.)

z 5value 2 mean

standard deviation or z 5

X 2 m

s

0 1–1–2–3 2 3

100 115857055 130 145

z

Figure 6–17

Test Scores and Their

Corresponding z

Values

To solve the application problems in this section, transform the values of the variableto z values and then find the areas under the standard normal distribution, as shown inSection 6–1.

Section 6–2 Applications of the Normal Distribution 317

6–19

Example 6–6 Summer Spending

A survey found that women spend on average $146.21 on beauty products during thesummer months. Assume the standard deviation is $29.44. Find the percentage of womenwho spend less than $160.00. Assume the variable is normally distributed.

Solution

Step 1 Draw the figure and represent the area as shown in Figure 6–18.

$146.21 $160

Figure 6–18

Area Under a

Normal Curve for

Example 6–6

0 0.47

Figure 6–19

Area and z Values for

Example 6–6

Step 2 Find the z value corresponding to $160.00.

Hence $160.00 is 0.47 of a standard deviation above the mean of $146.21, asshown in the z distribution in Figure 6–19.

z 5X 2 m

s5

$160.00 2 $146.21

$29.445 0.47

Step 3 Find the area, using Table E. The area under the curve to the left of z 5 0.47is 0.6808.

Therefore 0.6808, or 68.08%, of the women spend less than $160.00 on beauty productsduring the summer months.

Example 6–7 Monthly Newspaper Recycling

Each month, an American household generates an average of 28 pounds of newspaperfor garbage or recycling. Assume the standard deviation is 2 pounds. If a household isselected at random, find the probability of its generating

a. Between 27 and 31 pounds per month

b. More than 30.2 pounds per month

Assume the variable is approximately normally distributed.

Source: Michael D. Shook and Robert L. Shook, The Book of Odds.


6–20

Solution a

Step 1 Draw the figure and represent the area. See Figure 6–20.

28 3127

Figure 6–20

Area Under a Normal

Curve for Part a of

Example 6–7

28 3127

0 1.5–0.5

Figure 6–21

Area and z Values for

Part a of Example 6–7

28 30.2

Figure 6–22

Area Under a Normal

Curve for Part b of

Example 6–7

Step 2 Find the two z values.

Step 3 Find the appropriate area, using Table E. The area to the left of z2 is 0.9332,

and the area to the left of z1 is 0.3085. Hence the area between z1 and z2 is

0.9332 2 0.3085 5 0.6247. See Figure 6–21.

z2 5

X 2 m

s5

31 2 28

25

3

25 1.5

z1 5

X 2 m

s5

27 2 28

25 2

1

25 20.5

Hence, the probability that a randomly selected household generates between 27 and

31 pounds of newspapers per month is 62.47%.

Solution b

Step 1 Draw the figure and represent the area, as shown in Figure 6–22.

Historical Note

Astronomers in the

late 1700s and the

1800s used the

principles underlying

the normal distribution

to correct

measurement errors

that occurred in

charting the positions

of the planets.

Step 2 Find the z value for 30.2.

z 5

X 2 m

s5

30.2 2 28

25

2.2

25 1.1


6–21

Step 3 Find the appropriate area. The area to the left of z 5 1.1 is 0.8643. Hence thearea to the right of z 5 1.1 is 1.0000 2 0.8643 5 0.1357.

Hence, the probability that a randomly selected household willaccumulate more than 30.2 pounds of newspapers is 0.1357, or 13.57%.

A normal distribution can also be used to answer questions of “How many?” Thisapplication is shown in Example 6–8.

1.641

Figure 6–23

Area Under a

Normal Curve for

Example 6–8

Example 6–8 Coffee Consumption

Americans consume an average of 1.64 cups of coffee per day. Assume the variable isapproximately normally distributed with a standard deviation of 0.24 cup. If 500individuals are selected, approximately how many will drink less than 1 cup of coffeeper day?

Source: Chicago Sun-Times.

Solution

Step 1 Draw a figure and represent the area as shown in Figure 6–23.

Step 2 Find the z value for 1.

Step 3 Find the area to the left of z 5 22.67. It is 0.0038.

Step 4 To find how many people drank less than 1 cup of coffee, multiply the samplesize 500 by 0.0038 to get 1.9. Since we are asking about people, round theanswer to 2 people. Hence, approximately 2 people will drink less than 1 cupof coffee a day.

Note: For problems using percentages, be sure to change the percentage to a decimalbefore multiplying. Also, round the answer to the nearest whole number, since it is notpossible to have 1.9 people.

Finding Data Values Given Specific Probabilities

A normal distribution can also be used to find specific data values for given percentages.This application is shown in Example 6–9.

z 5X 2 m

s5

1 2 1.64

0.245 22.67


6–22

Example 6–9 Police Academy Qualifications

To qualify for a police academy, candidates must score in the top 10% on a generalabilities test. The test has a mean of 200 and a standard deviation of 20. Find the lowestpossible score to qualify. Assume the test scores are normally distributed.

Solution

Since the test scores are normally distributed, the test value X that cuts off the upper 10%of the area under a normal distribution curve is desired. This area is shown in Figure 6–24.

200 X

10%, or 0.1000

Figure 6–24

Area Under a

Normal Curve for

Example 6–9

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

0.0

0.1

0.2

1.1

1.2

1.3

1.4

0.8997

......

0.9000

0.9015

Closest value

Specificvalue

Figure 6–25

Finding the z Value

from Table E

(Example 6–9)

Work backward to solve this problem.

Step 1 Subtract 0.1000 from 1.000 to get the area under the normal distribution to theleft of x: 1.0000 2 0.10000 5 0.9000.

Step 2 Find the z value that corresponds to an area of 0.9000 by looking up 0.9000 inthe area portion of Table E. If the specific value cannot be found, use the closestvalue—in this case 0.8997, as shown in Figure 6–25. The corresponding z

value is 1.28. (If the area falls exactly halfway between two z values, use thelarger of the two z values. For example, the area 0.9500 falls halfway between0.9495 and 0.9505. In this case use 1.65 rather than 1.64 for the z value.)

Objective

Find specific datavalues for givenpercentages, usingthe standard normaldistribution.

5

Interesting Fact

Americans are the

largest consumers of

chocolate. We spend

$16.6 billion annually.

Step 3 Substitute in the formula z 5 (X 2 m)/s and solve for X.

A score of 226 should be used as a cutoff. Anybody scoring 226 or higher qualifies.

226 5 X

225.60 5 X

25.60 1 200 5 X

_1.28+_20+ 1 200 5 X

1.28 5X 2 200

20


6–23

Instead of using the formula shown in step 3, you can use the formula X 5 z ? s1 m.This is obtained by solving

for X as shown.

Multiply both sides by s.

Add m to both sides.

Exchange both sides of the equation.X 5 z • s 1 m

z • s 1 m 5 X

z • s 5 X 2 m

z 5X 2 m

s

Formula for Finding X

When you must find the value of X, you can use the following formula:

X 5 z ? s 1 m

Example 6–10 Systolic Blood Pressure

For a medical study, a researcher wishes to select people in the middle 60% of thepopulation based on blood pressure. If the mean systolic blood pressure is 120 and thestandard deviation is 8, find the upper and lower readings that would qualify people toparticipate in the study.

Solution

Assume that blood pressure readings are normally distributed; then cutoff points are asshown in Figure 6–26.

120 X1X2

30%

60%

20%20%

Figure 6–26

Area Under a

Normal Curve for

Example 6–10

Figure 6–26 shows that two values are needed, one above the mean and one belowthe mean. To get the area to the left of the positive z value, add 0.5000 1 0.3000 50.8000 (30% 5 0.3000). The z value with area to the left closest to 0.8000 is 0.84.Substituting in the formula X 5 zs 1 m gives

X1 5 zs 1 m 5 (0.84)(8) 1 120 5 126.72

The area to the left of the negative z value is 20%, or 0.2000. The area closest to 0.2000is 20.84.

X2 5 (20.84)(8) 1 120 5 113.28

Therefore, the middle 60% will have blood pressure readings of 113.28 , X , 126.72.

As shown in this section, a normal distribution is a useful tool in answering manyquestions about variables that are normally or approximately normally distributed.


6–24

Determining Normality

A normally shaped or bell-shaped distribution is only one of many shapes that a distribu-tion can assume; however, it is very important since many statistical methods require thatthe distribution of values (shown in subsequent chapters) be normally or approximatelynormally shaped.

There are several ways statisticians check for normality. The easiest way is to drawa histogram for the data and check its shape. If the histogram is not approximately bell-shaped, then the data are not normally distributed.

Skewness can be checked by using the Pearson coefficient of skewness (PC) alsocalled Pearson’s index of skewness. The formula is

If the index is greater than or equal to 11 or less than or equal to 21, it can be concludedthat the data are significantly skewed.

In addition, the data should be checked for outliers by using the method shown inChapter 3. Even one or two outliers can have a big effect on normality.

Examples 6–11 and 6–12 show how to check for normality.

PC 53_X 2 median+

s

Example 6–11 Technology Inventories

A survey of 18 high-technology firms showed the number of days’ inventory theyhad on hand. Determine if the data are approximately normally distributed.

5 29 34 44 45 63 68 74 7481 88 91 97 98 113 118 151 158

Source: USA TODAY.

Solution

Step 1 Construct a frequency distribution and draw a histogram for the data, asshown in Figure 6–27.

Class Frequency

5–29 230–54 355–79 480–104 5

105–129 2130–154 1155–179 1

Freq

uenc

y

4.5

5

4

3

2

1

Days

29.5 79.554.5 104.5 129.5 154.5 179.5

Figure 6–27

Histogram for

Example 6–11


6–25

Since the histogram is approximately bell-shaped, we can say that the distribution isapproximately normal.

Step 2 Check for skewness. For these data, 5 79.5, median 5 77.5, and s 5 40.5.Using the Pearson coefficient of skewness gives

PC 5

5 0.148

In this case, the PC is not greater than 11 or less than 21, so it can beconcluded that the distribution is not significantly skewed.

Step 3 Check for outliers. Recall that an outlier is a data value that lies more than1.5(IQR) units below Q1 or 1.5(IQR) units above Q3. In this case, Q1 5 45and Q3 5 98; hence, IQR 5 Q3 2 Q1 5 98 2 45 5 53. An outlier would be a data value less than 45 2 1.5(53) 5 234.5 or a data value larger than 98 1 1.5(53) 5 177.5. In this case, there are no outliers.

Since the histogram is approximately bell-shaped, the data are not significantlyskewed, and there are no outliers, it can be concluded that the distribution isapproximately normally distributed.

3_79.5 2 77.5+

40.5

X

Example 6–12 Number of Baseball Games Played

The data shown consist of the number of games played each year in the career ofBaseball Hall of Famer Bill Mazeroski. Determine if the data are approximatelynormally distributed.

81 148 152 135 151 152

159 142 34 162 130 162

163 143 67 112 70

Source: Greensburg Tribune Review.

Solution

Step 1 Construct a frequency distribution and draw a histogram for the data. SeeFigure 6–28.

Freq

uenc

y

33.5

8

7

6

5

4

3

2

1

Games

58.5 83.5 108.5 133.5 158.5 183.5

Figure 6–28

Histogram for

Example 6–12

Class Frequency

34–58 159–83 384–108 0

109–133 2134–158 7159–183 4


6–26

The histogram shows that the frequency distribution is somewhat negativelyskewed.

Step 2 Check for skewness; 5 127.24, median 5 143, and s 5 39.87.

PC 5

5

5 21.19

Since the PC is less than 21, it can be concluded that the distribution issignificantly skewed to the left.

Step 3 Check for outliers. In this case, Q1 5 96.5 and Q3 5 155.5. IQR 5 Q3 2

Q1 5 155.5 2 96.5 5 59. Any value less than 96.5 2 1.5(59) 5 8 or above155.5 1 1.5(59) 5 244 is considered an outlier. There are no outliers.

In summary, the distribution is somewhat negatively skewed.

Another method that is used to check normality is to draw a normal quantile plot.

Quantiles, sometimes called fractiles, are values that separate the data set into approxi-mately equal groups. Recall that quartiles separate the data set into four approximatelyequal groups, and deciles separate the data set into 10 approximately equal groups. A nor-mal quantile plot consists of a graph of points using the data values for the x coordinatesand the z values of the quantiles corresponding to the x values for the y coordinates.(Note: The calculations of the z values are somewhat complicated, and technology is usu-ally used to draw the graph. The Technology Step by Step section shows how to draw anormal quantile plot.) If the points of the quantile plot do not lie in an approximatelystraight line, then normality can be rejected.

There are several other methods used to check for normality. A method using normalprobability graph paper is shown in the Critical Thinking Challenge section at the end ofthis chapter, and the chi-square goodness-of-fit test is shown in Chapter 11. Two othertests sometimes used to check normality are the Kolmogorov-Smikirov test and theLilliefors test. An explanation of these tests can be found in advanced textbooks.


Smart People

Assume you are thinking about starting a Mensa chapter in your hometown of Visiala,California, which has a population of about 10,000 people. You need to know how manypeople would qualify for Mensa, which requires an IQ of at least 130. You realize that IQ isnormally distributed with a mean of 100 and a standard deviation of 15. Complete thefollowing.

1. Find the approximate number of people in Visiala who are eligible for Mensa.

2. Is it reasonable to continue your quest for a Mensa chapter in Visiala?

3. How could you proceed to find out how many of the eligible people would actually jointhe new chapter? Be specific about your methods of gathering data.

4. What would be the minimum IQ score needed if you wanted to start an Ultra-Mensa clubthat included only the top 1% of IQ scores?

See page 354 for the answers.

3_127.24 2 143+

39.87

3_X 2 median+

s

XUnusual Stats

The average amount

of money stolen by a

pickpocket each time

is $128.


6–27

1. Admission Charge for Movies The average early-birdspecial admission price for a movie is $5.81. If thedistribution of movie admission charges is approximatelynormal with a standard deviation of $0.81, what is theprobability that a randomly selected admission charge isless than $3.50? 0.0022

2. Teachers’Salaries The average annual salary for allU.S. teachers is $47,750. Assume that the distribution isnormal and the standard deviation is $5680. Find theprobability that a randomly selected teacher earns

a. Between $35,000 and $45,000 a year 0.3031

b. More than $40,000 a year 0.9131

c. If you were applying for a teaching position andwere offered $31,000 a year, how would you feel(based on this information)? Not too happy—it’s reallyat the bottom of the heap! (prob. 5 0.0016)

Source: New York Times Almanac.

3. Population in U.S. Jails The average daily jailpopulation in the United States is 706,242. If thedistribution is normal and the standard deviation is52,145, find the probability that on a randomly selectedday, the jail population is

a. Greater than 750,000 0.2005 (TI: 0.2007)

b. Between 600,000 and 700,000 0.4315 (TI: 0.4316)


4. SAT Scores The national average SAT score (forVerbal and Math) is 1028. If we assume a normaldistribution with s5 92, what is the 90th percentilescore? What is the probability that a randomly selectedscore exceeds 1200? 1146; 0.0307


5. Chocolate Bar Calories The average number ofcalories in a 1.5-ounce chocolate bar is 225. Supposethat the distribution of calories is approximately normalwith s 5 10. Find the probability that a randomlyselected chocolate bar will have

a. Between 200 and 220 calories 0.3023

b. Less than 200 calories 0.0062

Source: The Doctor’s Pocket Calorie, Fat, and Carbohydrate Counter.

6. Monthly Mortgage Payments The average monthlymortgage payment including principal and interest is$982 in the United States. If the standard deviation isapproximately $180 and the mortgage payments areapproximately normally distributed, find the probabilitythat a randomly selected monthly payment is

a. More than $1000 0.4602

b. More than $1475 0.0031

c. Between $800 and $1150 0.6676

Source: World Almanac.

7. Professors’ Salaries The average salary for a QueensCollege full professor is $85,900. If the average salaries

are normally distributed with a standard deviation of$11,000, find these probabilities. a. 0.3557 (TI: 0.3547)

a. The professor makes more than $90,000.

b. The professor makes more than $75,000.

Source: AAUP, Chronicle of Higher Education. b. 0.8389 (TI: 0.8391)

8. Doctoral Student Salaries Full-time Ph.D. studentsreceive an average of $12,837 per year. If the averagesalaries are normally distributed with a standarddeviation of $1500, find these probabilities.

a. The student makes more than $15,000. 0.0749

b. The student makes between $13,000 and $14,000. 0.2385

Source: U.S. Education Dept., Chronicle of Higher Education.

9. Miles Driven Annually The mean number of milesdriven per vehicle annually in the United States is12,494 miles. Choose a randomly selected vehicle, andassume the annual mileage is normally distributed witha standard deviation of 1290 miles. What is theprobability that the vehicle was driven more than 15,000miles? Less than 8000 miles? Would you buy a vehicleif you had been told that it had been driven less than6000 miles in the past year?

Source: World Almanac.

10. Commute Time to Work The average commute to work(one way) is 25 minutes according to the 2005 AmericanCommunity Survey. If we assume that commuting timesare normally distributed and that the standard deviation is6.1 minutes, what is the probability that a randomlyselected commuter spends more than 30 minutescommuting one way? Less than 18 minutes? 0.2061; 0.1251

Source: www.census.gov

11. Credit Card Debt The average credit card debt forcollege seniors is $3262. If the debt is normallydistributed with a standard deviation of $1100, findthese probabilities.

a. That the senior owes at least $1000 0.9803 (TI: 0.9801)

b. That the senior owes more than $4000

c. That the senior owes between $3000 and $4000

Source: USA TODAY. b. 0.2514 (TI: 0.2511) c. 0.3434 (TI: 0.3430)

12. Price of Gasoline The average retail price of gasoline(all types) for the first half of 2009 was 236.5 cents. Whatwould the standard deviation have to be in order for a15% probability that a gallon of gas costs less than $2.00?

Source: World Almanac. 35.1 cents

13. Waiting Time at a Bank Drive-in Window Theaverage waiting time at a drive-in window of a localbank is 10.3 minutes, with a standard deviation of2.7 minutes. Assume the variable is normallydistributed. If a customer arrives at the bank, find theprobability that the customer will have to wait

Exercises 6–2


a. Between 4 and 9 minutes 0.3057

b. Less than 5 minutes or more than 10 minutes 0.5688

c. How might a customer estimate his or herapproximate waiting time?

14. Newborn Elephant Weights Newborn elephant calvesusually weigh between 200 and 250 pounds—untilOctober 2006, that is. An Asian elephant at the Houston(Texas) Zoo gave birth to a male calf weighing in at awhopping 384 pounds! Mack (like the truck) is believedto be the heaviest elephant calf ever born at a facilityaccredited by the Association of Zoos and Aquariums.If, indeed, the mean weight for newborn elephant calvesis 225 pounds with a standard deviation of 45 pounds,what is the probability of a newborn weighing at least384 pounds? Assume that the weights of newbornelephants are normally distributed. Less than 0.0001

Source: www.houstonzoo.org

15. Waiting to Be Seated The average waiting time to beseated for dinner at a popular restaurant is 23.5 minutes,with a standard deviation of 3.6 minutes. Assume thevariable is normally distributed. When a patron arrivesat the restaurant for dinner, find the probability that thepatron will have to wait the following time.

a. Between 15 and 22 minutes 0.3281

b. Less than 18 minutes or more than 25 minutes 0.4002

c. Is it likely that a person will be seated in less than15 minutes? Not usually

16. Salary of Full-Time Male Professors The averagesalary of a male full professor at a public four-yearinstitution offering classes at the doctoral level is$99,685. For a female full professor at the same kind ofinstitution, the salary is $90,330. If the standarddeviation for the salaries of both genders isapproximately $5200 and the salaries are normallydistributed, find the 80th percentile salary for maleprofessors and for female professors.

Source: World Almanac. Men: $104,053 Women: $94,698

17. Lake Temperatures During September, the averagetemperature in Keystone Lake is 71.2°, and the standarddeviation is 3.4°. For a randomly selected day, find theprobability that the temperature of the lake is less than63°. Based on your answer, would this be a likely orunlikely occurrence? Assume the variable is normallydistributed. 0.0080 or 0.8%. A temperature of 63° is unlikelysince the probability is about 0.8%.

18. Itemized Charitable Contributions The averagecharitable contribution itemized per income tax return in Pennsylvania is $792. Suppose that thedistribution of contributions is normal with a standarddeviation of $103. Find the limits for the middle 50%of contributions. $722.99 and $861.01

Source: IRS, Statistics of Income Bulletin.

19. New Home Sizes A contractor decided to buildhomes that will include the middle 80% of the market.If the average size of homes built is 1810 square feet,

find the maximum and minimum sizes of the homes thecontractor should build. Assume that the standarddeviation is 92 square feet and the variable is normallydistributed.


20. New Home Prices If the average price of a new one-family home is $246,300 with a standard deviation of$15,000, find the minimum and maximum prices of thehouses that a contractor will build to satisfy the middle80% of the market. Assume that the variable is normallydistributed. $227,100 to $265,500


21. Cost of Personal Computers The average price of apersonal computer (PC) is $949. If the computer pricesare approximately normally distributed and s5 $100,what is the probability that a randomly selected PC costsmore than $1200? The least expensive 10% of personalcomputers cost less than what amount? 0.006; $821


22. Reading Improvement Program To help studentsimprove their reading, a school district decides toimplement a reading program. It is to be administered tothe bottom 5% of the students in the district, based onthe scores on a reading achievement exam. If theaverage score for the students in the district is 122.6,find the cutoff score that will make a student eligible forthe program. The standard deviation is 18. Assume thevariable is normally distributed. 92.99 or 93

23. Used Car Prices An automobile dealer finds that theaverage price of a previously owned vehicle is $8256.He decides to sell cars that will appeal to the middle60% of the market in terms of price. Find the maximumand minimum prices of the cars the dealer will sell. Thestandard deviation is $1150, and the variable is normallydistributed. The maximum price is $9222, and the minimumprice is $7290. (TI: $7288.14 minimum, $9223.86 maximum)

24. Ages of Amtrak Passenger Cars The average age ofAmtrak passenger train cars is 19.4 years. If thedistribution of ages is normal and 20% of the cars areolder than 22.8 years, find the standard deviation. 4.05


25. Lengths of Hospital Stays The average length ofa hospital stay for all diagnoses is 4.8 days. If weassume that the lengths of hospital stays are normallydistributed with a variance of 2.1, then 10% of hospitalstays are longer than how many days? Thirty percentof stays are less than how many days?

Source: www.cdc.gov 6.7; 4.05 (TI: for 10%, 6.657; for 30%, 4.040)

26. High School Competency Test A mandatorycompetency test for high school sophomores has anormal distribution with a mean of 400 and a standarddeviation of 100.

a. The top 3% of students receive $500. What is theminimum score you would need to receive thisaward? 588

6–28


6–29

b. The bottom 1.5% of students must go to summerschool. What is the minimum score you would needto stay out of this group? 183

27. Product Marketing An advertising company plans tomarket a product to low-income families. A study statesthat for a particular area, the average income per familyis $24,596 and the standard deviation is $6256. If thecompany plans to target the bottom 18% of the familiesbased on income, find the cutoff income. Assume thevariable is normally distributed. $18,840.48 (TI: $18,869.48)

28. Bottled Drinking Water Americans drank an averageof 23.2 gallons of bottled water per capita in 2008. If thestandard deviation is 2.7 gallons and the variable isnormally distributed, find the probability that a randomlyselected American drank more than 25 gallons of bottledwater. What is the probability that the selected persondrank between 22 and 30 gallons? 0.0968; 0.6641


29. Wristwatch Lifetimes The mean lifetime of awristwatch is 25 months, with a standard deviation of5 months. If the distribution is normal, for how manymonths should a guarantee be made if the manufacturerdoes not want to exchange more than 10% of the watches?Assume the variable is normally distributed. 18.6 months

30. Police Academy Acceptance Exams To qualify for apolice academy, applicants are given a test of physicalfitness. The scores are normally distributed with amean of 64 and a standard deviation of 9. If only thetop 20% of the applicants are selected, find the cutoffscore. 71.6 or 72

31. In the distributions shown, state the mean andstandard deviation for each. Hint: See Figures 6–5 and 6–6. Also the vertical lines are 1 standard deviationapart. a. m5 120,s5 20; b. m5 15,s5 2.5; c. m5 30,s5 5

15 17.512.5107.5 20 22.5

b.

120 1401008060 160 180

a.

32. SAT Scores Suppose that the mathematics SAT scoresfor high school seniors for a specific year have a meanof 456 and a standard deviation of 100 and areapproximately normally distributed. If a subgroup ofthese high school seniors, those who are in the NationalHonor Society, is selected, would you expect thedistribution of scores to have the same mean andstandard deviation? Explain your answer.

33. Given a data set, how could you decide if thedistribution of the data was approximately normal?There are several mathematics tests that can be used.

34. If a distribution of raw scores were plotted and then thescores were transformed to z scores, would the shape ofthe distribution change? Explain your answer.No. The shape of the distribution would be the same.

35. In a normal distribution, find s when m 5 105 and5.48% of the area lies to the right of 110. 3.125

36. In a normal distribution, find m when s is 6 and 3.75%of the area lies to the left of 85. 95.68

37. In a certain normal distribution, 1.25% of the area liesto the left of 42, and 1.25% of the area lies to the rightof 48. Find m and s. m 5 45, s 5 1.34

38. Exam Scores An instructor gives a 100-pointexamination in which the grades are normallydistributed. The mean is 60 and the standard deviationis 10. If there are 5% A’s and 5% F’s, 15% B’s and15% D’s, and 60% C’s, find the scores that divide thedistribution into those categories.

39. Drive-in Movies The data shown represent thenumber of outdoor drive-in movies in the United States

for a 14-year period. Check for normality. Not normal

2084 1497 1014 910 899 870 837 859848 826 815 750 637 737

Source: National Association of Theater Owners.

40. Cigarette Taxes The data shown represent thecigarette tax (in cents) for 50 selected states. Check

for normality. Not normal

200 160 156 200 30 300 224 346 170 55160 170 270 60 57 80 37 153 200 60100 178 302 84 251 125 44 435 79 16668 37 153 252 300 141 57 42 134 136

200 98 45 118 200 87 103 250 17 62

Source: http://www.tobaccofreekids.org

30 35252015 40 45

c.

41. Box Office Revenues The data shown representthe box office total revenue (in millions of dollars) for

a randomly selected sample of the top-grossing films in2009. Check for normality. Not normal

37 32 155 277146 80 66 11371 29 166 3628 72 32 3230 32 52 8437 402 42 109

Source: http://boxofficemojo.com

42. Number of Runs Made The data shownrepresent the number of runs made each year during

Bill Mazeroski’s career. Check for normality. Not normal

30 59 69 50 58 71 55 43 66 52 56 6236 13 29 17 3

Source: Greensburg Tribune Review.


6–30

Determining Normality

There are several ways in which statisticians test a data set for normality. Four are shown here.

Construct a Histogram

Inspect the histogram forshape.

1. Enter the data in the firstcolumn of a newworksheet. Name thecolumn Inventory.

2. Use Stat>Basic

Statistics>Graphical

Summary presented inSection 3–3 to createthe histogram. Is itsymmetric? Is there asingle peak?

Check for Outliers

Inspect the boxplot for outliers. There are no outliers in this graph. Furthermore, the box is inthe middle of the range, and the median is in the middle of the box. Most likely this is not askewed distribution either.

Calculate The Pearson Coefficient of Skewness

The measure of skewness in the graphical summary is not the same as the Pearson coefficient.Use the calculator and the formula.

3. Select Calc>Calculator, then type PC in the text box for Store result in:.

4. Enter the expression: 3*(MEAN(C1)2MEDI(C1))/(STDEV(C1)). Make sure you get allthe parentheses in the right place!

5. Click [OK]. The result, 0.148318, will be stored in the first row of C2 named PC. Since itis smaller than 11, the distribution is not skewed.

Construct a Normal Probability Plot

6. Select Graph>Probability Plot, then Single and click [OK].

7. Double-click C1 Inventory to select the data to be graphed.

8. Click [Distribution] and make sure that Normal is selected. Click [OK].

PC 53_X 2 median+

s

Technology Step by Step

MINITABStep by Step

Data

5 29 34 44 45

63 68 74 74 81

88 91 97 98 113

118 151 158


6–31

9. Click [Labels] and enter the title for thegraph: Quantile Plot for Inventory. Youmay also put Your Name in the subtitle.

10. Click [OK] twice. Inspect the graph to see ifthe graph of the points is linear.

These data are nearly normal.

What do you look for in the plot?

a) An “S curve” indicates a distribution thatis too thick in the tails, a uniformdistribution, for example.

b) Concave plots indicate a skeweddistribution.

c) If one end has a point that is extremelyhigh or low, there may be outliers.

This data set appears to be nearly normal byevery one of the four criteria!

TI-83 Plus orTI-84 PlusStep by Step

Normal Random Variables

To find the probability for a normal random variable:Press 2nd [DISTR], then 2 for normalcdf(The form is normalcdf(lower x value, upper x value, m, s)Use E99 for ` (infinity) and 2E99 for 2` (negative infinity). Press 2nd [EE] to get E.

Example: Find the probability that x is between 27 and 31 when m 5 28 and s 5 2 (Example 6–7a from the text).normalcdf(27,31,28,2)

To find the percentile for a normal random variable:Press 2nd [DISTR], then 3 for invNorm(The form is invNorm(area to the left of x value, m, s)

Example: Find the 90th percentile when m 5 200 and s 5 20 (Example 6–9 from text).invNorm(.9,200,20)

To construct a normal quantile plot:

1. Enter the data values into L1.

2. Press 2nd [STAT PLOT] to get the STAT PLOT menu.

3. Press 1 for Plot 1.

4. Turn on the plot by pressing ENTER while the cursor is flashing over ON.

5. Move the cursor to the normal quantile plot (6th graph).

6. Make sure L1 is entered for the Data List and X is highlighted for the Data Axis.

7. Press WINDOW for the Window menu. Adjust Xmin and Xmax according to the datavalues. Adjust Ymin and Ymax as well, Ymin 5 23 and Ymax 5 3 usually work fine.

8. Press GRAPH.

Using the data from the previous example gives

Since the points in the normal quantile plot lie close to a straight line, the distribution isapproximately normal.

Normal Quantile Plot

Excel can be used to construct a normal quantile plot in order to examine if a set of data isapproximately normally distributed.

1. Enter the data from the MINITAB example into column A of a new worksheet. The datashould be sorted in ascending order. If the data are not already sorted in ascending order,highlight the data to be sorted and select the Sort & Filter icon from the toolbar. Thenselect Sort Smallest to Largest.

2. After all the data are entered and sorted in column A, select cell B1. Type:=NORMSINV(1/(2*18)). Since the sample size is 18, each score represents , orapproximately 5.6%, of the sample. Each data value is assumed to subdivide the data intoequal intervals. Each data value corresponds to the midpoint of a particular subinterval.Thus, this procedure will standardize the data by assuming each data value represents themidpoint of a subinterval of width .

3. Repeat the procedure from step 2 for each data value in column A. However, for eachsubsequent value in column A, enter the next odd multiple of in the argument for theNORMSINV function. For example, in cell B2, type: =NORMSINV(3/(2*18)). In cellB3, type: =NORMSINV(5/(2*18)), and so on until all the data values have correspondingz scores.

4. Highlight the data from columns A and B, and select Insert, then Scatter chart. Select theScatter with only markers (the first Scatter chart).

5. To insert a title to the chart: Left-click on any region of the chart. Select Chart Tools andLayout from the toolbar. Then select Chart Title.

6. To insert a label for the variable on the horizontal axis: Left-click on any region of the chart.Select Chart Tools and Layout form the toolbar. Then select Axis Titles>Primary HorizontalAxis Title.

136

118

118


6–32

ExcelStep by Step

The points on the chart appear to lie close to a straight line. Thus, we deduce that the data areapproximately normally distributed.

Section 6–3 The Central Limit Theorem 331

6–33

6–3 The Central Limit Theorem

Objective

Use the central limittheorem to solveproblems involvingsample means forlarge samples.

6

Properties of the Distribution of Sample Means

1. The mean of the sample means will be the same as the population mean.

2. The standard deviation of the sample means will be smaller than the standard deviation ofthe population, and it will be equal to the population standard deviation divided by thesquare root of the sample size.

The following example illustrates these two properties. Suppose a professor gave an8-point quiz to a small class of four students. The results of the quiz were 2, 6, 4, and 8.For the sake of discussion, assume that the four students constitute the population. Themean of the population is

The standard deviation of the population is

The graph of the original distribution is shown in Figure 6–29. This is called a uniform

distribution.

s 5 Ï _2 2 5+2 1 _6 2 5+2 1 _4 2 5+2 1 _8 2 5+2

45 2.236

m 52 1 6 1 4 1 8

45 5

In addition to knowing how individual data values vary about the mean for a population,statisticians are interested in knowing how the means of samples of the same size takenfrom the same population vary about the population mean.

Distribution of Sample Means

Suppose a researcher selects a sample of 30 adult males and finds the mean of themeasure of the triglyceride levels for the sample subjects to be 187 milligrams/deciliter.Then suppose a second sample is selected, and the mean of that sample is found to be192 milligrams/deciliter. Continue the process for 100 samples. What happens then is thatthe mean becomes a random variable, and the sample means 187, 192, 184, . . . , 196 con-stitute a sampling distribution of sample means.

A sampling distribution of sample means is a distribution using the meanscomputed from all possible random samples of a specific size taken from a population.

If the samples are randomly selected with replacement, the sample means, for themost part, will be somewhat different from the population mean m. These differences arecaused by sampling error.

Sampling error is the difference between the sample measure and the correspondingpopulation measure due to the fact that the sample is not a perfect representation of thepopulation.

When all possible samples of a specific size are selected with replacement from apopulation, the distribution of the sample means for a variable has two important prop-erties, which are explained next.


6–34

Now, if all samples of size 2 are taken with replacement and the mean of each sam-ple is found, the distribution is as shown.

Sample Mean Sample Mean

2, 2 2 6, 2 42, 4 3 6, 4 52, 6 4 6, 6 62, 8 5 6, 8 74, 2 3 8, 2 54, 4 4 8, 4 64, 6 5 8, 6 74, 8 6 8, 8 8

A frequency distribution of sample means is as follows.

f

2 13 24 35 46 37 28 1

For the data from the example just discussed, Figure 6–30 shows the graph of thesample means. The histogram appears to be approximately normal.

The mean of the sample means, denoted by , is

mX_

52 1 3 1 . . . 1 8

165

80

165 5

mX

X

Figure 6–29

Distribution of

Quiz Scores

Freq

uenc

y

2

1

Score

4 6 8Historical Notes

Two mathematicians

who contributed to

the development

of the central limit

theorem were

Abraham DeMoivre

(1667–1754) and

Pierre Simon Laplace

(1749–1827).

DeMoivre was once

jailed for his religious

beliefs. After his

release, DeMoivre

made a living by

consulting on the

mathematics of

gambling and

insurance. He wrote

two books, Annuities

Upon Lives and The

Doctrine of Chance.

Laplace held a

government position

under Napoleon and

later under Louis XVIII.

He once computed

the probability of the

sun rising to be

18,226,214/

18,226,215.

Freq

uenc

y

2

5

4

3

2

1

Sample mean

3 4 5 6 7 8

Figure 6–30

Distribution of Sample

Means


6–35

which is the same as the population mean. Hence,

The standard deviation of sample means, denoted by , is

which is the same as the population standard deviation, divided by :

(Note: Rounding rules were not used here in order to show that the answers coincide.)In summary, if all possible samples of size n are taken with replacement from the

same population, the mean of the sample means, denoted by , equals the populationmean m; and the standard deviation of the sample means, denoted by , equals .The standard deviation of the sample means is called the standard error of the mean.

Hence,

A third property of the sampling distribution of sample means pertains to the shapeof the distribution and is explained by the central limit theorem.

sX_ 5

s

Ïn

s/ÏnsX_

mX_

sX_ 5

2.236

Ï25 1.581

Ï2

sX_ 5 Ï _2 2 5+2 1 _3 2 5+2 1 . . . 1 _8 2 5+2

165 1.581

sX_

mX_ 5 m

Unusual Stats

Each year a person

living in the United

States consumes on

average 1400 pounds

of food.

The Central Limit Theorem

As the sample size n increases without limit, the shape of the distribution of the sample meanstaken with replacement from a population with mean m and standard deviation s willapproach a normal distribution. As previously shown, this distribution will have a mean m anda standard deviation .s/Ïn

If the sample size is sufficiently large, the central limit theorem can be used toanswer questions about sample means in the same manner that a normal distribution canbe used to answer questions about individual values. The only difference is that a newformula must be used for the z values. It is

Notice that is the sample mean, and the denominator must be adjusted since meansare being used instead of individual data values. The denominator is the standard devia-tion of the sample means.

If a large number of samples of a given size are selected from a normally distributedpopulation, or if a large number of samples of a given size that is greater than or equal to30 are selected from a population that is not normally distributed, and the sample meansare computed, then the distribution of sample means will look like the one shown inFigure 6–31. Their percentages indicate the areas of the regions.

It’s important to remember two things when you use the central limit theorem:

1. When the original variable is normally distributed, the distribution of the samplemeans will be normally distributed, for any sample size n.

2. When the distribution of the original variable might not be normal, a sample size of30 or more is needed to use a normal distribution to approximate the distribution ofthe sample means. The larger the sample, the better the approximation will be.

X

z 5X 2 m

s/Ïn


6–36

Examples 6–13 through 6–15 show how the standard normal distribution can be usedto answer questions about sample means.

Figure 6–31

Distribution of Sample

Means for a Large

Number of Samples

13.59% 13.59%2.28%2.28%

34.13% 34.13%

m – 3 X– m – 2

X– mm – 1

X– m + 1

X– m + 2

X– m + 3

X–

Example 6–13 Hours That Children Watch Television

A. C. Neilsen reported that children between the ages of 2 and 5 watch an average of25 hours of television per week. Assume the variable is normally distributed and thestandard deviation is 3 hours. If 20 children between the ages of 2 and 5 are randomlyselected, find the probability that the mean of the number of hours they watch televisionwill be greater than 26.3 hours.


Solution

Since the variable is approximately normally distributed, the distribution of samplemeans will be approximately normal, with a mean of 25. The standard deviation of thesample means is

The distribution of the means is shown in Figure 6–32, with the appropriate areashaded.

sX_ 5

s

Ïn5

3

Ï205 0.671

Figure 6–32

Distribution of

the Means for

Example 6–13

25 26.3

The z value is

The area to the right of 1.94 is 1.000 2 0.9738 5 0.0262, or 2.62%.One can conclude that the probability of obtaining a sample mean larger than

26.3 hours is 2.62% [i.e., P( . 26.3) 5 2.62%].X

z 5X 2 m

s/Ïn5

26.3 2 25

3/Ï205

1.3

0.6715 1.94


6–37

The two z values are

To find the area between the two z values of 22.25 and 1.50, look up the correspondingarea in Table E and subtract one from the other. The area for z 5 22.25 is 0.0122,and the area for z 5 1.50 is 0.9332. Hence the area between the two values is0.9332 2 0.0122 5 0.9210, or 92.1%.

Hence, the probability of obtaining a sample mean between 90 and 100 months is92.1%; that is, P(90 , , 100) 5 92.1%.

Students sometimes have difficulty deciding whether to use

The formula

should be used to gain information about a sample mean, as shown in this section. Theformula

is used to gain information about an individual data value obtained from the population.Notice that the first formula contains , the symbol for the sample mean, while the sec-ond formula contains X, the symbol for an individual data value. Example 6–15 illus-trates the uses of the two formulas.

X]

z 5X 2 m

s

z 5X]

2 m

s/Ïn

z 5X]

2 m

s/Ïn or z 5

X 2 m

s

X]

z2 5100 2 96

16/Ï365 1.50

z1 590 2 96

16/Ï365 22.25

Example 6–14 The average age of a vehicle registered in the United States is 8 years, or 96 months.Assume the standard deviation is 16 months. If a random sample of 36 vehicles isselected, find the probability that the mean of their age is between 90 and 100 months.

Source: Harper’s Index.

Solution

Since the sample is 30 or larger, the normality assumption is not necessary. The desiredarea is shown in Figure 6–33.

9690 100

Figure 6–33

Area Under a

Normal Curve for

Example 6–14


6–38

The z value is

The area to the left of z 5 0.22 is 0.5871. Hence, the probability of selecting anindividual who consumes less than 224 pounds of meat per year is 0.5871, or58.71% [i.e., P(X , 224) 5 0.5871].

b. Since the question concerns the mean of a sample with a size of 40, the formulaz 5 ( 2 m)y( ) is used. The area is shown in Figure 6–35.s/ÏnX

]

z 5X 2 m

s5

224 2 218.4

255 0.22

Example 6–15 Meat Consumption

The average number of pounds of meat that a person consumes per year is 218.4 pounds.Assume that the standard deviation is 25 pounds and the distribution is approximatelynormal.


a. Find the probability that a person selected at random consumes less than224 pounds per year.

b. If a sample of 40 individuals is selected, find the probability that the mean ofthe sample will be less than 224 pounds per year.

Solution

a. Since the question asks about an individual person, the formula z 5 (X 2 m)ys isused. The distribution is shown in Figure 6–34.

Figure 6–34

Area Under a Normal

Curve for Part a of

Example 6–15

218.4Distribution of individual data values for the population

224

Figure 6–35

Area Under a Normal

Curve for Part b of

Example 6–15

218.4Distribution of means for all samples of size 40 taken from the population

224

The z value is

The area to the left of z 5 1.42 is 0.9222.

z 5X]

2 m

s/Ïn5

224 2 218.4

25/Ï405 1.42


6–39

Hence, the probability that the mean of a sample of 40 individuals is less than224 pounds per year is 0.9222, or 92.22%. That is, P( , 224) 5 0.9222.

Comparing the two probabilities, you can see that the probability of selectingan individual who consumes less than 224 pounds of meat per year is 58.71%,but the probability of selecting a sample of 40 people with a mean consumptionof meat that is less than 224 pounds per year is 92.22%. This rather largedifference is due to the fact that the distribution of sample means is much lessvariable than the distribution of individual data values. (Note: An individualperson is the equivalent of saying n 5 1.)

Finite Population Correction Factor (Optional)

The formula for the standard error of the mean is accurate when the samples aredrawn with replacement or are drawn without replacement from a very large or infinite pop-ulation. Since sampling with replacement is for the most part unrealistic, a correction factor

is necessary for computing the standard error of the mean for samples drawn withoutreplacement from a finite population. Compute the correction factor by using the expression

where N is the population size and n is the sample size.This correction factor is necessary if relatively large samples are taken from a small

population, because the sample mean will then more accurately estimate the populationmean and there will be less error in the estimation. Therefore, the standard error of themean must be multiplied by the correction factor to adjust for large samples taken froma small population. That is,

Finally, the formula for the z value becomes

When the population is large and the sample is small, the correction factor is gener-ally not used, since it will be very close to 1.00.

The formulas and their uses are summarized in Table 6–1.

z 5X]

2 m

s

Ïn? ÏN 2 n

N 2 1

sX_ 5

s

Ïn? ÏN 2 n

N 2 1

ÏN 2 n

N 2 1

s/Ïn

X]

Interesting Fact

The bubonic plague

killed more than

25 million people in

Europe between

1347 and 1351.

Table 6–1 Summary of Formulas and Their Uses

Formula Use

1.Used to gain information about an individual data value when the variableis normally distributed.

2.Used to gain information when applying the central limit theorem about asample mean when the variable is normally distributed or when thesample size is 30 or more.

z 5X]

2 m

s/Ïn

z 5X 2 m

s


Central Limit Theorem

Twenty students from a statistics class each collected a random sample of times on how long ittook students to get to class from their homes. All the sample sizes were 30. The resultingmeans are listed.

Student Mean Std. Dev. Student Mean Std. Dev.

1 22 3.7 11 27 1.42 31 4.6 12 24 2.23 18 2.4 13 14 3.14 27 1.9 14 29 2.45 20 3.0 15 37 2.86 17 2.8 16 23 2.77 26 1.9 17 26 1.88 34 4.2 18 21 2.09 23 2.6 19 30 2.2

10 29 2.1 20 29 2.8

1. The students noticed that everyone had different answers. If you randomly sample over andover from any population, with the same sample size, will the results ever be the same?

2. The students wondered whose results were right. How can they find out what thepopulation mean and standard deviation are?

3. Input the means into the computer and check to see if the distribution is normal.

4. Check the mean and standard deviation of the means. How do these values compare to thestudents’ individual scores?

5. Is the distribution of the means a sampling distribution?

6. Check the sampling error for students 3, 7, and 14.

7. Compare the standard deviation of the sample of the 20 means. Is that equal to the standarddeviation from student 3 divided by the square of the sample size? How about for student7, or 14?



6–40

1. If samples of a specific size are selected from apopulation and the means are computed, what is thisdistribution of means called? The distribution is called thesampling distribution of sample means.

2. Why do most of the sample means differ somewhatfrom the population mean? What is this differencecalled? The sample is not a perfect representation of thepopulation. The difference is due to what is called sampling error.

3. What is the mean of the sample means? The mean of thesample means is equal to the population mean.

4. What is the standard deviation of the sample meanscalled? What is the formula for this standard deviation?The standard error of the mean: s––

X 5 sy .

5. What does the central limit theorem say about the shapeof the distribution of sample means? The distribution willbe approximately normal when the sample size is large.

6. What formula is used to gain information about anindividual data value when the variable is normallydistributed? z 5

X 2 m

s

Ïn

7. What formula is used to gain information about asample mean when the variable is normally distributedor when the sample size is 30 or more?

For Exercises 8 through 25, assume that the sample is

taken from a large population and the correction factor

can be ignored.

8. Glass Garbage Generation A survey found that theAmerican family generates an average of 17.2 pounds ofglass garbage each year. Assume the standard deviation ofthe distribution is 2.5 pounds. Find the probability that themean of a sample of 55 families will be between 17 and18 pounds. 0.7135


9. College Costs The mean undergraduate cost fortuition, fees, room, and board for four-year institutionswas $26,489 for a recent academic year. Suppose

z 5 X 2 m

s/Ïn

Exercises 6–3


6–41

that s5 $3204 and that 36 four-year institutions arerandomly selected. Find the probability that the samplemean cost for these 36 schools is

a. Less than $25,000 0.0026 (TI: 0.0026)

b. Greater than $26,000 0.8212 (TI: 0.8201)

c. Between $24,000 and $26,000 0.1787 (TI: 0.1799)

Source: www.nces.ed.gov

10. Teachers’ Salaries in Connecticut The averageteacher’s salary in Connecticut (ranked first amongstates) is $57,337. Suppose that the distribution ofsalaries is normal with a standard deviation of $7500.

a. What is the probability that a randomly selectedteacher makes less than $52,000 per year? 0.2389

b. If we sample 100 teachers’ salaries, what is theprobability that the sample mean is less than$56,000? 0.0375


11. Serum Cholesterol Levels The mean serum cholesterollevel of a large population of overweight children is220 milligrams per deciliter (mg/dl), and the standarddeviation is 16.3 mg/dl. If a random sample of 35overweight children is selected, find the probability thatthe mean will be between 220 and 222 mg/dl. Assume theserum cholesterol level variable is normally distributed.0.2673

12. Teachers’ Salaries in North Dakota The averageteacher’s salary in North Dakota is $37,764. Assume anormal distribution with s 5 $5100.

a. What is the probability that a randomly selectedteacher’s salary is greater than $45,000? 0.0778

b. For a sample of 75 teachers, what is the probabilitythat the sample mean is greater than $38,000? 0.3466


13. Fuel Efficiency for U.S. Light Vehicles The averagefuel efficiency of U.S. light vehicles (cars, SUVs,minivans, vans, and light trucks) for 2005 was 21 mpg.If the standard deviation of the population was 2.9 andthe gas ratings were normally distributed, what is theprobability that the mean mpg for a random sample of25 light vehicles is under 20? Between 20 and 25?

Source: World Almanac. 0.0427; 0.9572 (TI: 0.0423; 0.9577)

14. SAT Scores The national average SAT score (forVerbal and Math) is 1028. Suppose that nothing isknown about the shape of the distribution and that thestandard deviation is 100. If a random sample of 200scores were selected and the sample mean werecalculated to be 1050, would you be surprised? Explain.Yes—the probability of such is less than 0.0001.


15. Sodium in Frozen Food The average number ofmilligrams (mg) of sodium in a certain brand of low-saltmicrowave frozen dinners is 660 mg, and the standarddeviation is 35 mg. Assume the variable is normallydistributed. a. 0.3859 (TI: 0.3875)

a. If a single dinner is selected, find the probability that thesodium content will be more than 670 mg.

b. If a sample of 10 dinners is selected, find theprobability that the mean of the sample will belarger than 670 mg. 0.1841 (TI: 0.1831)

c. Why is the probability for part a greater than thatfor part b? Individual values are more variable than means.

16. Cell Phone Lifetimes A recent study of the lifetimesof cell phones found the average is 24.3 months.The standard deviation is 2.6 months. If a companyprovides its 33 employees with a cell phone, find theprobability that the mean lifetime of these phones willbe less than 23.8 months. Assume cell phone life is anormally distributed variable. 0.1357

17. Water Use The Old Farmer’s Almanac reports that theaverage person uses 123 gallons of water daily. If thestandard deviation is 21 gallons, find the probability thatthe mean of a randomly selected sample of 15 peoplewill be between 120 and 126 gallons. Assume thevariable is normally distributed. 0.4176 (TI: 0.4199)

18. Medicare Hospital Insurance The average yearlyMedicare Hospital Insurance benefit per person was$4064 in a recent year. If the benefits are normallydistributed with a standard deviation of $460, find theprobability that the mean benefit for a random sampleof 20 patients is

a. Less than $3800 0.0051

b. More than $4100 0.3632


19. Amount of Laundry Washed Each Year Procter &Gamble reported that an American family of fourwashes an average of 1 ton (2000 pounds) of clotheseach year. If the standard deviation of the distribution is187.5 pounds, find the probability that the mean of arandomly selected sample of 50 families of four will bebetween 1980 and 1990 pounds. 0.1254 (TI: 0.12769)

Source: The Harper’s Index Book.

20. Per Capita Income of Delaware Residents In arecent year, Delaware had the highest per capitaannual income with $51,803. If s 5 $4850, what isthe probability that a random sample of 34 stateresidents had a mean income greater than $50,000?Less than $48,000?

Source: New York Times Almanac. 0.9850; Less than 0.0001

21. Annual Precipitation The average annual precipitationfor a large Midwest city is 30.85 inches with a standarddeviation of 3.6 inches. Assume the variable is normallydistributed.

a. Find the probability that a randomly selected monthwill have less than 30 inches. 0.4052 or 40.52%

b. Find the probability that the mean of a randomselection of 32 months will have a mean less than30 inches. 0.0901 or 9.01%

c. Does it seem reasonable that one month could havea rainfall amount less than 30 inches? Yes, theprobability is slightly more than 40%.

d. Does it seem reasonable that the mean of a sampleof 32 months could be less than 30 inches? It’spossible since the probability is about 9%.

22. Systolic Blood Pressure Assume that the mean systolicblood pressure of normal adults is 120 millimeters ofmercury (mm Hg) and the standard deviation is 5.6.Assume the variable is normally distributed.

a. If an individual is selected, find the probability thatthe individual’s pressure will be between 120 and121.8 mm Hg. 0.1255

b. If a sample of 30 adults is randomly selected, findthe probability that the sample mean will bebetween 120 and 121.8 mm Hg. 0.4608

c. Why is the answer to part a so much smaller thanthe answer to part b? Means are less variable thanindividual data.

23. Cholesterol Content The average cholesterol contentof a certain brand of eggs is 215 milligrams, and thestandard deviation is 15 milligrams. Assume thevariable is normally distributed.

a. If a single egg is selected, find the probabilitythat the cholesterol content will be greater than220 milligrams. 0.3707 (TI: 0.3694)

b. If a sample of 25 eggs is selected, find theprobability that the mean of the sample will belarger than 220 milligrams. 0.0475 (TI: 0.04779)

Source: Living Fit.

24. Ages of Proofreaders At a large publishing company,the mean age of proofreaders is 36.2 years, and thestandard deviation is 3.7 years. Assume the variable isnormally distributed.

a. If a proofreader from the company is randomlyselected, find the probability that his or her age willbe between 36 and 37.5 years. 0.1567

b. If a random sample of 15 proofreaders is selected,find the probability that the mean age of theproofreaders in the sample will be between 36 and37.5 years. 0.4963

25. Weekly Income of Private Industry Information

Workers The average weekly income of informationworkers in private industry is $777. If the standarddeviation is $77, what is the probability that a randomsample of 50 information workers will earn, on average,more than $800 per week? Do we need to assume anormal distribution? Explain.

Source: World Almanac. 0.0174 No—the central limit theorem applies.


6–42

For Exercises 26 and 27, check to see whether the

correction factor should be used. If so, be sure to include

it in the calculations.

26. Life Expectancies In a study of the life expectancy of500 people in a certain geographic region, the mean ageat death was 72.0 years, and the standard deviation was5.3 years. If a sample of 50 people from this region isselected, find the probability that the mean lifeexpectancy will be less than 70 years. 0.0025

27. Home Values A study of 800 homeowners in a certainarea showed that the average value of the homes was$82,000, and the standard deviation was $5000. If 50homes are for sale, find the probability that the mean ofthe values of these homes is greater than $83,500. 0.0143


28. Breaking Strength of Steel Cable The averagebreaking strength of a certain brand of steel cable is2000 pounds, with a standard deviation of 100 pounds.A sample of 20 cables is selected and tested. Find thesample mean that will cut off the upper 95% of allsamples of size 20 taken from the population. Assumethe variable is normally distributed. 1963.10 pounds

29. The standard deviation of a variable is 15. If a sample of100 individuals is selected, compute the standard errorof the mean. What size sample is necessary to doublethe standard error of the mean? 5 1.5, n 5 25

30. In Exercise 29, what size sample is needed to cut thestandard error of the mean in half? 400

sX

6–4 The Normal Approximation to the BinomialDistribution

A normal distribution is often used to solve problems that involve the binomial distribu-tion since when n is large (say, 100), the calculations are too difficult to do by hand usingthe binomial distribution. Recall from Chapter 5 that a binomial distribution has the fol-lowing characteristics:

1. There must be a fixed number of trials.

2. The outcome of each trial must be independent.

Section 6–4 The Normal Approximation to the Binomial Distribution 341

6–43

3. Each experiment can have only two outcomes or outcomes that can be reduced totwo outcomes.

4. The probability of a success must remain the same for each trial.

Also, recall that a binomial distribution is determined by n (the number of trials) andp (the probability of a success). When p is approximately 0.5, and as n increases, theshape of the binomial distribution becomes similar to that of a normal distribution. Thelarger n is and the closer p is to 0.5, the more similar the shape of the binomial distribu-tion is to that of a normal distribution.

But when p is close to 0 or 1 and n is relatively small, a normal approximation isinaccurate. As a rule of thumb, statisticians generally agree that a normal approxima-tion should be used only when n ? p and n ? q are both greater than or equal to 5. (Note:

q 5 1 2 p.) For example, if p is 0.3 and n is 10, then np 5 (10)(0.3) 5 3, and a normaldistribution should not be used as an approximation. On the other hand, if p 5 0.5 andn 5 10, then np 5 (10)(0.5) 5 5 and nq 5 (10)(0.5) 5 5, and a normal distribution canbe used as an approximation. See Figure 6–36.

Objective

Use the normalapproximation tocompute probabilitiesfor a binomial variable.

7

0123456789

10

Binomial probabilities for n = 10, p = 0.3[n ? p = 10(0.3) = 3; n ? q = 10(0.7) = 7]

0.0280.1210.2330.2670.2000.1030.0370.0090.0010.0000.000

0.3

P(X )

P(X )X

X

0.2

0.1

0 1 2 3 4 5 6 7 8 9 10

0123456789

10

Binomial probabilities for n = 10, p = 0.5[n ? p = 10(0.5) = 5; n ? q = 10(0.5) = 5]

0.0010.0100.0440.1170.2050.2460.2050.1170.0440.0100.001

0.3

P(X )

P(X )X

X

0.2

0.1

0 1 2 3 4 5 6 7 8 9 10

Figure 6–36

Comparison of the

Binomial Distribution

and a Normal

Distribution


6–44

In addition to the previous condition of np $ 5 and nq $ 5, a correction for conti-nuity may be used in the normal approximation.

A correction for continuity is a correction employed when a continuous distribution isused to approximate a discrete distribution.

The continuity correction means that for any specific value of X, say 8, the bound-aries of X in the binomial distribution (in this case, 7.5 to 8.5) must be used. (See Sec-tion 1–2.) Hence, when you employ a normal distribution to approximate the binomial,you must use the boundaries of any specific value X as they are shown in the binomialdistribution. For example, for P(X 5 8), the correction is P(7.5 , X , 8.5). For P(X # 7),the correction is P(X , 7.5). For P(X $ 3), the correction is P(X . 2.5).

Students sometimes have difficulty deciding whether to add 0.5 or subtract 0.5 fromthe data value for the correction factor. Table 6–2 summarizes the different situations.

Table 6–2 Summary of the Normal Approximation to the Binomial Distribution

Binomial Normal

When finding: Use:

1. P(X 5 a) P(a 2 0.5 , X , a 1 0.5)

2. P(X $ a) P(X . a 2 0.5)

3. P(X . a) P(X . a 1 0.5)

4. P(X # a) P(X , a 1 0.5)

5. P(X , a) P(X , a 2 0.5)

For all cases, m5 n ? p, s 5 , n ? p $ 5, and n ? q $ 5.Ïn ? p ? q

Procedure Table

Procedure for the Normal Approximation to the Binomial Distribution

Step 1 Check to see whether the normal approximation can be used.

Step 2 Find the mean m and the standard deviation s.

Step 3 Write the problem in probability notation, using X.

Step 4 Rewrite the problem by using the continuity correction factor, and show thecorresponding area under the normal distribution.

Step 5 Find the corresponding z values.

Step 6 Find the solution.

Interesting Fact

Of the 12 months,

August ranks first in

the number of births

for Americans.

The formulas for the mean and standard deviation for the binomial distribution arenecessary for calculations. They are

m 5 n ? p and s 5

The steps for using the normal distribution to approximate the binomial distributionare shown in this Procedure Table.

Ïn ? p ? q


6–45

Example 6–16 Reading While Driving

A magazine reported that 6% of American drivers read the newspaper while driving. If300 drivers are selected at random, find the probability that exactly 25 say they read thenewspaper while driving.

Source: USA Snapshot, USA TODAY.

Solution

Here, p 5 0.06, q 5 0.94, and n 5 300.

Step 1 Check to see whether a normal approximation can be used.

np 5 (300)(0.06) 5 18 nq 5 (300)(0.94) 5 282

Since np $ 5 and nq $ 5, the normal distribution can be used.

Step 2 Find the mean and standard deviation.

m 5 np 5 (300)(0.06) 5 18

s 5 5 5 5 4.11

Step 3 Write the problem in probability notation: P(X 5 25).

Step 4 Rewrite the problem by using the continuity correction factor. Seeapproximation number 1 in Table 6–2: P(25 2 0.5 , X , 25 1 0.5) 5P(24.5 , X , 25.5). Show the corresponding area under the normaldistribution curve. See Figure 6–37.

Ï16.92Ï_300+_0.06+_0.94+Ïnpq

25

1824.5 25.5

Figure 6–37

Area Under a Normal

Curve and X Values for

Example 6–16

Step 5 Find the corresponding z values. Since 25 represents any value between 24.5and 25.5, find both z values.

Step 6 The area to the left of z 5 1.82 is 0.9656, and the area to the left of z 5 1.58 is0.9429. The area between the two z values is 0.9656 2 0.9429 5 0.0227, or2.27%. Hence, the probability that exactly 25 people read the newspaperwhile driving is 2.27%.

z1 525.5 2 18

4.115 1.82 z2 5

24.5 2 18

4.115 1.58

Example 6–17 Widowed Bowlers

Of the members of a bowling league, 10% are widowed. If 200 bowling leaguemembers are selected at random, find the probability that 10 or more will be widowed.

Solution

Here, p 5 0.10, q 5 0.90, and n 5 200.

Step 1 Since np 5 (200)(0.10) 5 20 and nq 5 (200)(0.90) 5 180, the normalapproximation can be used.

Step 2 m 5 np 5 (200)(0.10) 5 20

s 5 5 5 5 4.24

Step 3 P(X $ 10)

Step 4 See approximation number 2 in Table 6–2: P(X . 10 2 0.5) 5 P(X . 9.5).

The desired area is shown in Figure 6–38.

Ï18Ï_200+_0.10+_0.90+Ïnpq


6–46

20109.5

Figure 6–38

Area Under a Normal

Curve and X Value for

Example 6–17

Step 5 Since the problem is to find the probability of 10 or more positive responses,

a normal distribution graph is as shown in Figure 6–38.

The z value is

Step 6 The area to the left of z 5 22.48 is 0.0066. Hence the area to the right of

z 5 22.48 is 1.0000 2 0.0066 5 0.9934, or 99.34%.

It can be concluded, then, that the probability of 10 or more widowed people in a

random sample of 200 bowling league members is 99.34%.

z 59.5 2 20

4.245 22.48

Example 6–18 Batting Averages

If a baseball player’s batting average is 0.320 (32%), find the probability that the player

will get at most 26 hits in 100 times at bat.

Solution

Here, p 5 0.32, q 5 0.68, and n 5 100.

Step 1 Since np 5 (100)(0.320) 5 32 and nq 5 (100)(0.680) 5 68, the normal

distribution can be used to approximate the binomial distribution.

Step 2 m 5 np 5 (100)(0.320) 5 32

s 5 5 5 5 4.66

Step 3 P(X # 26)

Step 4 See approximation number 4 in Table 6–2: P(X , 26 1 0.5) 5 P(X , 26.5).

The desired area is shown in Figure 6–39.

Step 5 The z value is

z 526.5 2 32

4.665 21.18

Ï21.76Ï_100+_0.32+_0.68+Ïnpq


6–47

Step 6 The area to the left of z 5 21.18 is 0.1190. Hence the probability is 0.1190,or 11.9%.

The closeness of the normal approximation is shown in Example 6–19.

32.026 26.5

Figure 6–39

Area Under a

Normal Curve for

Example 6–18

Example 6–19 When n 5 10 and p 5 0.5, use the binomial distribution table (Table B in Appendix C)to find the probability that X 5 6. Then use the normal approximation to find theprobability that X 5 6.

Solution

From Table B, for n 5 10, p 5 0.5, and X 5 6, the probability is 0.205.For a normal approximation,

m 5 np 5 (10)(0.5) 5 5

s 5 5 5 1.58

Now, X 5 6 is represented by the boundaries 5.5 and 6.5. So the z values are

The corresponding area for 0.95 is 0.8289, and the corresponding area for 0.32 is0.6255. The area between the two z values of 0.95 and 0.32 is 0.8289 2 0.6255 50.2034, which is very close to the binomial table value of 0.205. See Figure 6–40.

z1 56.5 2 5

1.585 0.95 z2 5

5.5 2 5

1.585 0.32

Ï_10+_0.5+_0.5+Ïnpq

55.5

6

6.5

Figure 6–40

Area Under a

Normal Curve for

Example 6–19

The normal approximation also can be used to approximate other distributions, suchas the Poisson distribution (see Table C in Appendix C).


How Safe Are You?

Assume one of your favorite activities is mountain climbing. When you go mountain climbing,

you have several safety devices to keep you from falling. You notice that attached to one of

your safety hooks is a reliability rating of 97%. You estimate that throughout the next year you

will be using this device about 100 times. Answer the following questions.

1. Does a reliability rating of 97% mean that there is a 97% chance that the device will not

fail any of the 100 times?

2. What is the probability of at least one failure?

3. What is the complement of this event?

4. Can this be considered a binomial experiment?

5. Can you use the binomial probability formula? Why or why not?

6. Find the probability of at least two failures.

7. Can you use a normal distribution to accurately approximate the binomial distribution?

Explain why or why not.

8. Is correction for continuity needed?

9. How much safer would it be to use a second safety hook independently of the first?



6–48

1. Explain why a normal distribution can be used as an

approximation to a binomial distribution. What

conditions must be met to use the normal distribution

to approximate the binomial distribution? Why is a

correction for continuity necessary?

2. (ans) Use the normal approximation to the binomial to

find the probabilities for the specific value(s) of X.

a. n 5 30, p 5 0.5, X 5 18 0.0811

b. n 5 50, p 5 0.8, X 5 44 0.0516

c. n 5 100, p 5 0.1, X 5 12 0.1052

d. n 5 10, p 5 0.5, X $ 7 0.1711

e. n 5 20, p 5 0.7, X # 12 0.2327

f. n 5 50, p 5 0.6, X # 40 0.9988

3. Check each binomial distribution to see whether it can

be approximated by a normal distribution (i.e., are

np $ 5 and nq $ 5?).

a. n 5 20, p 5 0.5 Yes d. n 5 50, p 5 0.2 Yes

b. n 5 10, p 5 0.6 No e. n 5 30, p 5 0.8 Yes

c. n 5 40, p 5 0.9 No f. n 5 20, p 5 0.85 No

4. School Enrollment Of all 3- to 5-year-old children,

56% are enrolled in school. If a sample of 500 such

children is randomly selected, find the probability that

at least 250 will be enrolled in school. 0.9970

Source: Statistical Abstract of the United States.

5. Youth Smoking Two out of five adult smokers

acquired the habit by age 14. If 400 smokers are

randomly selected, find the probability that 170 or

fewer acquired the habit by age 14. 0.8577

Source: Harper’s Index.

6. Mail Order A mail order company has an 8% success

rate. If it mails advertisements to 600 people, find the

probability of getting less than 40 sales. 0.1003

7. Voter Preference A political candidate estimates that

30% of the voters in her party favor her proposed tax

reform bill. If there are 400 people at a rally, find the

probability that at least 100 voters will favor her tax bill.

Based on your answer, is it likely that 100 or more

people will favor the bill? 0.9875

8. Household Computers According to recent surveys,

60% of households have personal computers. If a

random sample of 180 households is selected, what is

the probability that more than 60 but fewer than 100

have a personal computer?

Source: New York Times Almanac. 0.0984

9. Female Americans Who Have Completed 4 Years of

College The percentage of female Americans 25 years

old and older who have completed 4 years of college

or more is 26.1. In a random sample of 200 American

women who are at least 25, what is the probability that at

most 50 have completed 4 years of college or more?

Source: New York Times Almanac. 0.3936

Exercises 6–4


6–49

10. Population of College Cities College studentsoften make up a substantial portion of the population ofcollege cities and towns. State College, Pennsylvania,ranks first with 71.1% of its population made up ofcollege students. What is the probability that in arandom sample of 150 people from State College,more than 50 are not college students? 0.0985

Source: www.infoplease.com

11. Elementary School Teachers Women comprise 80.3%of all elementary school teachers. In a random sample of300 elementary teachers, what is the probability that lessthan three-fourths are women? 0.0087


12. Telephone Answering Devices Seventy-eight percentof U.S. homes have a telephone answering device. In arandom sample of 250 homes, what is the probability

that fewer than 50 do not have a telephone answeringdevice? 0.2005


13. Parking Lot Construction The mayor of a small townestimates that 35% of the residents in the town favorthe construction of a municipal parking lot. If there are350 people at a town meeting, find the probability thatat least 100 favor construction of the parking lot. Basedon your answer, is it likely that 100 or more peoplewould favor the parking lot? 0.9951; yes (TI: 0.9950)

14. Residences of U.S. Citizens According to the U.S.Census, 67.5% of the U.S. population were born intheir state of residence. In a random sample of 200Americans, what is the probability that fewer than 125were born in their state of residence? 0.0559


15. Recall that for use of a normal distribution as anapproximation to the binomial distribution, theconditions np $ 5 and nq $ 5 must be met. For eachgiven probability, compute the minimum sample sizeneeded for use of the normal approximation.


a. p 5 0.1 n $ 50 d. p 5 0.8 n $ 25

b. p 5 0.3 n $ 17 e. p 5 0.9 n $ 50

c. p 5 0.5 n $ 10

Summary

• A normal distribution can be used to describe a variety of variables, such asheights, weights, and temperatures. A normal distribution is bell-shaped, unimodal,symmetric, and continuous; its mean, median, and mode are equal. Since eachnormally distributed variable has its own distribution with mean m and standarddeviation s, mathematicians use the standard normal distribution, which has a meanof 0 and a standard deviation of 1. Other approximately normally distributedvariables can be transformed to the standard normal distribution with the formulaz 5 (X 2 m)ys. (6–1)

• A normal distribution can be used to solve a variety of problems in which thevariables are approximately normally distributed. (6–2)

• A sampling distribution of sample means is a distribution using the meanscomputed from all possible random samples of a specific size taken from apopulation. The difference between a sample measure and the correspondingpopulation measure is due to what is called sampling error. The mean of the samplemeans will be the same as the population mean. The standard deviation of thesample mean will be equal to the population standard deviation divided by thesquare root of the sample size. The central limit theorem states that as the samplesize increases without limit, the shape of the distribution of the sample means takenwith replacement from a population will approach a normal distribution. (6–3)

• A normal distribution can be used to approximate other distributions, such as abinomial distribution. For a normal distribution to be used as an approximation, theconditions np $ 5 and nq $ 5 must be met. Also, a correction for continuity may beused for more accurate results. (6–4)


6–50

Important Terms

central limit theorem 333

correction for continuity 342

negatively or left-skeweddistribution 301

normal distribution 303

positively or right-skeweddistribution 301

sampling distribution ofsample means 331

sampling error 331

standard error of the mean 333

standard normal distribution 304

symmetric distribution 301

z value (score) 304

Formula for the z score (or standard score):

Formula for finding a specific data value:

X 5 z ? S 1 M

Formula for the mean of the sample means:

5 MMX_

z 5X 2 M

S

Formula for the standard error of the mean:

Formula for the z value for the central limit theorem:

Formulas for the mean and standard deviation for thebinomial distribution:

M 5 n ? p S 5 Ïn ? p ? q

z 5X]

2 M

S/Ïn

SX_ 5

S

Ïn

Important Formulas

Review Exercises

1. Find the area under the standard normal distributioncurve for each. (6–1)

a. Between z 5 0 and z 5 1.95 0.4744

b. Between z 5 0 and z 5 0.37 0.1443

c. Between z 5 1.32 and z 5 1.82 0.0590

d. Between z 5 21.05 and z 5 2.05 0.8329 (TI: 0.8330)

e. Between z 5 20.03 and z 5 0.53 0.2139

f. Between z 5 11.10 and z 5 21.80 0.8284

g. To the right of z 5 1.99 0.0233

h. To the right of z 5 21.36 0.9131

i. To the left of z 5 22.09 0.0183

j. To the left of z 5 1.68 0.9535

2. Using the standard normal distribution, find eachprobability. (6–1)

a. P(0 , z , 2.07) 0.4808

b. P(21.83 , z , 0) 0.0336

c. P(21.59 , z , 12.01) 0.9219

d. P(1.33 , z , 1.88) 0.0617

e. P(22.56 , z , 0.37) 0.6391

f. P(z . 1.66) 0.0485

g. P(z , 22.03) 0.0212

h. P(z . 21.19) 0.8830

i. P(z , 1.93) 0.9732

j. P(z . 21.77) 0.9616

3. Per Capita Spending on Health Care The averageper capita spending on health care in the United Statesis $5274. If the standard deviation is $600 and thedistribution of health care spending is approximatelynormal, what is the probability that a randomlyselected person spends more than $6000? Find thelimits of the middle 50% of individual health careexpenditures. (6–2)

Source: World Almanac. 0.1131; $4872 and $5676 (TI: $4869.31minimum, $5678.69 maximum)

4. Salaries for Actuaries The average salary forgraduates entering the actuarial field is $40,000. If thesalaries are normally distributed with a standarddeviation of $5000, find the probability that

a. An individual graduate will have a salary over$45,000. 0.1587

b. A group of nine graduates will have a group averageover $45,000. (6–2) 0.0013

Source: www.BeAnActuary.org

5. Commuter Train Passengers On a certain run of acommuter train, the average number of passengers is 476and the standard deviation is 22. Assume the variable isnormally distributed. If the train makes the run, find theprobability that the number of passengers will be

Review Exercises 349

6–51

a. Between 476 and 500 passengers 0.3621 or 36.21%

b. Less than 450 passengers 0.1190 or 11.9%

c. More than 510 passengers (6–2) 0.0606 or 6.06%

6. Monthly Spending for Paging and Messaging

Services The average individual monthly spending inthe United States for paging and messaging servicesis $10.15. If the standard deviation is $2.45 and theamounts are normally distributed, what is theprobability that a randomly selected user of theseservices pays more than $15.00 per month? Between$12.00 and $14.00 per month? (6–2) 0.0239; 0.1654


7. Cost of iPod Repair The average cost of repairing aniPod is $120 with a standard deviation of $10.50. Thecosts are normally distributed. If 15% of the costs areconsidered excessive, find the cost in dollars that wouldbe considered excessive. (6–2) $130.92

8. Heights of Active Volcanoes The heights (in feetabove sea level) of a random sample of the world’s

active volcanoes are shown here. Check for normality. (6–2) Not normal

13,435 5,135 11,339 12,224 7,470

9,482 12,381 7,674 5,223 5,631

3,566 7,113 5,850 5,679 15,584

5,587 8,077 9,550 8,064 2,686

5,250 6,351 4,594 2,621 9,348

6,013 2,398 5,658 2,145 3,038


9. Private Four-Year College Enrollment Arandom sample of enrollments in Pennsylvania’s

private four-year colleges is listed here. Check fornormality. (6–2) Not normal

1350 1886 1743 1290 1767

2067 1118 3980 1773 4605

1445 3883 1486 980 1217

3587


10. Average Precipitation For the first 7 months of theyear, the average precipitation in Toledo, Ohio, is19.32 inches. If the average precipitation is normallydistributed with a standard deviation of 2.44 inches,find these probabilities.

a. A randomly selected year will have precipitationgreater than 18 inches for the first 7 months.

b. Five randomly selected years will have an averageprecipitation greater than 18 inches for the first7 months. (6–3)

Source: Toledo Blade. a. 0.7054 (TI: 0.7057) b. 0.8869 (TI: 0.8868)

11. Confectionary Products Americans ate an average of25.7 pounds of confectionary products each last yearand spent an average of $61.50 per person doing so. Ifthe standard deviation for consumption is 3.75 poundsand the standard deviation for the amount spent is$5.89, find the following:

a. The probability that the sample mean confectionaryconsumption for a random sample of 40 Americanconsumers was greater than 27 pounds

b. The probability that for a random sample of 50, thesample mean for confectionary spending exceeded$60.00 (6–3)

Source: www.census.gov a. 0.0143 (TI: 0.0142) b. 0.9641

12. Portable CD Player Lifetimes A recent study of thelife span of portable compact disc players found theaverage to be 3.7 years with a standard deviation of0.6 year. If a random sample of 32 people who own CDplayers is selected, find the probability that the meanlifetime of the sample will be less than 3.4 years. If themean is less than 3.4 years, would you consider that3.7 years might be incorrect? (6–3) 0.0023; yes, since the

probability is less than 1%.

13. Retirement Income Of the total population ofAmerican households, including older Americans andperhaps some not so old, 17.3% receive retirementincome. In a random sample of 120 households, whatis the probability that more than 20 households but lessthan 35 households receive a retirement income? (6–4)

Source: www.bls.gov 0.5234

14. Slot Machines The probability of winning on a slotmachine is 5%. If a person plays the machine 500 times,find the probability of winning 30 times. Use the normalapproximation to the binomial distribution. (6–4) 0.0496

15. Multiple-Job Holders According to the government5.3% of those employed are multiple-job holders. In arandom sample of 150 people who are employed, whatis the probability that fewer than 10 hold multiple jobs?What is the probability that more than 50 are notmultiple-job holders? (6–4)

Source: www.bls.gov 0.7123; 0.9999 (TI: 0.7139; 0.9999)

16. Enrollment in Personal Finance Course In a largeuniversity, 30% of the incoming first-year students electto enroll in a personal finance course offered by theuniversity. Find the probability that of 800 randomlyselected incoming first-year students, at least 260 haveelected to enroll in the course. (6–4) 0.0668

17. U.S. Population Of the total population of the UnitedStates, 20% live in the northeast. If 200 residents of theUnited States are selected at random, find the probabilitythat at least 50 live in the northeast. (6–4) 0.0465

Source: Statistical Abstract of the United States.


6–52

Determine whether each statement is true or false. If the

statement is false, explain why.

1. The total area under a normal distribution is infinite. False

2. The standard normal distribution is a continuousdistribution. True

3. All variables that are approximately normally distributedcan be transformed to standard normal variables. True

4. The z value corresponding to a number below the meanis always negative. True

5. The area under the standard normal distribution to theleft of z 5 0 is negative. False

6. The central limit theorem applies to means of samplesselected from different populations. False

Select the best answer.

7. The mean of the standard normal distribution is

a. 0 c. 100b. 1 d. Variable

8. Approximately what percentage of normally distributeddata values will fall within 1 standard deviation aboveor below the mean?

a. 68% c. 99.7%b. 95% d. Variable

9. Which is not a property of the standard normaldistribution?

a. It’s symmetric about the mean.b. It’s uniform.c. It’s bell-shaped.d. It’s unimodal.

10. When a distribution is positively skewed, therelationship of the mean, median, and mode from left toright will be

a. Mean, median, mode c. Median, mode, meanb. Mode, median, mean d. Mean, mode, median

11. The standard deviation of all possible sample meansequals

a. The population standard deviationb. The population standard deviation divided by the

population mean

c. The population standard deviation divided by thesquare root of the sample size

d. The square root of the population standard deviation

Complete the following statements with the best answer.

12. When one is using the standard normal distribution, P(z , 0) 5 . 0.5

13. The difference between a sample mean and a populationmean is due to . Sampling error

14. The mean of the sample means equals .

15. The standard deviation of all possible sample means iscalled the . standard error of the mean

16. The normal distribution can be used to approximate thebinomial distribution when n ? p and n ? q are bothgreater than or equal to . 5

17. The correction factor for the central limit theoremshould be used when the sample size is greater than

of the size of the population. 5%

18. Find the area under the standard normal distributionfor each.

a. Between 0 and 1.50 0.4332

b. Between 0 and 21.25 0.3944

c. Between 1.56 and 1.96 0.0344

d. Between 21.20 and 22.25 0.1029

e. Between 20.06 and 0.73 0.2912

f. Between 1.10 and 21.80 0.8284

g. To the right of z 5 1.75 0.0401

h. To the right of z 5 21.28 0.8997

i. To the left of z 5 22.12 0.017

j. To the left of z 5 1.36 0.9131

19. Using the standard normal distribution, find eachprobability.

a. P(0 , z , 2.16) 0.4846

b. P(21.87 , z , 0) 0.4693

c. P(21.63 , z , 2.17) 0.9334

d. P(1.72 , z , 1.98) 0.0188

e. P(22.17 , z , 0.71) 0.7461

f. P(z . 1.77) 0.0384

g. P(z , 22.37) 0.0089

h. P(z . 21.73) 0.9582

the population mean

Chapter Quiz

StatisticsToday

What Is Normal?—RevisitedMany of the variables measured in medical tests—blood pressure, triglyceride level, etc.—are

approximately normally distributed for the majority of the population in the United States. Thus,

researchers can find the mean and standard deviation of these variables. Then, using these two

measures along with the z values, they can find normal intervals for healthy individuals. For

example, 95% of the systolic blood pressures of healthy individuals fall within 2 standard

deviations of the mean. If an individual’s pressure is outside the determined normal range (either

above or below), the physician will look for a possible cause and prescribe treatment if necessary.

Chapter Quiz 351

6–53

i. P(z , 2.03) 0.9788

j. P(z . 21.02) 0.8461

20. Amount of Rain in a City The average amount ofrain per year in Greenville is 49 inches. The standarddeviation is 8 inches. Find the probability that next yearGreenville will receive the following amount of rainfall.Assume the variable is normally distributed.

a. At most 55 inches of rain 0.7734

b. At least 62 inches of rain 0.0516

c. Between 46 and 54 inches of rain 0.3837

d. How many inches of rain would you consider to bean extremely wet year?

21. Heights of People The average height of a certain agegroup of people is 53 inches. The standard deviation is4 inches. If the variable is normally distributed, find theprobability that a selected individual’s height will be

a. Greater than 59 inches 0.0668

b. Less than 45 inches 0.0228

c. Between 50 and 55 inches 0.4649

d. Between 58 and 62 inches 0.0934

22. Lemonade Consumption The average number ofgallons of lemonade consumed by the football teamduring a game is 20, with a standard deviation of3 gallons. Assume the variable is normally distributed.When a game is played, find the probability of using

a. Between 20 and 25 gallons 0.4525

b. Less than 19 gallons 0.3707

c. More than 21 gallons 0.3707

d. Between 26 and 28 gallons 0.019

23. Years to Complete a Graduate Program The averagenumber of years a person takes to complete a graduatedegree program is 3. The standard deviation is4 months. Assume the variable is normally distributed.If an individual enrolls in the program, find theprobability that it will take

a. More than 4 years to complete the program 0.0013

b. Less than 3 years to complete the program 0.5

c. Between 3.8 and 4.5 years to complete the program 0.0081

d. Between 2.5 and 3.1 years to complete the program 0.5511

24. Passengers on a Bus On the daily run of an expressbus, the average number of passengers is 48. Thestandard deviation is 3. Assume the variable is normallydistributed. Find the probability that the bus will have

a. Between 36 and 40 passengers 0.0037

b. Fewer than 42 passengers 0.0228

c. More than 48 passengers 0.5

d. Between 43 and 47 passengers 0.3232

25. Thickness of Library Books The average thickness ofbooks on a library shelf is 8.3 centimeters. The standarddeviation is 0.6 centimeter. If 20% of the books areoversized, find the minimum thickness of the oversizedbooks on the library shelf. Assume the variable isnormally distributed. 8.804 centimeters

26. Membership in an Organization Membership in anelite organization requires a test score in the upper 30%range. If m 5 115 and s 5 12, find the lowestacceptable score that would enable a candidate to applyfor membership. Assume the variable is normallydistributed. 121.24 is the lowest acceptable score.

27. Repair Cost for Microwave Ovens The average repaircost of a microwave oven is $55, with a standarddeviation of $8. The costs are normally distributed. If12 ovens are repaired, find the probability that the meanof the repair bills will be greater than $60. 0.015

28. Electric Bills The average electric bill in a residentialarea is $72 for the month of April. The standarddeviation is $6. If the amounts of the electric bills arenormally distributed, find the probability that the meanof the bill for 15 residents will be less than $75. 0.9738

29. Sleep Survey According to a recent survey, 38% ofAmericans get 6 hours or less of sleep each night. If 25people are selected, find the probability that 14 or morepeople will get 6 hours or less of sleep each night. Doesthis number seem likely? 0.0495; no

Source: Amazing Almanac.

30. Unemployment If 8% of all people in a certaingeographic region are unemployed, find the probabilitythat in a sample of 200 people, there are fewer than 10people who are unemployed. 0.0455 or 4.55%

31. Household Online Connection The percentage ofU.S. households that have online connections is44.9%. In a random sample of 420 households, whatis the probability that fewer than 200 have onlineconnections? 0.8577


32. Computer Ownership Fifty-three percent of U.S.households have a personal computer. In a randomsample of 250 households, what is the probability thatfewer than 120 have a PC? 0.0495


33. Calories in Fast-Food Sandwiches The number ofcalories contained in a selection of fast-food sandwiches

is shown here. Check for normality. Not normal

390 405 580 300 320540 225 720 470 560535 660 530 290 440390 675 530 1010 450320 460 290 340 610430 530

Source: The Doctor’s Pocket Calorie, Fat, and Carbohydrate Counter.

34. GMAT Scores The average GMAT scores for thetop-30 ranked graduate schools of business are listed

here. Check for normality. Approximately normal

718 703 703 703 700 690 695 705 690 688676 681 689 686 691 669 674 652 680 670651 651 637 662 641 645 645 642 660 636

Source: U.S. News & World Report Best Graduate Schools.


6–54

Table 6–3 Normal Probability Paper

89.5 104.5 119.5 134.5 149.5 164.5 179.5

12

510

2030

4050

6070

8090

9598

99

3. Find the cumulative percents for each class by dividingeach cumulative frequency by 200 (the total frequencies)and multiplying by 100%. (For the first class, it would be24y200 3 100% 5 12%.) Place these values in the lastcolumn.

4. Using the normal probability paper shown in Table 6–3,label the x axis with the class boundaries as shown andplot the percents.

5. If the points fall approximately in a straight line, it canbe concluded that the distribution is normal. Do you feelthat this distribution is approximately normal? Explainyour answer.

6. To find an approximation of the mean or median, draw ahorizontal line from the 50% point on the y axis over tothe curve and then a vertical line down to the x axis.Compare this approximation of the mean with thecomputed mean.

Critical Thinking Challenges

Sometimes a researcher must decide whether a variable isnormally distributed. There are several ways to do this. Onesimple but very subjective method uses special graph paper,which is called normal probability paper. For the distributionof systolic blood pressure readings given in Chapter 3 of thetextbook, the following method can be used:

1. Make a table, as shown.Cumulative

Cumulative percent

Boundaries Frequency frequency frequency

89.5–104.5 24104.5–119.5 62119.5–134.5 72134.5–149.5 26149.5–164.5 12164.5–179.5 4

200

2. Find the cumulative frequencies for each class, andplace the results in the third column.

7. To find an approximation of the standard deviation,locate the values on the x axis that correspond to the16 and 84% values on the y axis. Subtract these twovalues and divide the result by 2. Compare this

approximate standard deviation to the computedstandard deviation.

8. Explain why the method used in step 7 works.

Answers to Applying the Concepts 353

6–55

1. Business and Finance Use the data collected in dataproject 1 of Chapter 2 regarding earnings per share tocomplete this problem. Use the mean and standarddeviation computed in data project 1 of Chapter 3 asestimates for the population parameters. What valueseparates the top 5% of stocks from the others?

2. Sports and Leisure Find the mean and standarddeviation for the batting average for a player in themost recently completed MBL season. What battingaverage would separate the top 5% of all hittersfrom the rest? What is the probability that a randomlyselected player bats over 0.300? What is theprobability that a team of 25 players has a mean thatis above 0.275?

3. Technology Use the data collected in data project 3 ofChapter 2 regarding song lengths. If the sampleestimates for mean and standard deviation are used asreplacements for the population parameters for this dataset, what song length separates the bottom 5% and top5% from the other values?

4. Health and Wellness Use the data regarding heartrates collected in data project 4 of Chapter 2 for thisproblem. Use the sample mean and standard deviationas estimates of the population parameters. For thebefore-exercise data, what heart rate separates the top

10% from the other values? For the after-exercise data,what heart rate separates the bottom 10% from the othervalues? If a student was selected at random, what is theprobability that her or his mean heart rate beforeexercise was less than 72? If 25 students were selectedat random, what is the probability that their mean heartrate before exercise was less than 72?

5. Politics and Economics Use the data collected in dataproject 6 of Chapter 2 regarding Math SAT scores tocomplete this problem. What are the mean and standarddeviation for statewide Math SAT scores? What SATscore separates the bottom 10% of states from theothers? What is the probability that a randomly selectedstate has a statewide SAT score above 500?

6. Your Class Confirm the two formulas hold true for thecentral limit theorem for the population containing theelements {1, 5, 10}. First, compute the population meanand standard deviation for the data set. Next, create alist of all 9 of the possible two-element samples thatcan be created with replacement: {1, 1}, {1, 5}, etc.For each of the 9 compute the sample mean. Nowfind the mean of the sample means. Does it equal thepopulation mean? Compute the standard deviationof the sample means. Does it equal the populationstandard deviation, divided by the square root of n?

Data Projects

Section 6–1 Assessing Normality

1. Answers will vary. One possible frequency distributionis the following:

Branches Frequency

0–9 110–19 1420–29 1730–39 740–49 350–59 260–69 270–79 180–89 290–99 1

2. Answers will vary according to the frequencydistribution in question 1. This histogram matchesthe frequency distribution in question 1.

3. The histogram is unimodal and skewed to the right(positively skewed).

4. The distribution does not appear to be normal.

5

18

16

14

12

10

8

6

4

2

025

Libraries

Freq

uenc

y

Histogram of Libraries

45 65 85

Answers to Applying the Concepts


6–56

5. The mean number of branches is and thestandard deviation is

6. Of the data values, 80% fall within 1 standard deviationof the mean (between 10.8 and 52).

7. Of the data values, 92% fall within 2 standarddeviations of the mean (between 0 and 72.6).

8. Of the data values, 98% fall within 3 standarddeviations of the mean (between 0 and 93.2).

9. My values in questions 6–8 differ from the 68, 95, and100% that we would see in a normal distribution.

10. These values support the conclusion that the distributionof the variable is not normal.

Section 6–2 Smart People

1. The area to the right of 2 in thestandard normal table is about 0.0228, so I wouldexpect about 10,000(0.0228) 5 228 people in Visialato qualify for Mensa.

2. It does seem reasonable to continue my quest to start aMensa chapter in Visiala.

3. Answers will vary. One possible answer would be torandomly call telephone numbers (both home and cellphones) in Visiala, ask to speak to an adult, and askwhether the person would be interested in joining Mensa.

4. To have an Ultra-Mensa club, I would need to find thepeople in Visiala who have IQs that are at least 2.326standard deviations above average. This means that Iwould need to recruit those with IQs that are at least 135:

Section 6–3 Central Limit Theorem

1. It is very unlikely that we would ever get the sameresults for any of our random samples. While it is aremote possibility, it is highly unlikely.

2. A good estimate for the population mean would be tofind the average of the students’ sample means.Similarly, a good estimate for the population standarddeviation would be to find the average of the students’sample standard deviations.

3. The distribution appears to be somewhat left-skewed(negatively skewed).

15

5

4

3

2

1

020

Central Limit Theorem Means

Freq

uenc

y

Histogram of Central Limit Theorem Means

25 30 35

2.326 5x 2 100

151 x 5 100 1 2.326_15+ 5 134.89

z 5130 – 100

15 5 2.

s 5 20.6.x 5 31.4, 4. The mean of the students’ means is 25.4, and the

standard deviation is 5.8.

5. The distribution of the means is not a samplingdistribution, since it represents just 20 of all possiblesamples of size 30 from the population.

6. The sampling error for student 3 is 18 2 25.4 5 27.4;the sampling error for student 7 is 26 2 25.4 5 10.6;the sampling error for student 14 is 29 2 25.4 5 13.6.

7. The standard deviation for the sample of the 20 meansis greater than the standard deviations for each ofthe individual students. So it is not equal to thestandard deviation divided by the square root of thesample size.

Section 6–4 How Safe Are You?

1. A reliability rating of 97% means that, on average, thedevice will not fail 97% of the time. We do not knowhow many times it will fail for any particular set of100 climbs.

2. The probability of at least 1 failure in 100 climbs is 1 2 (0.97)100 5 1 2 0.0476 5 0.9524 (about 95%).

3. The complement of the event in question 2 is the eventof “no failures in 100 climbs.”

4. This can be considered a binomial experiment. We havetwo outcomes: success and failure. The probability ofthe equipment working (success) remains constant at97%. We have 100 independent climbs. And we arecounting the number of times the equipment works inthese 100 climbs.

5. We could use the binomial probability formula, but itwould be very messy computationally.

6. The probability of at least two failures cannot beestimated with the normal distribution (see below). Sothe probability is 1 2 [(0.97)100 1 100(0.97)99(0.03)] 51 2 0.1946 5 0.8054 (about 80.5%).

7. We should not use the normal approximation to thebinomial since nq , 10.

8. If we had used the normal approximation, we wouldhave needed a correction for continuity, since we wouldhave been approximating a discrete distribution with acontinuous distribution.

9. Since a second safety hook will be successful or failindependently of the first safety hook, the probabilityof failure drops from 3% to (0.03)(0.03) 5 0.0009,or 0.09%.

Documents

298 Chapter 5 Discrete Probability Distributions · 298 Chapter 5 Discrete Probability Distributions 5–48 Section 5–2 Expected Value 1. The expected value is the mean in a discrete