Normal Probability Distributions -...

C H A P T E R 5Normal Probability

Distributions

The North Carolina Zoo is the largest walk-through natural-habitat zoo in the United States. It is one of only two state zoos in the United States, with the other located in Minnesota. ©1992 Susan Middleton & David Liittschwager

5.1 Introduction toNormal Distributionsand the StandardNormal Distribution

5.2 Normal Distributions:Finding Probabilities

5.3 Normal Distributions:Finding Values� CASE STUDY

5.4 SamplingDistributions and theCentral Limit Theorem� ACTIVITY

5.5 NormalApproximations toBinomial Distributions� USES AND ABUSES

� REAL STATISTICS–REAL DECISIONS

� TECHNOLOGY

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W H E R E Y O U ’ R E G O I N G

W H E R E Y O U ’ V E B E E N

In Chapters 1 through 4, you learned how to

collect and describe data, find the probability

of an event, and analyze discrete probability

distributions. You also learned that if a sample

is used to make inferences about a population,

then it is critical that the sample not be biased.

Suppose, for instance, that you wanted to

determine the rate of clinical mastitis

(infections caused by bacteria that can alter

milk production) in dairy herds. How would

you organize the study? When the Animal

Health Service performed this study, it used

random sampling and then classified the results

according to breed, housing, hygiene, health,

milking management, and milking machine.

One conclusion from the study was that herds

with Red and White cows as the predominant

breed had a higher rate of clinical mastitis than

herds with Holstein-Friesian cows as the

main breed.

In Chapter 5, you will learn how to recognize

normal (bell-shaped) distributions and how to

use their properties in real-life applications.

Suppose that you worked for the North Carolina

Zoo and were collecting data about various

physical traits of Eastern Box Turtles at the zoo.

Which of the following would you expect to have

bell-shaped, symmetric distributions: carapace

(top shell) length, plastral (bottom shell) length,

carapace width, plastral width, weight, total

length? For instance, the four graphs below show

the carapace length and plastral length of male

and female Eastern Box Turtles in the North

Carolina Zoo. Notice that the male Eastern Box

Turtle carapace length distribution is bell

shaped, but the other three distributions are

skewed left.

239

Carapace length(in millimeters)

Perc

ent

Female Eastern Box TurtleCarapace Length

70 90 110 130 150

3

6

9

12

15

18

Carapace length(in millimeters)

Perc

ent

Male Eastern Box TurtleCarapace Length

10080 120 140 160

5

10

15

20

25

Plastral length(in millimeters)

Perc

ent

Female Eastern Box TurtlePlastral Length

70 90 110 130 150

4

8

12

16

20

Plastral length(in millimeters)

Perc

ent

Male Eastern Box TurtlePlastral Length

70 90 110 130

3

6

9

12

15

18

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240 C H A P T E R 5 NORMAL PROBABILITY DISTRIBUTIONS

� Properties of a Normal DistributionIn Section 4.1, you distinguished between discrete and continuous randomvariables, and learned that a continuous random variable has an infinite numberof possible values that can be represented by an interval on the number line. Itsprobability distribution is called a continuous probability distribution. In thischapter, you will study the most important continuous probability distribution instatistics—the normal distribution. Normal distributions can be used to modelmany sets of measurements in nature, industry, and business. For instance, thesystolic blood pressure of humans, the lifetime of television sets, and evenhousing costs are all normally distributed random variables.

You have learned that a discrete probability distribution can be graphed witha histogram. For a continuous probability distribution, you can use a probabilitydensity function (pdf). A normal curve with mean and standard deviation can be graphed using the normal probability density function.

A normal curve depends completely on thetwo parameters and because and are constants.p L 3.14

e L 2.718smy =

1

s22p e-1x -m22>2s2

.

sm

Properties of a Normal Distribution � The Standard NormalDistribution

5.1 Introduction to Normal Distributions and the StandardNormal Distribution

What You SHOULD LEARN

� How to interpret graphs of normal probability distributions

� How to find areas under the standard normal curve

G U I D E L I N E SProperties of a Normal DistributionA normal distribution is a continuous probability distribution for arandom variable The graph of a normal distribution is called thenormal curve. A normal distribution has the following properties.

1. The mean, median, and mode are equal.

2. The normal curve is bell-shaped and is symmetric about the mean.

3. The total area under the normal curve is equal to one.

4. The normal curve approaches, but never touches, the -axis as itextends farther and farther away from the mean.

5. Between and (in the center of the curve), the graphcurves downward. The graph curves upward to the left of andto the right of The points at which the curve changes fromcurving upward to curving downward are called inflection points.

− μ σ +μ σμ

Inflection points

Total area = 1

− μ σ2− μ σ3 +μ σ2 +μ σ3x

m + s .m - s

m + sm - s

x

x.

G U I D E L I N E S

Insight

A probability density function has two requirements.

1. The total areaunder the curveis equal to one.

2. The function cannever be negative.

Insight

To learn how todetermine if a random sample is taken from a normal distribution, see Appendix C.

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A normal distribution can have any mean and any positive standarddeviation. These two parameters, and completely determine the shape ofthe normal curve. The mean gives the location of the line of symmetry, and thestandard deviation describes how much the data are spread out.

Notice that curve and curve above have the same mean, and curve and curve have the same standard deviation. The total area under each curveis 1.

CBBA

s ,m

S E C T I O N 5 . 1 INTRODUCTION TO NORMAL DISTRIBUTIONS AND THE STANDARD NORMAL DISTRIBUTIONS 241

0 1 2 3 4 5 6 7x

Inflectionpoints A

Mean:Standard deviation:

s = 1.5

m = 3.5 Mean:Standard deviation:

s = 0.7

m = 3.5 Mean:Standard deviation:

s = 0.7

m = 1.5

0 1 2 3 4 5 6 7x

Inflectionpoints

B

0 1 2 3 4 5 6 7x

Inflectionpoints

C

Perc

ent

5 C

A

B10

15

40 50 6030 70x

20

E X A M P L E 1

Understanding Mean and Standard Deviation

1. Which normal curve has a greater mean?

2. Which normal curve has a greater standard deviation?

Solution

1. The line of symmetry of curve occurs at The line of symmetry ofcurve occurs at So, curve has a greater mean.

2. Curve is more spread out than curve so, curve has a greater standarddeviation.

� Try It Yourself 1Consider the normal curves shown at the left. Which normal curve has thegreatest mean? Which normal curve has the greatest standard deviation?Justify your answers.

a. Find the location of the line of symmetry of each curve. Make a conclusionabout which mean is greatest.

b. Determine which normal curve is more spread out. Make a conclusion aboutwhich standard deviation is greatest.

Answer: Page A40

BA;B

Ax = 12.Bx = 15.A

6 9 12 15 18 21x

40

30 A

B20

10

Perc

ent

Study TipHere are instructions for graphing a normal distribution on a TI-83/84.

DISTR

1: normalpdf(

Enter and the values of and separated by commas.

GRAPH

sm

x

2ndY=

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80 100

0.2

0

Once you determine the meanand standard deviation, you canuse a TI-83/84 to graph thenormal curve in Example 2.

E X A M P L E 2

Interpreting Graphs of Normal DistributionsThe heights (in feet) of fully grown white oak trees are normally distributed.The normal curve shown below represents this distribution. What is the meanheight of a fully grown white oak tree? Estimate the standard deviation of thisnormal distribution.

Solution

Interpretation The heights of the oak trees are normally distributed with amean of about 90 feet and a standard deviation of about 3.5 feet.

� Try It Yourself 2The diameters (in feet) of fully grown white oak trees are normally distributed.The normal curve shown below represents this distribution. What is the meandiameter of a fully grown white oak tree? Estimate the standard deviation ofthis normal distribution.

a. Find the line of symmetry and identify the mean.b. Estimate the inflection points and identify the standard deviation.

Answer: Page A40

2.72.5 3.7 3.9 4.1 4.3 4.53.52.9 3.1 3.3

Diameter (in feet)

x

80 85 90 95 100x

Height (in feet)

Because a normal curve issymmetric about the mean, youcan estimate that ≈ 90 feet.μ

Because the inflection pointsare one standard deviation fromthe mean, you can estimate that ≈ 3.5 feet.σ

80 85 90 95 100x

Height (in feet)

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� The Standard Normal DistributionThere are infinitely many normal distributions, each with its own mean andstandard deviation. The normal distribution with a mean of 0 and a standarddeviation of 1 is called the standard normal distribution. The horizontal scale ofthe graph of the standard normal distribution corresponds to scores. In Section2.5, you learned that a score is a measure of position that indicates the numberof standard deviations a value lies from the mean. Recall that you can transforman value to a score using the formula

Round to the nearest hundredth.

If each data value of a normally distributed random variable is transformedinto a score, the result will be the standard normal distribution. When thistransformation takes place, the area that falls in the interval under thenonstandard normal curve is the same as that under the standard normal curvewithin the corresponding boundaries.

In Section 2.4, you learned to use the Empirical Rule to approximate areasunder a normal curve when the values of the random variable corresponded to

0, 1, 2, or 3 standard deviations from the mean. Now, you will learnto calculate areas corresponding to other values. After you use the formulagiven above to transform an value to a score, you can use the StandardNormal Table in Appendix B. The table lists the cumulative area under thestandard normal curve to the left of for scores from to 3.49. As youexamine the table, notice the following.

-3.49z-z

z-x-x-

-1,-2,-3,x

z-

z-x

=

x - m

s.

z =

Value - MeanStandard deviation

z-x-

z-z-


Study TipIt is important that you know the difference between and The random variable is sometimes called a raw score and represents values in a nonstandard normal distribution, whereas

represents values in the standard normal distribution.

z

xz.x

Because every normal distributioncan be transformed to the standard normal distribution, you can use -scores and the standard normal curve to find areas (and therefore probability) under any normal curve.

z

Insight

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

STANDARD NORMAL DISTRIBUTION

0 1 2 3z

−3 −2 −1

Area = 1

D E F I N I T I O N

P R O P E R T I E S O F T H E S TA N D A R D N O R M A L D I S T R I B U T I O N

1. The cumulative area is close to 0 for scores close to

2. The cumulative area increases as the scores increase.

3. The cumulative area for is 0.5000.

4. The cumulative area is close to 1 for scores close to z = 3.49.z-

z = 0

z-

z = -3.49.z-

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When the -score is not in the table, use the entry closest to it. If the given -score is exactly midway between two -scores, then use the area midway

between the corresponding areas.zz

z


1.15z

Area =0.8749

0

−0.24 0z

Area =0.4052

Study TipHere are instructions for finding the area that corresponds to

on a TI-83/84.

To specify the lower bound in thiscase, use

DISTR

2: normalcdf(

ENTER

)- .24-10000,

2nd

-10,000.

z = -0.24

E X A M P L E 3

Using the Standard Normal Table

1. Find the cumulative area that corresponds to a -score of 1.15.

2. Find the cumulative area that corresponds to a -score of 0.24.

Solution

1. Find the area that corresponds to by finding 1.1 in the left columnand then moving across the row to the column under 0.05. The number inthat row and column is 0.8749. So, the area to the left of is 0.8749.

2. Find the area that corresponds to by finding in the leftcolumn and then moving across the row to the column under 0.04. Thenumber in that row and column is 0.4052. So, the area to the left of

is 0.4052.

You can also use a computer or calculator to find the cumulative area thatcorresponds to a -score, as shown in the margin.

� Try It Yourself 3

1. Find the area under the curve to the left of a -score of

2. Find the area under the curve to the left of a -score of 2.17.

Locate the given -score and find the area that corresponds to it in theStandard Normal Table. Answer: Page A40

z

z

-2.19.z

z

�0.5�0.4�0.3�0.2�0.1�0.0

z .09 .08 .07 .06 .05 .03�3.4�3.3�3.2

.0002 .0003 .0003 .0003 .0003 .0003

.0003 .0004 .0004 .0004 .0004 .0004

.0005 .0005 .0005 .0006 .0006 .0006

.0003

.0006

.2776 .2810 .2843 .2877 .2912 .2946 .2981

.3121 .3156 .3192 .3228 .3264 .3300 .3336

.3483 .3520 .3557 .3594 .3632 .3669 .3707

.3859 .3897 .3936 .3974 .4013 .4052 .4090

.4247 .4286 .4325 .4364 .4404 .4443 .4483

.4641 .4681 .4721 .4761 .4801 .4840 .4880

.0004

.04

z = -0.24

-0.2z = -0.24

0.9 .8159 .8186 .8212 .8238 .8264 .8289 .83151.0 .8413 .8438 .8461 .8485 .8508 .8531 .85541.1 .8643 .8665 .8686 .8708 .8729 .8749 .87701.2 .8849 .8869 .8888 .8907 .8925 .8944 .89621.3 .9032 .9049 .9066 .9082 .9099 .9115 .91311.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279

z .00 .01 .02 .03 .04 .05 .060.0 .5000 .5040 .5080 .5120 .5160 .5199 .52390.1 .5398 .5438 .5478 .5517 .5557 .5596 .56360.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026

z = 1.15

z = 1.15

-z

z

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You can use the following guidelines to find various types of areas under thestandard normal curve.


G U I D E L I N E SFinding Areas Under the Standard Normal Curve1. Sketch the standard normal curve and shade the appropriate area

under the curve.

2. Find the area by following the directions for each case shown.

a. To find the area to the left of find the area that corresponds to in the Standard Normal Table.

b. To find the area to the right of use the Standard Normal Table to find the area that corresponds to Then subtract the area from 1.

c. To find the area between two scores, find the area correspondingto each score in the Standard Normal Table. Then subtract thesmaller area from the larger area.

z1.23−0.75

Subtract to find the areaof the region between thetwo z-scores: 0.8907 − 0.2266 = 0.6641.

The area to the leftof z = −0.75 is 0.2266.

Use the table to findthe area for the z-scores.

0

1.

3.

4.The area to the leftof z = 1.23 is 0.8907.

2.

z-z-

z1.230Use the table to

find the area for the z-score.1.

Subtract to find the areato the right of z = 1.23: 1 − 0.8907 = 0.1093.

3.The area to the leftof z = 1.23 is 0.8907.

2.

z.z,

z1.230Use the table to

find the area for the z-score.1.

The area to theleft of z = 1.23 is 0.8907.

2.

zz,

G U I D E L I N E S

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Because the normal distribution is a continuous probability distribution, the area under the standard normal curve to the left of a -score gives theprobability that is less than that -score. For instance,in Example 4, the area to the left of

is0.1611. So,

which is read as “theprobability that is lessthan is 0.1611.”-0.99

z

P1z 6 -0.992 = 0.1611,

z = -0.99

zz

z

Insight

Using a TI-83/84, you can findthe area automatically

Use 10,000 for the upper bound.

E X A M P L E 4

Finding Area Under the Standard Normal CurveFind the area under the standard normal curve to the left of

SolutionThe area under the standard normal curve to the left of is shown.

From the Standard Normal Table, this area is equal to 0.1611.

� Try It Yourself 4Find the area under the standard normal curve to the left of

a. Draw the standard normal curve and shade the area under the curve and tothe left of

b. Use the Standard Normal Table to find the area that corresponds toAnswer: Page A40z = 2.13.

z = 2.13.

z = 2.13.

−0.99z

0

z = -0.99

z = -0.99.

E X A M P L E 5

Finding Area Under the Standard Normal CurveFind the area under the standard normal curve to the right of

SolutionThe area under the standard normal curve to the right of is shown.

From the Standard Normal Table, the area to the left of is 0.8554. Because the total area under the curve is 1, the area to the right of is

= 0.1446.

Area = 1 - 0.8554

z = 1.06

z = 1.06

1.06z

0

Area = 0.8554 Area = 1 − 0.8554

z = 1.06

z = 1.06.

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Recall in Section 2.4 you learned, using the Empirical Rule, that values lyingmore than two standard deviations from the mean are considered unusual.Values lying more than three standard deviations from the mean are consideredvery unusual. So if a score is greater than 2 or less than it is unusual. If a

score is greater than 3 or less than it is very unusual.-3,z--2,z-


When using technology, youranswers may differ slightly fromthose found using the StandardNormal Table.

According to onepublication, thenumber of birthsin a recent year

was 4,112,052. The weights of the newborns can be approximated by a normal distribution, as shown by the following graph. (Source: NationalCenter for Health Statistics)

Find the z-scores that correspond to weights of 2000,3000, and 4000 grams. Are any of these unusually heavy or light?

Weight (in grams)

3282

2666

2050

1434

3898

4514

5130

Weights of Newborns

� Try It Yourself 5Find the area under the standard normal curve to the right of

a. Draw the standard normal curve and shade the area below the curve and tothe right of

b. Use the Standard Normal Table to find the area to the left of c. Subtract the area from 1. Answer: Page A40

z = -2.16.z = -2.16.

z = -2.16.

E X A M P L E 6

Finding Area Under the Standard Normal CurveFind the area under the standard normal curve between and

SolutionThe area under the standard normal curve between and isshown.

From the Standard Normal Table, the area to the left of is 0.8944 andthe area to the left of is 0.0668. So, the area between and

is

Interpretation So, 82.76% of the area under the curve falls betweenand

� Try It Yourself 6Find the area under the standard normal curve between and

a. Use the Standard Normal Table to find the area to the left of b. Use the Standard Normal Table to find the area to the left of c. Subtract the smaller area from the larger area. Answer: Page A40

z = -2.16.z = -1.35.

z = -1.35.z = -2.16

z = 1.25.z = -1.5

= 0.8276.

Area = 0.8944 - 0.0668

z = 1.25z = -1.5z = -1.5

z = 1.25

1.25z

−1.5 0

z = 1.25z = -1.5

z = 1.25.z = -1.5

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� Building Basic Skills and Vocabulary1. Find three real-life examples of a continuous variable. Which do you think

may be normally distributed? Why?

2. What is the total area under the normal curve?

3. Draw two normal curves that have the same mean but different standarddeviations. Describe the similarities and differences.

4. Draw two normal curves that have different means but the same standarddeviations. Describe the similarities and differences.

5. What is the mean of the standard normal distribution? What is the standarddeviation of the standard normal distribution?

6. Describe how you can transform a nonstandard normal distribution to astandard normal distribution.

7. Getting at the Concept Why is it correct to say “a” normal distribution and“the” standard normal distribution?

8. Getting at the Concept If a -score is zero, which of the following must betrue? Explain your reasoning.

(a) The mean is zero.

(b) The corresponding -value is zero.

(c) The corresponding -value is equal to the mean.

Graphical Analysis In Exercises 9–14, determine whether the graph couldrepresent a variable with a normal distribution. Explain your reasoning.

9. 10.

11. 12.

13. 14.x

x

xx

x

x

x

x

z

For Extra Help

5.1 EXERCISES

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Graphical Analysis In Exercises 15 and 16, determine whether the histogramrepresents data with a normal distribution. Explain your reasoning.

15. 16.

Graphical Analysis In Exercises 17–20, find the area of the indicated regionunder the standard normal curve. If convenient, use technology to find the area.

17. 18.

19. 20.

Finding Area In Exercises 21–40, find the indicated area under the standardnormal curve. If convenient, use technology to find the area.

21. To the left of 22. To the left of


25. To the right of 26. To the right of




33. Between and 34. Between and



39. To the left of or to 40. To the left of or tothe right of the right of z = 1.96z = 1.28

z = -1.96z = -1.28

z = 2.33z = 1.96z = -2.33z = -1.96

z = 0z = -0.51z = 0z = -1.53

z = 2.86z = 0z = 1.54z = 0

z = 2.51z = 1.615

z = -3.16z = -2.575

z = 3.25z = 1.28

z = -1.95z = -0.65

z = 1.28z = 1.96

z = 0.08z = 1.36

2z

0−0.5 1.5z

0

−2.25 0z

1.2z

0

10

Pounds lost

Rel

ativ

e fr

eque

ncy

Weight Loss

0.05

0.10

0.15

0.20

20 30 40 50 60 70 80

4 12 20 28 36

Time (in minutes)

Rel

ativ

e fr

eque

ncy

Waiting Time in aDentist’s Office

0.1

0.2

0.3

0.4

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� Using and Interpreting Concepts41. Manufacturer Claims You work for a consumer watchdog publication

and are testing the advertising claims of a light bulb manufacturer. Themanufacturer claims that the life span of the bulb is normallydistributed, with a mean of 2000 hours and a standard deviation of 250hours. You test 20 light bulbs and get the following life spans.

2210, 2406, 2267, 1930, 2005, 2502, 1106, 2140, 1949, 1921,2217, 2121, 2004, 1397, 1659, 1577, 2840, 1728, 1209, 1639

(a) Draw a frequency histogram to display these data. Use five classes.Is it reasonable to assume that the life span is normally distributed?Why?

(b) Find the mean and standard deviation of your sample.

(c) Compare the mean and standard deviation of your sample withthose in the manufacturer’s claim. Discuss the differences.

42. Heights of Men You are performing a study about the height of 20- to29-year-old men. A previous study found the height to be normallydistributed, with a mean of 69.6 inches and a standard deviation of 3.0 inches. You randomly sample 30 men and find their heights to be as follows. (Adapted from National Center for Health Statistics)

72.1, 71.2, 67.9, 67.3, 69.5, 68.6, 68.8, 69.4, 73.5, 67.1,69.2, 75.7, 71.1, 69.6, 70.7, 66.9, 71.4, 62.9, 69.2, 64.9,68.2, 65.2, 69.7, 72.2, 67.5, 66.6, 66.5, 64.2, 65.4, 70.0

(a) Draw a frequency histogram to display these data. Use seven classes with midpoints of 63.85, 65.85, 67.85, 69.85, 71.85, 73.85, and75.85. Is it reasonable to assume that the heights are normally distributed? Why?

(b) Find the mean and standard deviation of your sample.

(c) Compare the mean and standard deviation of your sample withthose in the previous study. Discuss the differences.

Computing and Interpreting z-Scores of Normal Distributions InExercises 43–46, you are given a normal distribution, the distribution’s mean andstandard deviation, four values from that distribution, and a graph of the StandardNormal Distribution. (a) Without converting to z-scores, match each value with theletters A, B, C, and D on the given graph of the Standard Normal Distribution.(b) Find the z-score that corresponds to each value and check your answers to part(a). (c) Determine whether any of the values are unusual.

43. Piston Rings Your company manufactures piston rings for cars. The insidediameters of the piston rings are normally distributed, with a mean of 93.01 millimeters and a standard deviation of 0.005 millimeter. The insidediameters of four piston rings selected at random are 93.014 millimeters,93.018 millimeters, 93.004 millimeters, and 92.994 millimeters.

FIGURE FOR EXERCISE 43 FIGURE FOR EXERCISE 44

z

A CB D

z

A CB D

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44. House Wren Eggs The incubation period for the eggs of a House Wren isnormally distributed with a mean time of 336 hours and a standarddeviation of 3.5 hours. The incubation times for four eggs selected atrandom are 328 hours, 338 hours, 330 hours, and 341 hours.

45. SAT I Scores The SAT is an exam used by colleges and universities toevaluate undergraduate applicants. The test scores are normally distributed.In a recent year, the mean test score was 1518 and the standard deviation was308. The test scores of four students selected at random are 1406, 1848, 2177,and 1186. (Source: College Board Online)


46. ACT Scores The ACT is an exam used by colleges and universities toevaluate undergraduate applicants. The test scores are normally distributed.In a recent year, the mean test score was 21.0 and the standard deviation was4.8. The test scores of four students selected at random are 18, 32, 14, and 25.(Source: ACT, Inc.)

Graphical Analysis In Exercises 47–52, find the probability of z occurring inthe indicated region. If convenient, use technology to find the probability.

47. 48.

49. 50.

51. 52.

0.5 2z

0−0.5 1z

0

zz0−1.281.645

z0

−1.0z

00.5z

0

z

A CB D

z

A CB D

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Finding Probabilities In Exercises 53–62, find the indicated probability using the standard normal distribution. If convenient, use technology to find theprobability.

53. 54. 55.

56. 57. 58.

59. 60.

61. 62.

� Extending Concepts63. Writing Draw a normal curve with a mean of 60 and a standard deviation

of 12. Describe how you constructed the curve and discuss its features.

64. Writing Draw a normal curve with a mean of 450 and a standard deviationof 50. Describe how you constructed the curve and discuss its features.

65. Uniform Distribution Another continuous distribution is the uniformdistribution. An example is for The mean of thisdistribution for this example is 0.5 and the standard deviation is approximately0.29. The graph of this distribution for this example is a square with the heightand width both equal to 1 unit. In general, the density function for a uniformdistribution on the interval from to is given by

The mean is

and the variance is

(a) Verify that the area under the curve is 1.

(b) Find the probability that x falls between 0.25 and 0.5.

(c) Find the probability that x falls between 0.3 and 0.7.

66. Uniform Distribution Consider the uniform density function for The mean of this distribution is 15 and the standarddeviation is about 2.89.

(a) Draw a graph of the distribution and show that the area under the curveis 1.

(b) Find the probability that x falls between 12 and 15.

(c) Find the probability that x falls between 13 and 18.

10 … x … 20.f1x2 = 0.1

1

1

x

= 0.5μ

f(x)

1b - a22

12.

a + b

2

f1x2 =

1b - a

.

x = bx = a

0 … x … 1.f1x2 = 1

P1z 6 -1.54 or z 7 1.542P1z 6 -2.58 or z 7 2.582

P1-1.54 6 z 6 1.542P1-1.65 6 z 6 1.652

P1-2.08 6 z 6 02P1-0.89 6 z 6 02P1z 7 -1.852

P1z 7 -0.952P1z 6 0.452P1z 6 1.452

41831S4_0501.qxd 11/5/07 12:39 PM Page 252

S E C T I O N 5 . 2 NORMAL DISTRIBUTIONS: FINDING PROBABILITIES 253

� Probability and Normal DistributionsIf a random variable x is normally distributed, you can find the probability that xwill fall in a given interval by calculating the area under the normal curve for thegiven interval. To find the area under any normal curve, you can first convert theupper and lower bounds of the interval to z-scores. Then use the standard normaldistribution to find the area. For instance, consider a normal curve with and as shown at the upper left. The value of x one standard deviationabove the mean is Now consider the standard normalcurve shown at the lower left. The value of z one standard deviation above themean is Because a z-score of 1 corresponds to an x-value of 600, and areas are not changed with a transformation to a standardnormal curve, the shaded areas in the graphs are equal.

m + s = 0 + 1 = 1.

m + s = 500 + 100 = 600.s = 100,

m = 500

Probability and Normal Distributions

253

5.2 Normal Distributions: Finding ProbabilitiesWhat You

SHOULD LEARN� How to find

probabilities for normally distributed variables using a table and using technology

500400300200 600 700 800x

0−1−2−3 1 2 3z

= 0μSamearea

= 500μ

In Example 1, you can use a TI-83/84 to find the probabilityonce the upper bound is convertedto a -score.z

Study TipAnother way to write theanswer to Example 1 isP1x 6 22 = 0.2119 .

E X A M P L E 1

Finding Probabilities for Normal DistributionsA survey indicates that people use their computers an average of 2.4 years beforeupgrading to a new machine. The standard deviation is 0.5 year. A computerowner is selected at random. Find the probability that he or she will use it for fewer than 2 years before upgrading. Assume that the variable x is normallydistributed.

Solution The graph shows a normalcurve with and and ashaded area for x less than 2. The z-scorethat corresponds to 2 years is

The Standard Normal Table shows thatThe probability

that the computer will be upgraded infewer than 2 years is 0.2119.

Interpretation So, 21.19% of computer owners will upgrade in fewer than 2 years.

� Try It Yourself 1A Ford Focus manual transmission gets an average of 24 miles per gallon(mpg) in city driving with a standard deviation of 1.6 mpg. A Focus is selectedat random. What is the probability that it will get more than 28 mpg? Assumethat gas mileage is normally distributed. (Adapted from U.S. Department of Energy)

a. Sketch a graph.b. Find the z-score that corresponds to 28 miles per gallon.c. Find the area to the right of that z-score.d. Write the result as a sentence. Answer: Page A40

= 0.2119.P1z 6 -0.82

z =

x - m

s=

2 - 2.40.5

= -0.80.

s = 0.5m = 2.4

0 1 2 3 4 5

= 2.4μ

x

Age of computer (in years)

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10 20 30 40 6050 70 80

μ = 45

x

Time (in minutes)

10 20 30 40 6050 70 80

μ = 45

x

Time (in minutes)

E X A M P L E 2

Finding Probabilities for Normal DistributionsA survey indicates that for each trip to the supermarket, a shopper spends anaverage of 45 minutes with a standard deviation of 12 minutes in the store. Thelength of time spent in the store is normally distributed and is represented bythe variable A shopper enters the store. (a) Find the probability that theshopper will be in the store for each interval of time listed below. (b) Interpretyour answer if 200 shoppers enter the store. How many shoppers would youexpect to be in the store for each interval of time listed below?

1. Between 24 and 54 minutes 2. More than 39 minutes

Solution

1. (a) The graph at the left shows a normal curve with minutes andminutes. The area for between 24 and 54 minutes is shaded.

The -scores that correspond to 24 minutes and to 54 minutes are

and

So, the probability that a shopper will be in the store between 24 and 54minutes is

(b) Interpretation So, 73.33% of the shoppers will be in the store between24 and 54 minutes. If 200 shoppers enter the store, then you would expect

(or about 147) shoppers to be in the store between24 and 54 minutes.

2. (a) The graph at the left shows a normal curve with minutes andminutes. The area for x greater than 39 minutes is shaded. The

z-score that corresponds to 39 minutes is

So, the probability that a shopper will be in the store more than 39 minutes is

(b) Interpretation If 200 shoppers enter the store, then you would expect(or about 138) shoppers to be in the store more than

39 minutes.

� Try It Yourself 2What is the probability that the shopper in Example 2 will be in thesupermarket between 33 and 60 minutes?

a. Sketch a graph.b. Find z-scores that correspond to 60 minutes and 33 minutes.c. Find the cumulative area for each -score and subtract the smaller area from

the larger.d. Interpret your answer if 150 shoppers enter the store. How many shoppers

would you expect to be in the store between 33 and 60 minutes?Answer: Page A40

z

20010.69152 = 138.3

= 0.6915.= 1 - 0.3085 P1x 7 392 = P1z 7 -0.52 = 1 - P1z 6 -0.52

z =

39 - 4512

= -0.5.

s = 12m = 45

20010.73332 = 146.66

= 0.7333. = 0.7734 - 0.0401

= P1z 6 0.752 - P1z 6 -1.752

P124 6 x 6 542 = P1-1.75 6 z 6 0.752

z2 =

54 - 4512

= 0.75.z1 =

24 - 4512

= -1.75

zxs = 12

m = 45

x.

41831S4_0502.qxd 11/5/07 12:42 PM Page 254

Another way to find normal probabilities is to use a calculator or a computer.You can find normal probabilities using MINITAB, Excel, and the TI-83/84.

Example 3 shows only one of several ways to find normal probabilities usingMINITAB, Excel, and the TI-83/84.


In baseball, abatting average is the number ofhits divided by the

number of at-bats. The battingaverages of the more than 750 Major League Baseball players in a recent year can be approximated by a normaldistribution, as shown in the following graph. The mean ofthe batting averages is 0.269 andthe standard deviation is 0.009.

What percent of the playershave a batting average of 0.275 or greater? If there are40 players on a roster, howmany would you expect to have a batting average of 0.275 or greater?

Batting average

μ = 0.269

Major League Baseball

0.250 0.270 0.290

E X A M P L E 3

Using Technology to Find Normal ProbabilitiesAssume that cholesterol levels of men in the United States are normallydistributed, with a mean of 215 milligrams per deciliter and a standarddeviation of 25 milligrams per deciliter. You randomly select a man from theUnited States.What is the probability that his cholesterol level is less than 175?Use a technology tool to find the probability.

Solution MINITAB, Excel, and the TI-83/84 each have features that allowyou to find normal probabilities without first converting to standard z-scores.For each, you must specify the mean and standard deviation of the population,as well as the x-value(s) that determine the interval.

From the displays, you can see that the probability that his cholesterol level isless than 175 is about 0.0548, or 5.48%.

� Try It Yourself 3A man from the United States is selected at random. What is the probabilitythat his cholesterol is between 190 and 225? Use a technology tool.

a. Read the user’s guide for the technology tool you are using.b. Enter the appropriate data to obtain the probability.c. Write the result as a sentence. Answer: Page A40

T I - 8 3 / 8 4

normalcdf(0,175,215,25).0547992894

M I N I TA B

Cumulative Distribution Function

Normal with mean 215.000 and standard deviation 25.0000

x P(X <= x)175.0000 0.0548

==

E X C E L

0.054799

CBANORMDIST(175,215,25,TRUE)1

2

41831S4_0502.qxd 11/5/07 12:42 PM Page 255


� Building Basic Skills and Vocabulary

Computing Probabilities In Exercises 1–6, assume the random variable isnormally distributed with mean and standard deviation Find theindicated probability.

1. 2.

3. 4.

5. 6.

Graphical Analysis In Exercises 7–12, assume a member is selected at randomfrom the population represented by the graph. Find the probability that themember selected at random is from the shaded area of the graph. Assume thevariable x is normally distributed.

7. 8.

(Source: College Board Online) (Source: College Board Online)

9. 10.

(Adapted from National Center (Adapted from National Center for Health Statistics) for Health Statistics)

11. 12.

(Adapted from Consumer Reports) (Adapted from Consumer Reports)

x

= 149μ= 5.18σ

133 140 149 165

Braking distance (in feet)

Toyota Camry: BrakingDistance on a Wet Surface

140 < x < 149

x

= 137μ= 4.76σ

151

Braking distance (in feet)

Toyota Camry: BrakingDistance on a Dry Surface

141 < x < 151

141122

x

= 219μ= 41.6σ

95 239 340200

Total cholesterol level (in mg/dL)

U.S. Women Ages 55–64:Total Cholesterol

200 < x < 239

x

= 186μ= 35.8σ

75 200 239 300

Total cholesterol level (in mg/dL)

U.S. Women Ages 20–34:Total Cholesterol

200 < x < 239

x200 670 800

= 518μ= 115σ

Score

SAT Math Scores

670 < x < 800x

Score

= 503μ200 < x < 450

= 113σ

SAT Critical Reading Scores

200 450 800

P185 6 x 6 952P170 6 x 6 802

P1x 7 752P1x 7 922

P1x 6 1002P1x 6 802

s = 5.m = 86x

For Extra Help

5.2 EXERCISES

41831S4_0502.qxd 11/5/07 12:42 PM Page 256


� Using and Interpreting Concepts

Finding Probabilities In Exercises 13–20, find the indicated probabilities.If convenient, use technology to find the probabilities.

13. Heights of Men A survey was conducted to measure the height of U.S.men. In the survey, respondents were grouped by age. In the 20–29 age group,the heights were normally distributed, with a mean of 69.6 inches and astandard deviation of 3.0 inches. A study participant is randomly selected.(Adapted from U.S. National Center for Health Statistics)

(a) Find the probability that his height is less than 66 inches.

(b) Find the probability that his height is between 66 and 72 inches.

(c) Find the probability that his height is more than 72 inches.

14. Fish Lengths The lengths of Atlantic croaker fish are normally distributed,with a mean of 10 inches and a standard deviation of 2 inches. An Atlanticcroaker fish is randomly selected. (Adapted from National Marine FisheriesService, Fisheries Statistics and Economics Division)

(a) Find the probability that the length of the fish is less than 7 inches.

(b) Find the probability that the length of the fish is between 7 and 15 inches.

(c) Find the probability that the length of the fish is more than 15 inches.

15. ACT Scores In a recent year, the ACT scores for high school students witha 3.50 to 4.00 grade point average were normally distributed, with a mean of24.2 and a standard deviation of 4.3.A student with a 3.50 to 4.00 grade pointaverage who took the ACT during this time is randomly selected. (Source:ACT, Inc.)

(a) Find the probability that the student’s ACT score is less than 17.

(b) Find the probability that the student’s ACT score is between 20 and 29.

(c) Find the probability that the student’s ACT score is more than 32.

16. Beagles The weights of adult male beagles are normally distributed, with amean of 25 pounds and a standard deviation of 3 pounds. A beagle is ran-domly selected.

(a) Find the probability that the beagle’s weight is less than 23 pounds.

(b) Find the probability that the weight is between 23 and 25 pounds.

(c) Find the probability that the beagle’s weight is more than 27 pounds.

17. Computer Usage A survey was conducted to measure the number of hoursper week adults in the United States spend on home computers. In thesurvey, the number of hours were normally distributed, with a mean of 7 hours and a standard deviation of 1 hour. A survey participant is randomlyselected.

(a) Find the probability that the hours spent on the home computer by theparticipant are less than 5 hours per week.

(b) Find the probability that the hours spent on the home computer by theparticipant are between 5.5 and 9.5 hours per week.

(c) Find the probability that the hours spent on the home computer by theparticipant are more than 10 hours per week.

41831S4_0502.qxd 11/5/07 12:42 PM Page 257


18. Utility Bills The monthly utility bills in a city are normally distributed,with a mean of $100 and a standard deviation of $12. A utility bill israndomly selected.

(a) Find the probability that the utility bill is less than $70.

(b) Find the probability that the utility bill is between $90 and $120.

(c) Find the probability that the utility bill is more than $140.

19. Computer Lab Schedule The time per week a student uses a lab computeris normally distributed, with a mean of 6.2 hours and a standard deviation of0.9 hour. A student is randomly selected.

(a) Find the probability that the student uses a lab computer less than 4 hours per week.

(b) Find the probability that the student uses a lab computer between 5 and 7 hours per week.

(c) Find the probability that the student uses a lab computer more than 8 hours per week.

20. Health Club Schedule The time per workout an athlete uses a stairclimberis normally distributed, with a mean of 20 minutes and a standard deviationof 5 minutes. An athlete is randomly selected.

(a) Find the probability that the athlete uses a stairclimber for less than 17 minutes.

(b) Find the probability that the athlete uses a stairclimber between 20 and 28 minutes.

(c) Find the probability that the athlete uses a stairclimber for more than 30minutes.

Using Normal Distributions In Exercises 21–30, answer the questions aboutthe specified normal distribution.

21. SAT Critical Reading Scores Use the normal distribution of SAT criticalreading scores in Exercise 7 for which the mean is 503 and the standarddeviation is 113.

(a) What percent of the SAT verbal scores are less than 600?

(b) If 1000 SAT verbal scores are randomly selected, about how many wouldyou expect to be greater than 550?

22. SAT Math Scores Use the normal distribution of SAT math scores inExercise 8 for which the mean is 518 and the standard deviation is 115.

(a) What percent of the SAT math scores are less than 500?

(b) If 1500 SAT math scores are randomly selected, about how many wouldyou expect to be greater than 600?

23. Cholesterol Use the normal distribution of women’s total cholesterol levels in Exercise 9 for which the mean is 186 milligrams per deciliter and thestandard deviation is 35.8 milligrams per deciliter.

(a) What percent of the women have a total cholesterol level less than 200 milligrams per deciliter of blood?

(b) If 250 U.S. women in the 20–34 age group are randomly selected, abouthow many would you expect to have a total cholesterol level greaterthan 240 milligrams per deciliter of blood?

41831S4_0502.qxd 11/5/07 12:42 PM Page 258


24. Cholesterol Use the normal distribution of women’s total cholesterol levels in Exercise 10 for which the mean is 219 milligrams per deciliter andthe standard deviation is 41.6 milligrams per deciliter.

(a) What percent of the women have a total cholesterol level less than 239 milligrams per deciliter of blood?

(b) If 200 U.S. women in the 55–64 age group are randomly selected, abouthow many would you expect to have a total cholesterol level greaterthan 200 milligrams per deciliter of blood?

25. Fish Lengths Use the normal distribution of fish lengths in Exercise 14 forwhich the mean is 10 inches and the standard deviation is 2 inches.

(a) What percent of the fish are longer than 11 inches?

(b) If 200 Atlantic croakers are randomly selected, about how many wouldyou expect to be shorter than 8 inches?

26. Beagles Use the normal distribution of beagle weights in Exercise 16 forwhich the mean is 25 pounds and the standard deviation is 3 pounds.

(a) What percent of the beagles have a weight that is greater than 30 pounds?

(b) If 50 beagles are randomly selected, about how many would you expectto weigh less than 22 pounds?

27. Computer Usage Use the normal distribution of computer usage inExercise 17 for which the mean is 7 hours and the standard deviation is 1 hour.

(a) What percent of the adults spend more than 4 hours per week on a homecomputer?

(b) If 35 adults in the United States are randomly selected, about howmany would you expect to say they spend less than 5 hours per weekon a home computer?

28. Utility Bills Use the normal distribution of utility bills in Exercise 18 forwhich the mean is $100 and the standard deviation is $12.

(a) What percent of the utility bills are more than $125?

(b) If 300 utility bills are randomly selected, about how many would youexpect to be less than $90?

29. Battery Life Spans The life span of a battery is normally distributed, witha mean of 2000 hours and a standard deviation of 30 hours. What percentof batteries have a life span that is more than 2065 hours? Would it beunusual for a battery to have a life span that is more than 2065 hours?Explain your reasoning.

30. Peanuts Assume the mean annual consumption of peanuts is normallydistributed, with a mean of 5.9 pounds per person and a standard deviationof 1.8 pounds per person.What percent of people annually consume less than3.1 pounds of peanuts per person? Would it be unusual for a person to consume less than 3.1 pounds of peanuts in a year? Explain your reasoning.

41831S4_0502.qxd 11/5/07 12:42 PM Page 259


� Extending Concepts

Control Charts Statistical process control (SPC) is the use of statistics tomonitor and improve the quality of a process, such as manufacturing an enginepart. In SPC, information about a process is gathered and used to determine if aprocess is meeting all of the specified requirements. One tool used in SPC is acontrol chart. When individual measurements of a variable x are normallydistributed, a control chart can be used to detect processes that are possibly out ofstatistical control.Three warning signals that a control chart uses to detect a processthat may be out of control are as follows:

(1) A point lies beyond three standard deviations of the mean.

(2) There are nine consecutive points that fall on one side of the mean.

(3) At least two of three consecutive points lie more than two standard deviations from the mean.

In Exercises 31–34, a control chart is shown. Each chart has horizontal lines drawnat the mean at and at Determine if the process shown is incontrol or out of control. Explain.

m ; 3s.m ; 2s,m,

31. A gear has been designed to havea diameter of 3 inches. Thestandard deviation of the processis 0.2 inch.

33. A liquid-dispensing machine hasbeen designed to fill bottles with 1 liter of liquid. The standarddeviation of the process is 0.1liter.

32. A nail has been designed to havea length of 4 inches. The standarddeviation of the process is 0.12inch.

34. An engine part has beendesigned to have a diameter of 55 millimeters. The standarddeviation of the process is 0.001millimeter.

2 4 6 8 10 12

55.0000

55.0025

55.0050

54.9975

Dia

met

er (

in m

illim

eter

s)

Observation number

Engine Part

2 4 6 8 1011121 3 5 7 9

3.75

4.00

4.25

4.50

Observation number

Len

gth

(in

inch

es)

Nails

2 4 6 8 10 11 121 3 5 7 9

0.5

1.0

1.5

Observation number

Liq

uid

disp

ense

d (i

n lit

ers)

Liquid Dispenser

2 4 6 8 101 3 5 7 9

1

2

3

4

Observation number

Dia

met

er (

in in

ches

)

Gears

41831S4_0502.qxd 11/5/07 12:42 PM Page 260

S E C T I O N 5 . 3 NORMAL DISTRIBUTIONS: FINDING VALUES 261

� Finding z-ScoresIn Section 5.2, you were given a normally distributed random variable and youfound the probability that would fall in a given interval by calculating the areaunder the normal curve for the given interval.

But what if you are given a probability and want to find a value? Forinstance, a university might want to know what is the lowest test score a studentcan have on an entrance exam and still be in the top 10%, or a medical researchermight want to know the cutoff values to select the middle 90% of patients by age.In this section, you will learn how to find a value given an area under a normalcurve (or a probability), as shown in the following example.

xx

Finding -Scores � Transforming a -Score to an -Value �

Finding a Specific Data Value for a Given Probabilityxzz

5.3 Normal Distributions: Finding Values

z

Area =0.3632

−0.35 0

1.24z

0

Area =0.1075

Study TipHere are instructions for findingthe -score that corresponds to agiven area on a TI-83/84.

3: invNorm(

Enter the cumulative area.

ENTER

DISTR2nd

z

E X A M P L E 1

Finding a z-Score Given an Area1. Find the -score that corresponds to a cumulative area of 0.3632.2. Find the -score that has 10.75% of the distribution’s area to its right.

Solution1. Find the -score that corresponds to an area of 0.3632 by locating 0.3632 in

the Standard Normal Table. The values at the beginning of the correspondingrow and at the top of the corresponding column give the -score. For thisarea, the row value is and the column value is 0.05. So, the -score is

2. Because the area to the right is 0.1075, the cumulative area isFind the -score that corresponds to an area of 0.8925

by locating 0.8925 in the Standard Normal Table. For this area, the row valueis 1.2 and the column value is 0.04. So, the -score is 1.24.

You can also use a computer or calculator to find the -scores thatcorrespond to the given cumulative areas, as shown in the margin.

z

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .85541.1 .8643 .8665 .8686 .8708 .8729 .8749 .87701.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962

z .00 .01 .02 .03 .04 .05 .060.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239

1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131

z

z1 - 0.1075 = 0.8925.

�0.5�0.4�0.3�0.2

z .09 .08 .07 .06 .05 .03�3.4 .0002 .0003 .0003 .0003 .0003 .0003.0003

.2776 .2810 .2843 .2877 .2912 .2946 .2981

.3121 .3156 .3192 .3228 .3264 .3300 .3336

.3483 .3520 .3557 .3594 .3632 .3669 .3707

.3859 .3897 .3936 .3974 .4013 .4052 .4090

.04

-0.35.z-0.3

z

z

zz

What You SHOULD LEARN

� How to find a -score given the area under the normal curve

� How to transform a -score to an -value

� How to find a specific data value of a normal distribution given the probability

xz

z

41831S4_0503.qxd 11/5/07 12:42 PM Page 261

In Section 2.5, you learned that percentiles divide a data set into one hundredequal parts. To find a score that corresponds to a percentile, you can use the Standard Normal Table. Recall that if a value x represents the 83rd percentile

then 83% of the data values are below x and 17% of the data values areabove x.P83,

z-


z−1.645 0

Area = 0.05

0z

Area = 0.5

z1.280

Area = 0.8997

Study TipIn most cases, the given area willnot be found in the table, so usethe entry closest to it. If the givenarea is halfway between two areaentries, use the -scorehalfway between thecorresponding -scores. For instance, in part 1 ofExample 2, the -scorebetween and

is -1.645.-1.65-1.64

z

z

z

� Try It Yourself 11. Find the score that has 96.16% of the distribution’s area to the right.2. Find the score for which 95% of the distribution’s area lies between

and

a. Determine the cumulative area.b. Locate the area in the Standard Normal Table.c. Find the score that corresponds to the area. Answer: Page A40z-

z.-zz-

z-

E X A M P L E 2

Finding a z-Score Given a PercentileFind the z-score that corresponds to each percentile.

1. 2. 3.

Solution

1. To find the score that corresponds to find the score that correspondsto an area of 0.05 (see figure) by locating 0.05 in the Standard Normal Table.The areas closest to 0.05 in the table are 0.0495 and 0.0505

Because 0.05 is halfway between the two areas in the table,use the score that is halfway between and So, the scorethat corresponds to an area of 0.05 is

2. To find the score that corresponds to find the score thatcorresponds to an area of 0.5 (see figure) by locating 0.5 in the StandardNormal Table. The area closest to 0.5 in the table is 0.5000, so the scorethat corresponds to an area of 0.5 is 0.00.

3. To find the score that corresponds to find the score thatcorresponds to an area of 0.9 (see figure) by locating 0.9 in the StandardNormal Table. The area closest to 0.9 in the table is 0.8997, so the scorethat corresponds to an area of 0.9 is 1.28.

� Try It Yourself 2Find the score that corresponds to each percentile.

1. 2. 3.

a. Write the percentile as an area. If necessary, draw a graph of the area tovisualize the problem.

b. Locate the area in the Standard Normal Table. If the area is not in the table,use the closest area. (See Study Tip above.)

c. Identify the score that corresponds to the area. Answer: Page A40z-

P99P20P10

z-

z-

z-P90 ,z-

z-

z-P50 ,z-

-1.645.z--1.65.-1.64z-

1z = -1.642 .1z = -1.652

z-P5 ,z-

P90P50P5

41831S4_0503.qxd 11/5/07 12:42 PM Page 262

� Transforming a z-Score to an x-Value Recall that to transform an -value to a score, you can use the formula

This formula gives in terms of x. If you solve this formula for x, you get anew formula that gives x in terms of .

Formula for z in terms of x

Multiply each side by

Add to each side.

Interchange sides. x = m + zs

m m + zs = x

s. zs = x - m

z =

x - m

s

zz

z =

x - m

s.

z-x


To transform a standard score to a data value in a given population, usethe formula

x = m + zs .

xz-

T R A N S F O R M I N G A Z - S C O R E T O A N X - VA L U E

E X A M P L E 3

Finding an x-Value Corresponding to a z-ScoreThe speeds of vehicles along a stretch of highway are normally distributed, witha mean of 67 miles per hour and a standard deviation of 4 miles per hour. Findthe speeds corresponding to scores of 1.96, and 0. Interpret yourresults.

Solution The x-value that corresponds to each standard score is calculatedusing the formula

miles per hour

miles per hour

miles per hour

Interpretation You can see that 74.84 miles per hour is above the mean, 57.68is below the mean, and 67 is equal to the mean.

� Try It Yourself 3The monthly utility bills in a city are normally distributed, with a mean of $70and a standard deviation of $8. Find the x-values that correspond to scores of

4.29, and What can you conclude?

a. Identify and of the nonstandard normal distribution.b. Transform each score to an x-value.c. Interpret the results. Answer: Page A40

z-sm

-1.82.-0.75,z-

= 67 x = 67 + 0142z = 0:

= 57.68 x = 67 + 1-2.332142z = -2.33:

= 74.84 x = 67 + 1.96142z = 1.96:

x = m + zs.

-2.33,z-x

41831S4_0503.qxd 11/5/07 12:42 PM 3

� Finding a Specific Data Value for a GivenProbability

You can also use the normal distribution to find a specific data value (x-value)for a given probability, as shown in Example 4.


According to theAmerican MedicalAssociation, themean number of

hours all physicians spend inpatient care each week is about52.8 hours. The hours spent inpatient care each week byphysicians can be approximatedby a normal distribution. Assumethe standard deviation is 3 hours.

Between what two values doesthe middle 90% of the data lie?

Hours Physicians Spendin Patient Care

x

= 52.8μ

44 46 48 50 52 54 56

Hours

58 60 62

Study TipHere are instructions for finding a specific -value for a given probability on a TI-83/84.

3: invNorm(

Enter the values for the area under the normal distribution, the specified mean, and the specified standard deviationseparated by commas.

ENTER

DISTR2nd

x

E X A M P L E 4

Finding a Specific Data ValueScores for a civil service exam are normally distributed, with a mean of 75 anda standard deviation of 6.5. To be eligible for civil service employment, youmust score in the top 5%. What is the lowest score you can earn and still beeligible for employment?

Solution Exam scores in the top 5% correspond to the shaded regionshown.

An exam score in the top 5% is any score above the 95th percentile.To find thescore that represents the 95th percentile, you must first find the -score thatcorresponds to a cumulative area of 0.95. From the Standard Normal Table, youcan find that the areas closest to 0.95 are 0.9495 and 0.9505

Because 0.95 is halfway between the two areas in the table, use the-score that is halfway between 1.64 and 1.65. That is, Using the

equation you have

Interpretation The lowest score you can earn and still be eligible foremployment is 86.

� Try It Yourself 4The braking distances of a sample of Honda Accords are normally distributed.On a dry surface, the mean braking distance was 142 feet and the standarddeviation was 6.51 feet. What is the longest braking distance on a dry surfaceone of these Honda Accords could have and still be in the top 1%? (Adaptedfrom Consumer Reports)

a. Sketch a graph.b. Find the -score that corresponds to the given area.c. Find x using the equation d. Interpret the result. Answer: Page A40

x = m + zs .z

L 85.69.

= 75 + 1.64516.52

x = m + zs

x = m + zs ,z = 1.645.z

1z = 1.652 .1z = 1.642

z

0 1.645z

75 ?x (exam score)

5%

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Using a TI-83/84, you can findthe highest total cholesterollevel automatically.

E X A M P L E 5

Finding a Specific Data ValueIn a randomly selected sample of 1169 men ages 35–44, the mean totalcholesterol level was 210 milligrams per deciliter with a standard deviation of38.6 milligrams per deciliter. Assume the total cholesterol levels are normallydistributed. Find the highest total cholesterol level a man in this 35–44 age group can have and be in the lowest 1%. (Adapted from National Center forHealth Statistics)

SolutionTotal cholesterol levels in the lowest 1% correspond to the shaded regionshown.

A total cholesterol level in the lowest 1% is any level below the 1st percentile.To find the level that represents the 1st percentile, you must first find the -score that corresponds to a cumulative area of 0.01. From the Standard

Normal Table, you can find that the area closest to 0.01 is 0.0099. So, the -scorethat corresponds to an area of 0.01 is Using the equation

you have

Interpretation The value that separates the lowest 1% of total cholesterollevels for men in the 35–44 age group from the highest 99% is about 120.

� Try It Yourself 5The length of time employees have worked at a corporation is normallydistributed, with a mean of 11.2 years and a standard deviation of 2.1 years.In a company cutback, the lowest 10% in seniority are laid off. What is the maximum length of time an employee could have worked and still be laid off?

a. Sketch a graph.b. Find the -score that corresponds to the given area.c. Find x using the equation d. Interpret the result. Answer: Page A41

x = m + zs .z

L 120.06.

= 210 + 1-2.332138.62

x = m + zs

x = m + zs ,z = -2.33.

zz

−2.33 0z

? 210x (total cholesterol level, in mg/dL)

Total Cholesterol Levels inMen Ages 35–44

1%

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In Exercises 1–24, use the Standard Normal Table to find the -score thatcorresponds to the given cumulative area or percentile. If the area is not in thetable, use the entry closest to the area. If the area is halfway between two entries, usethe -score halfway between the corresponding -scores. If convenient, usetechnology to find the -score.

1. 0.7580 2. 0.2090 3. 0.6331 4. 0.0918

5. 0.4364 6. 0.0080 7. 0.9916 8. 0.7995

9. 0.05 10. 0.85 11. 0.94 12. 0.01

13. 14. 15. 16.

17. 18. 19. 20.

21. 22. 23. 24.

Graphical Analysis In Exercises 25–30, find the indicated -score(s) shown inthe graph. If convenient, use technology to find the -score(s).

25. 26.

27. 28.

29. 30.

In Exercises 31–38, find the indicated -score.

31. Find the -score that has 11.9% of the distribution’s area to its left.

32. Find the score that has 78.5% of the distribution’s area to its left.

33. Find the -score that has 11.9% of the distribution’s area to its right.

34. Find the score that has 78.5% of the distribution’s area to its right.

35. Find the -score for which 80% of the distribution’s area lies between and .z

-zz

z-

z

z-

z

z

z

Area =0.475

Area =0.475

z = ?z = ? 0

Area =0.05

Area =0.05

zz = ?z = ? 0

z

Area =0.0233

z = ?0z

Area =0.7190

z = ? 0

z

Area =0.5987

z = ?0z = ? 0z

Area =0.3520

zz

P65P35P90P75

P50P25P67P88

P55P20P15P1

zzz

z

For Extra Help

5.3 EXERCISES

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36. Find the -score for which 99% of the distribution’s area lies between and .




Using Normal Distributions In Exercises 39–44, answer the questions aboutthe specified normal distribution.

39. Heights of Women In a survey of women in the United States (ages20–29), the mean height was 64.1 inches with a standard deviation of 2.71inches. (Adapted from National Center for Health Statistics)

(a) What height represents the 95th percentile?

(b) What height represents the first quartile?

40. Heights of Men In a survey of men in the United States (ages 20–29), themean height was 69.6 inches with a standard deviation of 3.0 inches.(Adapted from National Center for Health Statistics)

(a) What height represents the 90th percentile?

(b) What height represents the first quartile?

41. Apples The annual per capita utilization of apples (in pounds) in theUnited States can be approximated by a normal distribution, as shown in the graph. (Adapted from U.S. Department of Agriculture)

(a) What annual per capita utilization of apples represents the 10th percentile?

(b) What annual per capita utilization of apples represents the thirdquartile?


42. Oranges The annual per capita utilization of oranges (in pounds) in theUnited States can be approximated by a normal distribution, as shown in thegraph. (Adapted from U.S. Department of Agriculture)

(a) What annual per capita utilization of oranges represents the 5th percentile?

(b) What annual per capita utilization of oranges represents the thirdquartile?

x

= 11.4 lbμ= 3 lbσ

2 5 8 11 14 2017

Utilization (in pounds)

Annual U.S. per CapitaOrange Utilization

x

= 17.1 lbμ= 4 lbσ

Utilization (in pounds)

Annual U.S. per CapitaApple Utilization

5 9 13 17 21 25 29

z-zz

z-zz

z-zz

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43. Heart Transplant Waiting Times The time spent (in days) waiting for aheart transplant in Ohio and Michigan for patients with type blood canbe approximated by a normal distribution, as shown in the graph. (Adaptedfrom Organ Procurement and Transplant Network)

(a) What is the shortest time spent waiting for a heart that would still placea patient in the top 30% of waiting times?

(b) What is the longest time spent waiting for a heart that would still placea patient in the bottom 10% of waiting times?

44. Ice Cream The annual per capita consumption of ice cream (in pounds) inthe United States can be approximated by a normal distribution, as shown inthe graph. (Adapted from U.S. Department of Agriculture)

(a) What is the smallest annual per capita consumption of ice cream that canbe in the top 25% of consumptions?

(b) What is the largest annual per capita consumption of ice cream that canbe in the bottom 15% of consumptions?

45. Cereal Boxes The weights of the contents of a cereal box are normallydistributed with a mean weight of 20 ounces and a standard deviation of 0.07 ounce. Boxes in the lower 5% do not meet the minimum weightrequirements and must be repackaged. What is the minimum weightrequirement for a cereal box?

46. Bags of Baby Carrots The weights of bags of baby carrots are normallydistributed with a mean of 32 ounces and a standard deviation of 0.36 ounce.Bags in the upper 4.5% are too heavy and must be repackaged. What is themost a bag of baby carrots can weigh and not need to be repackaged?

� Extending Concepts47. Writing a Guarantee You sell a brand of automobile tire that has a life

expectancy that is normally distributed, with a mean life of 30,000 miles and a standard deviation of 2500 miles. You want to give a guarantee for freereplacement of tires that don’t wear well. How should you word yourguarantee if you are willing to replace approximately 10% of the tires you sell?

48. Statistics Grades In a large section of a statistics class, the points for the finalexam are normally distributed with a mean of 72 and a standard deviation of 9. Grades are to be assigned according to the following rule.

• The top 10% receive As• The next 20% receive Bs• The middle 40% receive Cs• The next 20% receive Ds• The bottom 10% receive Fs

Find the lowest score on the final exam that would qualify a student for anA, a B, a C, and a D.

49. Vending Machine A vending machine dispenses coffee into an eight-ouncecup.The amount of coffee dispensed into the cup is normally distributed witha standard deviation of 0.03 ounce. You can allow the cup to overfill 1% ofthe time. What amount should you set as the mean amount of coffee to be dispensed?

A+

FIGURE FOR EXERCISE 44


x

Final Exam Grades

Points scored on final examD C B A

10%10%

20% 20%

40%

x

= 15.4 lbμ= 2.5 lbσ

Annual U.S. per CapitaIce Cream Consumption

Consumption (in pounds)8 10 12 14 16 18 20 22

x

= 127 daysμ= 23.5 daysσ

Time Spent Waiting fora Heart

60 87 114 195168141

Days


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� Exercises1. The distributions of birth weights for three gestation

periods are shown. Match the curves with thegestation periods. Explain your reasoning.

(a)

(b)

(c)

2. What percent of the babies born within each gestationperiod have a low birth weight (under 5.5 pounds)?Explain your reasoning.

(a) Under 28 weeks (b) 32 to 35 weeks

(c) 37 to 39 weeks (d) 42 weeks and over

3. Describe the weights of the top 10% of the babies bornwith each gestation period. Explain your reasoning.

(a) 37 to 39 weeks (b) 42 weeks and over

4. For each gestation period, what is the probability thata baby will weigh between 6 and 9 pounds at birth?

(a) 32 to 35 weeks (b) 37 to 39 weeks

(c) 42 weeks and over

5. A birth weight of less than 3.3 pounds is classified bythe NCHS as a “very low birth weight.” What is theprobability that a baby has a very low birth weightfor each gestation period?

(a) Under 28 weeks (b) 32 to 35 weeks

(c) 37 to 39 weeks

Pounds

μ

5 6 7 8 9 10 11

Pounds

μ

2 4 6 8 10

Pounds

μ

5 6 7 8 9 10 11

The National Center for Health Statistics (NCHS) keeps records ofmany health-related aspects of people, including the birth weightsof all babies born in the United States.

The birth weight of a baby is related to its gestation period(the time between conception and birth). For a given gestationperiod, the birth weights can be approximated by a normaldistribution. The means and standard deviations of the birthweights for various gestation periods are shown at the right.

One of the many goals of the NCHS is to reduce the percentageof babies born with low birth weights.As you can see from the graphat the upper right, the problem of low birth weights increased from1990 to 2004.

Year

Perc

ent

Preterm = under 37 weeksLow birth weight = under 5.5 pounds

1998 2000 2002 20041990 1992 1994 1996

6

8

10

12

14

16

Percent of preterm births

Percent of low birth weights

Birth Weights in America

269

Gestationperiod

Mean birthweight

Standarddeviation

Under 28 Weeks 1.88 lb 1.20 lb

28 to 31 Weeks 4.08 lb 1.85 lb

32 to 35 Weeks 5.71 lb 1.47 lb

36 Weeks 6.44 lb 1.19 lb

37 to 39 Weeks 7.31 lb 1.09 lb

40 Weeks 7.69 lb 1.04 lb

41 Weeks 7.79 lb 1.08 lb

42 Weeks and over 7.61 lb 1.11 lb

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� Sampling DistributionsIn previous sections, you studied the relationship between the mean of a population and values of a random variable. In this section, you will study the relationship between a population mean and the means of samples takenfrom the population.

For instance, consider the following Venn diagram. The rectangle representsa large population, and each circle represents a sample of size n. Because thesample entries can differ, the sample means can also differ. The mean of Sample1 is the mean of Sample 2 is and so on. The sampling distribution of thesample means for samples of size n for this population consists of andso on. If the samples are drawn with replacement, an infinite number of samplescan be drawn from the population.

Population with ,

Sample 1, x1

μ σ

Sample 2, x2

Sample 3, x3

Sample 4, x4

Sample 5, x5

x3,x2,x1,x2;x1;

Sampling Distributions � The Central Limit Theorem �

Probability and the Central Limit Theorem

5.4 Sampling Distributions and the Central Limit TheoremWhat You

SHOULD LEARN� How to find sampling

distributions and verifytheir properties

� How to interpretthe Central LimitTheorem

� How to apply theCentral LimitTheorem to findthe probability ofa sample mean

Sample means canvary from oneanother and canalso vary from thepopulation mean.This type of variation isto be expected and iscalled sampling error.

Insight

A sampling distribution is the probability distribution of a sample statisticthat is formed when samples of size are repeatedly taken from a population. If the sample statistic is the sample mean, then the distributionis the sampling distribution of sample means. Every sample statistic has a sampling distribution.

n

D E F I N I T I O N

P R O P E R T I E S O F S A M P L I N G D I S T R I B U T I O N S O F S A M P L E M E A N S

1. The mean of the sample means is equal to the population mean

2. The standard deviation of the sample means is equal to thepopulation standard deviation divided by the square root of

The standard deviation of the sampling distribution of the sample means iscalled the standard error of the mean.

sx =

s

2n

n.s

sx

mx = m

m .mx

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S E C T I O N 5 . 4 SAMPLING DISTRIBUTIONS AND THE CENTRAL LIMIT THEOREM 271

1 2 3 4 5 6 7

Prob

abili

ty

Sample mean

0.05

0.10

0.15

0.20

0.25

x

P(x)

Probability Distribution of Sample Means

Probability Histogram of Population of x

Probability Histogram of Sampling Distribution of x

Prob

abili

ty

Population values

0.25

1 2 3 4 5 6 7x

P(x)

Study TipReview Section 4.1 to find themean and standard deviationof a probabilitydistribution.

E X A M P L E 1

A Sampling Distribution of Sample MeansYou write the population values on slips of paper and put them ina box. Then you randomly choose two slips of paper, with replacement. List allpossible samples of size and calculate the mean of each. These meansform the sampling distribution of the sample means. Find the mean, variance,and standard deviation of the sample means. Compare your results with themean variance and standard deviation of thepopulation.

Solution List all 16 samples of size 2 from the population and the mean ofeach sample.

After constructing a probability distribution of the sample means, you can graphthe sampling distribution using a probability histogram as shown at the left.Notice that the shape of the histogram is bell shaped and symmetric, similar toa normal curve. The mean, variance, and standard deviation of the 16 samplemeans are

and

These results satisfy the properties of sampling distributions because

and

� Try It Yourself 1List all possible samples of with replacement, from the population

1, 3, 5, 7 Calculate the mean, variance, and standard deviation of the samplemeans. Compare these values with the corresponding population parameters.

a. Form all possible samples of size 3 and find the mean of each.b. Make a probability distribution of the sample means and find the mean,

variance, and standard deviation.c. Compare the mean, variance, and standard deviation of the sample means

with those for the population. Answer: Page A41

6.5n = 3,

sx =

s

2n=

25

22L

2.236

22L 1.581.mx = m = 4

sx =

A

52

= 22.5 L 1.581.1sx22

=

52

= 2.5

mx = 4

s = 25 L 2.236s2= 5,m = 4,

n = 2

51, 3, 5, 76

x f Probability

1 1 0.0625

2 2 0.1250

3 3 0.1875

4 4 0.2500

5 3 0.1875

6 2 0.1250

7 1 0.0625

Sample Sample mean, x

1, 1 1

1, 3 2

1, 5 3

1, 7 4

3, 1 2

3, 3 3

3, 5 4

3, 7 5

Sample Sample mean, x

5, 1 3

5, 3 4

5, 5 5

5, 7 6

7, 1 4

7, 3 5

7, 5 6

7, 7 7

To explore this topic further, see Activity 5.4 on page 284.

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� The Central Limit TheoremThe Central Limit Theorem forms the foundation for the inferential branch ofstatistics. This theorem describes the relationship between the samplingdistribution of sample means and the population that the samples are taken from.The Central Limit Theorem is an important tool that provides the informationyou’ll need to use sample statistics to make inferences about a population mean.


1. Any Population Distribution

Distribution of Sample Means,

2. Normal Population Distribution

Distribution of Sample Means(any )

μ μx

x =

Standarddeviation

Mean

σ σx =

n

n

xμ Mean

σ Standarddeviation

μ μx

x =

Standarddeviation

Mean

σ σx =

n

n Ú 30

μ

σ

x

Mean

Standarddeviation

The distribution of sample means has the same mean as the population. But its standarddeviation is less than the standarddeviation of the population. Thistells you that the distribution ofsample means has the same center as the population, but it is not as spread out.

Moreover, the distribution ofsample meansbecomes less andless spread out(tighter concentrationabout the mean) as thesample size n increases.

Insight

1. If samples of size n, where are drawn from any population witha mean and a standard deviation then the sampling distribution ofsample means approximates a normal distribution. The greater thesample size, the better the approximation.

2. If the population itself is normally distributed, the sampling distributionof sample means is normally distributed for any sample size n.

In either case, the sampling distribution of sample means has a mean equalto the population mean.

Mean

The sampling distribution of sample means has a variance equal to times the variance of the population and a standard deviation equal to thepopulation standard deviation divided by the square root of

Variance

Standard deviation

The standard deviation of the sampling distribution of the sample means,is also called the standard error of the mean.sx ,

sx =

s

1n

sx2

=

s2

n

n.

1>n

mx = m

s,m

n Ú 30,

T H E C E N T R A L L I M I T T H E O R E M

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In a recent year,there were morethan 5 millionparents in the

United States who received childsupport payments. The followinghistogram shows the distributionof children per custodial parent.The mean number of childrenwas 1.7 and the standarddeviation was 0.9. (Adapted from:U.S. Census Bureau)

You randomly select 35 parentswho receive child support andask how many children in theircustody are receiving child support payments. What is theprobability that the mean ofthe sample is between 1.5 and 1.9 children?

1 2 3 4 5 6 7

Prob

abili

ty

Number of children

0.1

0.2

0.3

0.5

0.4

x

Child SupportP(x)

E X A M P L E 2

Interpreting the Central Limit TheoremPhone bills for residents of a city have a mean of $64 and a standard deviationof $9, as shown in the following graph. Random samples of 36 phone bills aredrawn from this population and the mean of each sample is determined. Findthe mean and standard error of the mean of the sampling distribution. Thensketch a graph of the sampling distribution of sample means.

Solution The mean of the sampling distribution is equal to the populationmean, and the standard error of the mean is equal to the population standarddeviation divided by So,

and

Interpretation From the Central Limit Theorem, because the sample size isgreater than 30, the sampling distribution can be approximated by a normaldistribution with and as shown in the graph below.

� Try It Yourself 2Suppose random samples of size 100 are drawn from the population inExample 2. Find the mean and standard error of the mean of the samplingdistribution. Sketch a graph of the sampling distribution and compare it withthe sampling distribution in Example 2.

a. Find and b. Identify the sample size. If sketch a normal curve with mean and

standard deviation c. Compare the results with those in Example 2. Answer: Page A41

sx .mxn Ú 30,

sx .mx

64 73 8246 55x

Mean of 36 phone bills (in dollars)

Distribution ofSample Meanswith n = 36

s = $1.50,m = $64

sx =

s

1n=

9

236= 1.5.mx = m = 64

1n .

64 73 8246 55x

Individual phone bills (in dollars)

Distribution forAll Phone Bills

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E X A M P L E 3

Interpreting the Central Limit TheoremThe heights of fully grown white oak trees are normally distributed, with amean of 90 feet and standard deviation of 3.5 feet, as shown in the followinggraph. Random samples of size 4 are drawn from this population, and themean of each sample is determined. Find the mean and standard error of themean of the sampling distribution. Then sketch a graph of the samplingdistribution of sample means.

Solution The mean of the sampling distribution is equal to the populationmean, and the standard error of the mean is equal to the population standarddeviation divided by So,

feet and feet.

Interpretation From the Central Limit Theorem, because the population isnormally distributed, the sampling distribution of the sample means is alsonormally distributed, as shown in the graph below.

� Try It Yourself 3The diameters of fully grown white oak trees are normally distributed, with amean of 3.5 feet and a standard deviation of 0.2 foot, as shown in the graphbelow. Random samples of size 16 are drawn from this population, and the meanof each sample is determined. Find the mean and standard error of the mean ofthe sampling distribution. Then sketch a graph of the sampling distribution.

a. Find and b. Sketch a normal curve with mean and standard deviation

Answer: Page A41sx .mx

sx .mx

3.5

Diameter (in feet)

3.7 3.9 4.12.9 3.1 3.3x

Distribution ofPopulation Diameters

80 85 90

Mean height (in feet)

95 100x


sx =

s

1n=

3.5

24= 1.75mx = m = 90

1n .

80 85 90

Height (in feet)

95 100x

Distribution ofPopulation Heights

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� Probability and the Central Limit TheoremIn Section 5.2, you learned how to find the probability that a random variable xwill fall in a given interval of population values. In a similar manner, you can findthe probability that a sample mean will fall in a given interval of the samplingdistribution. To transform to a z-score, you can use the formula

z =

Value - MeanStandard error

=

x - mx

sx=

x - m

s>1n.

xxx


25.024.624.2 25.4 25.8

μ = 25

24.7 25.5

x

Mean time (in minutes)


2.36−1.41 0z

z-scoreDistributionof SampleMeans withn = 50

In Example 4, you can use a TI-83/84 to find the probabilityautomatically once the standarderror of the mean is calculated.

E X A M P L E 4

Finding Probabilities for Sampling DistributionsThe graph at the right showsthe length of time peoplespend driving each day. Yourandomly select 50 driversages 15 to 19. What is theprobability that the meantime they spend driving eachday is between 24.7 and 25.5 minutes? Assume that

minutes.

Solution The sample size is greater than 30, so you can use the Central LimitTheorem to conclude thatthe distribution of samplemeans is approximately normal with a mean and astandard deviation of

and

The graph of this distribution is shown at the left with a shaded area between24.7 and 25.5 minutes. The z-scores that correspond to sample means of 24.7and 25.5 minutes are

and

So, the probability that the mean time the 50 people spend driving each day isbetween 24.7 and 25.5 minutes is

Interpretation Of the samples of 50 drivers ages 15 to 19, 91.16% will have amean driving time that is between 24.7 and 25.5 minutes, as shown in the graphat the left. This implies that, assuming the value of is correct, only8.84% of such sample means will lie outside the given interval.

m = 25

= 0.9116. = 0.9909 - 0.0793

= P1z 6 2.362 - P1z 6 -1.412

P124.7 6 x 6 25.52 = P1-1.41 6 z 6 2.362

z2 =

25.5 - 25

1.5>250L

0.50.21213

L 2.36.

z1 =

24.7 - 25

1.5>250L

-0.30.21213

L -1.41

sx =

s

1n=

1.5

250L 0.21213 minute.mx = m = 25 minutes

s = 1.5

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μ = 6803

7088

x

Mean room and board (in dollars)

5600 6200 6800 7400 8000

Distributionof SampleMeans withn = 9

Study TipBefore you find probabilitiesfor intervals of the samplemean use the CentralLimit Theorem todetermine the meanand the standarddeviation of thesampling distribution ofthe sample means. Thatis, calculate and s x .m x

x,

In Example 5, you can use a TI-83/84 to find the probabilityautomatically.

E X A M P L E 5

Finding Probabilities for Sampling DistributionsThe mean room and board expense per year of four-year colleges is $6803.Yourandomly select 9 four-year colleges. What is the probability that the meanroom and board is less than $7088? Assume that the room and board expensesare normally distributed, with a standard deviation of $1125. (Source: NationalCenter for Education Statistics)

Solution Because the population is normally distributed, you can use theCentral Limit Theorem to conclude that the distribution of sample means isnormally distributed, with a mean of $6803 and a standard deviation of $375.

and

The graph of this distribution is shown at the left. The area to the left of $7088is shaded. The z-score that corresponds to $7088 is

So, the probability that the mean room and board expense is less than $7088 is

Interpretation So, 77.64% of such samples with will have a mean lessthan $7088 and 22.36% of these sample means will lie outside this interval.

� Try It Yourself 5The average sales price of a single-family house in the United States is$306,258. You randomly select 12 single-family houses. What is the probabilitythat the mean sales price is more than $280,000? Assume that the sales pricesare normally distributed with a standard deviation of $44,000. (Source: FederalHousing Finance Board)

a. Use the Central Limit Theorem to find and and sketch the samplingdistribution of the sample means.

b. Find the z-score that corresponds to $280,000.c. Find the cumulative area that corresponds to the z-score and calculate the

probability. Answer: Page A41

The Central Limit Theorem can also be used to investigate unusual events.An unusual event is one that occurs with a probability of less than 5%.

x =

sxmx

n = 9

P1x 6 70882 = P1z 6 0.762 = 0.7764.

z =

7088 - 6803

1125>29=

285375

= 0.76.

sx =

s

1n=

1125

29= 375mx = m = 6803

� Try It Yourself 4You randomly select 100 drivers ages 15 to 19 from Example 4. What is the probability that the mean time they spend driving each day is between 24.7 and 25.5 minutes? Use and minutes.

a. Use the Central Limit Theorem to find and and sketch the samplingdistribution of the sample means.

b. Find the z-scores that correspond to minutes and minutes.

c. Find the cumulative area that corresponds to each z-score and calculate theprobability. Answer: Page A41

x = 25.5x = 24.7

sxmx

s = 1.5m = 25

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Study TipTo find probabilities for individualmembers of a population with a normally distributed random variable x, use the formula

To find probabilities forthe mean of a samplesize n, use the formula

z =

x - m x

sx.

x

z =

x - m

s.

E X A M P L E 6

Finding Probabilities for x and A bank auditor claims that credit card balances are normally distributed, witha mean of $2870 and a standard deviation of $900.

1. What is the probability that a randomly selected credit card holder has acredit card balance less than $2500?

2. You randomly select 25 credit card holders.What is the probability that theirmean credit card balance is less than $2500?

3. Compare the probabilities from (1) and (2) and interpret your answer interms of the auditor’s claim.

Solution

1. In this case, you are asked to find the probability associated with a certainvalue of the random variable The -score that corresponds to is

So, the probability that the card holder has a balance less than $2500 is

2. Here, you are asked to find the probability associated with a sample meanThe -score that corresponds to is

So, the probability that the mean credit card balance of the 25 card holdersis less than $2500 is

3. Interpretation Although there is a 34% chance that an individual willhave a balance less than $2500, there is only a 2% chance that the mean ofa sample of 25 will have a balance less than $2500. Because there is only a2% chance that the mean of a sample of 25 will have a balance less than$2500, this is an unusual event. So, it is possible that the sample is unusual,or it is possible that the auditor’s claim that the mean is $2870 is incorrect.

� Try It Yourself 6A consumer price analyst claims that prices for sound-system receivers arenormally distributed, with a mean of $625 and a standard deviation of $150.(1) What is the probability that a randomly selected receiver costs less than$700? (2) You randomly select 10 receivers. What is the probability that theirmean cost is less than $700? (3) Compare these two probabilities.

a. Find the z-scores that correspond to x and b. Use the Standard Normal Table to find the probability associated with each

z-score.c. Compare the probabilities and interpret your answer. Answer: Page A41

x .

P1x 6 25002 = P1z 6 -2.062 = 0.0197.

L -2.06.=

-370180

=

2500 - 2870

900>225

z =

x - mx

sx=

x - m

s>1n

x = $2500zx.

= 0.3409.P1x 6 25002 = P1z 6 -0.412

=

2500 - 2870900

L -0.41.

z =

x - m

s

x = $2500zx.

x

41831S4_0504.qxd 11/5/07 12:43 PM Page 277



In Exercises 1–4, a population has a mean and a standard deviationFind the mean and standard deviation of a sampling distribution of

sample means with the given sample size

1. 2.

3. 4.

True or False? In Exercises 5–8, determine whether the statement is true orfalse. If it is false, rewrite it as a true statement.

5. As the size of a sample increases, the mean of the distribution of samplemeans increases.

6. As the size of a sample increases, the standard deviation of the distribution ofsample means increases.

7. A sampling distribution is normal only if the population is normal.

8. If the size of a sample is at least 30, you can use scores to determine theprobability that a sample mean falls in a given interval of the samplingdistribution.

Verifying Properties of Sampling Distributions In Exercises 9 and 10,find the mean and standard deviation of the population. List all samples (withreplacement) of the given size from that population. Find the mean and standarddeviation of the sampling distribution and compare them with the mean andstandard deviation of the population.

9. The number of movies that all four people in a family have seen in the pastmonth is 4, 2, 8, and 0. Use a sample size of 3.

10. Four people in a carpool paid the following amounts for textbooks thissemester: $120, $140, $180, and $220. Use a sample size of 2.

Graphical Analysis In Exercises 11 and 12, the graph of a populationdistribution is shown with its mean and standard deviation. Assume that a samplesize of 100 is drawn from each population. Decide which of the graphs labeled(a)–(c) would most closely resemble the sampling distribution of the samplemeans for each graph. Explain your reasoning.

11. The waiting time (in seconds) at a traffic signal during a red light

10

0.0050.0100.0150.0200.0250.0300.035

20

Time (in seconds)

Rel

ativ

e fr

eque

ncy

30 40 50

σ = 11.9

= 16.5μ

x

P(x)

z-

n = 1000n = 250

n = 100n = 50

n.s = 15.

m = 100

For Extra Help

5.4 EXERCISES

41831S4_0504.qxd 11/5/07 12:43 PM Page 278


(a) (b) (c)

12. The annual snowfall (in feet) for a central New York State county

(a) (b) (c)

Finding Probabilities In Exercises 13–16, the population mean and standarddeviation are given. Find the required probability and determine whether the givensample mean would be considered unusual. If convenient, use technology to findthe probability.

13. For a sample of find the probability of a sample mean being less than12.2 if and

14. For a sample of find the probability of a sample mean being greaterthan 12.2 if and

15. For a sample of find the probability of a sample mean being greaterthan 221 if and

16. For a sample of find the probability of a sample mean being less than12,750 or greater than 12,753 if and


Using the Central Limit Theorem In Exercises 17–22, use the Central LimitTheorem to find the mean and standard error of the mean of the indicatedsampling distribution. Then sketch a graph of the sampling distribution.

s = 1.7.m = 12,750n = 36,

s = 3.9.m = 220n = 75,

s = 0.95.m = 12n = 100,

s = 0.95.m = 12n = 36,

−2

0.4

0.8

1.2

1.6

2.0

4 620 8 10 12x

Snowfall (in feet)

Freq

uenc

y

f

σμx = 5.8

x = 2.3

2

0.3

0.6

0.9

1.2

1.5

1.8

4 6 8 10x

Snowfall (in feet)

Freq

uenc

y

fσ

μx = 5.8

x = 0.23

2

0.04

0.08

0.12

4 6 8 10x

Snowfall (in feet)

Rel

ativ

e fr

eque

ncy

P(x)

σ μx = 5.8x = 2.3

2

0.04

0.08

0.12

4

Snowfall (in feet)

Rel

ativ

e fr

eque

ncy

6 8 10

σ = 2.3

μ = 5.8

x

P(x)

0.2

0.1

0.3

10 20 30 40x

Time (in seconds)

Rel

ativ

e fr

eque

ncy

P(x)

σ

μx = 16.5

x = 1.19

10

0.0050.0100.0150.0200.0250.0300.035

20 30 40 50x

Time (in seconds)

Rel

ativ

e fr

eque

ncy

P(x)

σμx = 16.5

x = 11.9

100−10 20

0.02

0.01

0.03

30 40

σμ

x

Time (in seconds)

Rel

ativ

e fr

eque

ncy

P(x)

x = 16.5x = 11.9

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17. Heights of Trees The heights of fully grown sugar maple trees are normallydistributed, with a mean of 87.5 feet and a standard deviation of 6.25 feet.Random samples of size 12 are drawn from the population and the mean ofeach sample is determined.

18. Fly Eggs The number of eggs female house flies lay during their lifetimesis normally distributed, with a mean of 800 eggs and a standard deviationof 100 eggs. Random samples of size 15 are drawn from this population andthe mean of each sample is determined.

19. Digital Cameras The mean price of digital cameras at an electronics storeis $224, with a standard deviation of $8. Random samples of size 40 aredrawn from this population and the mean of each sample is determined.

20. Employees’ Ages The mean age of employees at a large corporation is 47.2years, with a standard deviation of 3.6 years. Random samples of size 36 aredrawn from this population and the mean of each sample is determined.

21. Red Meat Consumed The per capita consumption of red meat by people in the United States in a recent year was normally distributed, with a mean of 110 pounds and a standard deviation of 38.5 pounds. Random samples ofsize 20 are drawn from this population and the mean of each sample isdetermined. (Adapted from U.S. Department of Agriculture)

22. Soft Drinks The per capita consumption of soft drinks by people in theUnited States in a recent year was normally distributed, with a mean of 51.5gallons and a standard deviation of 17.1 gallons. Random samples of size 25are drawn from this population and the mean of each sample is determined.(Adapted from U.S. Department of Agriculture)

23. Repeat Exercise 17 for samples of size 24 and 36. What happens to the meanand standard deviation of the distribution of sample means as the size of thesample increases?

24. Repeat Exercise 18 for samples of size 30 and 45. What happens to the meanand to the standard deviation of the distribution of sample means as the sizeof the sample increases?

Finding Probabilities In Exercises 25–30, find the probabilities. If convenient,use technology to find the probabilities.

25. Plumber Salaries The population mean annual salary for plumbers is$46,700. A random sample of 42 plumbers is drawn from this population.What is the probability that the mean salary of the sample is less than$44,000? Assume (Adapted from Salary.com)

26. Nurse Salaries The population mean annual salary for registered nurses is $59,100 A random sample of 35 registered nurses is selected from thispopulation. What is the probability that the mean annual salary of the sample is less than $55,000? Assume (Adapted from Salary.com)

27. Gas Prices: New England During a certain week the mean price of gasolinein the New England region was $2.818 per gallon.A random sample of 32 gasstations is drawn from this population. What is the probability that the meanprice for the sample was between $2.768 and $2.918 that week? Assume

(Adapted from Energy Information Administration)

28. Gas Prices: California During a certain week the mean price of gasoline inCalifornia was $3.305 per gallon.A random sample of 38 gas stations is drawnfrom this population. What is the probability that the mean price for thesample was between $3.310 and $3.320 that week? Assume (Adapted from Energy Information Administration)

s = $0.049.

s = $0.045.

s = $1700.

s = $5600.

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29. Heights of Women The mean height of women in the United States (ages20–29) is 64.1 inches. A random sample of 60 women in this age group isselected. What is the probability that the mean height for the sample isgreater than 66 inches? Assume inches. (Source: National Center forHealth Statistics)

30. Heights of Men The mean height of men in the United States (ages 20 –29)is 69.6 inches. A random sample of 60 men in this age group is selected. Whatis the probability that the mean height for the sample is greater than 70 inch-es? Assume inches. (Source: National Center for Health Statistics)

31. Which Is More Likely? Assume that the heights given in Exercise 29 arenormally distributed. Are you more likely to randomly select 1 woman witha height less than 70 inches or are you more likely to select a sample of 20women with a mean height less than 70 inches? Explain.

32. Which Is More Likely? Assume that the heights given in Exercise 30 arenormally distributed. Are you more likely to randomly select 1 man with aheight less than 65 inches or are you more likely to select a sample of 15 men with a mean height less than 65 inches? Explain.

33. Make a Decision A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ouncesand a standard deviation of 0.20 ounce. You randomly select 40 cans andcarefully measure the contents. The sample mean of the cans is 127.9 ounces.Does the machine need to be reset? Explain your reasoning.

34. Make a Decision A machine used to fill half-gallon-sized milk containers isregulated so that the amount of milk dispensed has a mean of 64 ounces anda standard deviation of 0.11 ounce. You randomly select 40 containers andcarefully measure the contents. The sample mean of the containers is 64.05 ounces. Does the machine need to be reset? Explain your reasoning.

35. Lumber Cutter Your lumber company has bought a machine thatautomatically cuts lumber. The seller of the machine claims that the machinecuts lumber to a mean length of 8 feet (96 inches) with a standard deviationof 0.5 inch. Assume the lengths are normally distributed. You randomlyselect 40 boards and find that the mean length is 96.25 inches.

(a) Assuming the seller’s claim is correct, what is the probability the meanof the sample is 96.25 inches or more?

(b) Using your answer from part (a), what do you think of the seller’s claim?

(c) Would it be unusual to have an individual board with a length of 96.25 inches? Why or why not?

36. Ice Cream Carton Weights A manufacturer claims that the mean weight ofits ice cream cartons is 10 ounces with a standard deviation of 0.5 ounce.Assume the weights are normally distributed. You test 25 cartons and findtheir mean weight is 10.21 ounces.

(a) Assuming the manufacturer’s claim is correct, what is the probability themean of the sample is 10.21 ounces or more?

(b) Using your answer from part (a), what do you think of themanufacturer’s claim?

(c) Would it be unusual to have an individual carton with a weight of 10.21 ounces? Why or why not?

s = 3.0

s = 2.71

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37. Life of Tires A manufacturer claims that the life span of its tires is 50,000 miles. You work for a consumer protection agency and you are testingthis manufacturer’s tires. Assume the life spans of the tires are normallydistributed. You select 100 tires at random and test them. The mean life spanis 49,721 miles. Assume miles.

(a) Assuming the manufacturer’s claim is correct, what is the probability themean of the sample is 49,721 miles or less?

(b) Using your answer from part (a), what do you think of the manufacturer’sclaim?

(c) Would it be unusual to have an individual tire with a life span of 49,721 miles? Why or why not?

38. Brake Pads A brake pad manufacturer claims its brake pads will last for38,000 miles. You work for a consumer protection agency and you are testingthis manufacturer’s brake pads. Assume the life spans of the brake pads arenormally distributed. You randomly select 50 brake pads. In your tests, themean life of the brake pads is 37,650 miles. Assume miles.

(a) Assuming the manufacturer’s claim is correct, what is the probability themean of the sample is 37,650 miles or less?

(b) Using your answer from part (a), what do you think of the manufacturer’sclaim?

(c) Would it be unusual to have an individual brake pad last for 37,650 miles? Why or why not?

� Extending Concepts39. SAT Scores The average math SAT score is 518 with a standard deviation

of 115. A particular high school claims that its students have unusually highmath SAT scores. A random sample of 50 students from this school wasselected, and the mean math SAT score was 530. Is the high school justifiedin its claim? Explain. (Source: College Board Online)

40. Machine Calibrations A machine in a manufacturing plant is calibrated toproduce a bolt that has a mean diameter of 4 inches and a standard deviationof 0.5 inch. An engineer takes a random sample of 100 bolts from thismachine and finds the mean diameter is 4.2 inches. What are some possibleconsequences from these findings?

Finite Correction Factor The formula for the standard error of the mean

given in the Central Limit Theorem is based on an assumption that the populationhas infinitely many members. This is the case whenever sampling is done withreplacement (each member is put back after it is selected) because the samplingprocess could be continued indefinitely. The formula is also valid if the sample sizeis small in comparison to the population. However, when sampling is done withoutreplacement and the sample size n is more than 5% of the finite population of size

there is a finite number of possible samples. A finite correction factor,

A

N - n

N - 1

Nan

N7 0.05b ,

sx =

s

1n

s = 1000

s = 800

41831S4_0504.qxd 11/5/07 12:43 PM Page 282


should be used to adjust the standard error. The sampling distribution of thesample means will be normal with a mean equal to the population mean, and thestandard error of the mean will be

In Exercises 41 and 42, determine if the finite correction factor should be used.If so, use it in your calculations when you find the probability.

41. Gas Prices In a sample of 800 gas stations, the mean price for regulargasoline at the pump was $2.876 per gallon and the standard deviation was$0.009 per gallon. A random sample of size 55 is drawn from this population.What is the probability that the mean price per gallon is less than $2.871?(Adapted from U.S. Department of Energy)

42. Old Faithful In a sample of 500 eruptions of the Old Faithful geyser atYellowstone National Park, the mean duration of the eruptions was 3.32 minutes and the standard deviation was 1.09 minutes. A random sampleof size 30 is drawn from this population. What is the probability that themean duration of eruptions is between 2.5 minutes and 4 minutes? (Adaptedfrom Yellowstone National Park)

Sampling Distribution of Sample Proportions The sample mean is notthe only statistic with a sampling distribution. Every sample statistic, such as thesample median, the sample standard deviation, and the sample proportion, has asampling distribution. For a random sample of size n, the sample proportion is thenumber of individuals in the sample with a specified characteristic divided by thesample size. The sampling distribution of sample proportions is the distributionformed when sample proportions of size n are repeatedly taken from a populationwhere the probability of an individual with a specified characteristic is p.

In Exercises 43–45, suppose three births are randomly selected. There are twoequally possible outcomes for each birth, a boy (b) or a girl (g). The number ofboys can equal 0, 1, 2, or 3. These correspond to sample proportions of 0,and 1.

43. List the eight possible samples when randomly selecting three births. Forinstance, let bbb represents a sample of three boys. Make a table that showseach sample, the number of boys in each sample, and the proportion of boysin each sample.

44. Use the table from Exercise 43 to construct the sampling distribution of thesample proportion of boys from three births. What do you notice about thespread of the histogram as compared to the binomial probability distributionfor the number of boys in each sample?

45. Let represent a boy and represent a girl. Using these values, findthe sample mean for each sample. What do you notice?

46. Construct a sampling distribution of the sample proportion of boys from fourbirths.

47. Heart Transplants About 75% of all female heart transplant patients willsurvive for at least 3 years. Ninety female heart transplant patients arerandomly selected. What is the probability that the sample proportionsurviving for at least 3 years will be less than 70%? Assume the samplingdistribution of sample proportions is a normal distribution. The mean of thesample proportion is equal to the population proportion and the standard

deviation is equal to (Source: American Heart Association)A

pq

n.

p

x = 0x = 1

2>3,1>3,

sx =

s

1n A

N - n

N - 1.

41831S4_0504.qxd 11/5/07 12:43 PM Page 283

284

The sampling distributions applet allows you to investigate sampling distributionsby repeatedly taking samples from a population.The top plot displays the distributionof a population. Several options are available for the population distribution(Uniform, Bell-shaped, Skewed, Binary, and Custom). When SAMPLE is clicked,

random samples of size will be repeatedly selected from the population. Thesample statistics specified in the bottom two plots will be updated for each sample.If is set to 1 and is less than or equal to 50, the display will show, in an animatedfashion, the points selected from the population dropping into the second plot andthe corresponding summary statistic values dropping into the third and fourthplots. Click RESET to stop an animation and clear existing results. Summary statistics for each plot are shown in the panel to the left of the plot.

� ExploreStep 1 Specify a distribution.Step 2 Specify a value for Step 3 Specify a value for Step 4 Specify what to display in the bottom two graphs.Step 5 Click SAMPLE to generate the sampling distributions.

� Draw Conclusions1. Run the simulation using and for a uniform, a bell-shaped,

and a skewed distribution. What is the mean of the sampling distribution ofthe sample means for each distribution? For each distribution, is this whatyou would expect?

2. Run the simulation using and for a bell-shaped distribution.What is the standard deviation of the sampling distribution of the samplemeans? According to the formula, what should the standard deviation of thesampling distribution of the sample means be? Is this what you would expect?

N = 10n = 50

N = 10n = 30

N.n.

nN

nN

Sampling Distributions

5.4

Mean

Median

Std. Dev.

Mean

0

0 50

2

4

6Sample data

Population (can be changed with mouse)

Median

Std. Dev.

Mean

Median

N

Std. Dev.

Mean

Median

N

Std. Dev.

25

25

14.43380 50

0

0 50

2

4

6Sample Means

0

0 50

2

4

6Sample Medians

Sample

1

2

Uniform

Mean

n =

N =

Median

Reset

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S E C T I O N 5 . 5 NORMAL APPROXIMATIONS TO BINOMIAL DISTRIBUTIONS 285

� Approximating a Binomial DistributionIn Section 4.2, you learned how to find binomial probabilities. For instance, if asurgical procedure has an 85% chance of success and a doctor performs theprocedure on 10 patients, it is easy to find the probability of exactly twosuccessful surgeries.

But what if the doctor performs the surgical procedure on 150 patients andyou want to find the probability of fewer than 100 successful surgeries? To dothis using the techniques described in Section 4.2, you would have to use thebinomial formula 100 times and find the sum of the resulting probabilities. Thisapproach is not practical, of course. A better approach is to use a normaldistribution to approximate the binomial distribution.

To see why this result is valid, look at the following binomial distributions forand 10, 25, and 50. Notice that as n increases, the

histogram approaches a normal curve.

20 4 6 8 10 12 14 16 18 20 22 24

0.06

0.04

0.02

0.08

0.10

0.12

x

P(x)

n = 50np = 12.5nq = 37.5

20 4 6 8 10 12 14 16 18x

0.020.040.060.080.100.120.140.160.18

P(x)

n = 25np = 6.25nq = 18.75

10 2 3 4 5 6 7 8 9 10x

0.05

0.10

0.15

0.20

0.25

0.30 n = 10np = 2.5nq = 7.5

P(x)

10 2 3 4x

0.10

0.20

0.30

0.40

0.05

0.15

0.25

0.35

0.45 n = 4np = 1nq = 3

P(x)

n =n =n =n = 4,p = 0.25

Approximating a Binomial Distribution � Correction forContinuity � Approximating Binomial Probabilities

285

5.5 Normal Approximations to Binomial DistributionsWhat You

SHOULD LEARN� How to decide when the

normal distribution canapproximate the binomialdistribution

� How to find thecorrection for continuity

� How to use the normaldistribution toapproximatebinomial probabilities

N O R M A L A P P R O X I M AT I O N T O A B I N O M I A LD I S T R I B U T I O N

If and then the binomial random variable x is approximatelynormally distributed, with mean

and standard deviation

s = 1npq.

m = np

nq Ú 5,np Ú 5

Study TipProperties of a binomialexperiment

• n independent trials

• Two possible outcomes: success or failure

• Probability of success isp; probability of failureis

• p is constant for each trial

1 - p = q

41831S4_0505.qxd 11/5/07 12:41 PM Page 285


E X A M P L E 1

Approximating the Binomial DistributionTwo binomial experiments are listed. Decide whether you can use the normaldistribution to approximate x, the number of people who reply yes. If you can,find the mean and standard deviation. If you cannot, explain why. (Source:Opinion Research Corporation)

1. Fifty-one percent of adults in the United States whose New Year’sresolution was to exercise more achieved their resolution. You randomlyselect 65 adults in the United States whose resolution was to exercise moreand ask each if he or she achieved that resolution.

2. Fifteen percent of adults in the United States do not make New Year’sresolutions.You randomly select 15 adults in the United States and ask eachif he or she made a New Year’s resolution.

Solution

1. In this binomial experiment, and So,

and

Because and are greater than 5, you can use the normal distribution with

and

to approximate the distribution of

2. In this binomial experiment, and So,

and

Because you cannot use the normal distribution to approximate thedistribution of

� Try It Yourself 1Consider the following binomial experiment. Decide whether you can use thenormal distribution to approximate x, the number of people who reply yes. Ifyou can, find the mean and standard deviation. If you cannot, explain why.(Source: Opinion Research Corporation)

Over the past 5 years, 80% of adults in the United States have madeand kept 1 or more New Year’s resolutions. You randomly select 70adults in the United States who made a New Year’s resolution in thepast 5 years and ask each if he or she kept at least 1 resolution.

a. Identify n, p, and q.b. Find the products np and nq.c. Decide whether you can use the normal distribution to approximate x.d. Find the mean and standard deviation if appropriate.

Answer: Page A41s ,m

x.np 6 5,

np = 115210.852 = 12.75.

np = 115210.152 = 2.25.

q = 0.85.p = 0.15,n = 15,

x.

s = 1npq = 265 # 0.51 # 0.49 L 4.03

m = 33.15nqnp

nq = 165210.492 = 31.85.

np = 165210.512 = 33.15

q = 0.49.p = 0.51,n = 65,

41831S4_0505.qxd 11/5/07 12:41 PM Page 286

� Correction for ContinuityThe binomial distribution is discrete and can be represented by a probabilityhistogram. To calculate exact binomial probabilities, you can use the binomialformula for each value of x and add the results. Geometrically, this correspondsto adding the areas of bars in the probability histogram. Remember that eachbar has a width of one unit and x is the midpoint of the interval.

When you use a continuous normal distribution to approximate a binomialprobability, you need to move 0.5 unit to the left and right of the midpoint toinclude all possible x-values in the interval. When you do this, you are making acorrection for continuity.

c

P(x = c) P(c − 0.5 < x < c + 0.5)

c + 0.5c − 0.5x

cx

Normalapproximation

Exact binomialprobability


Study TipTo use a correction forcontinuity, simply subtract 0.5 from the lowest value and add 0.5 to thehighest.

E X A M P L E 2

Using a Correction for ContinuityUse a correction for continuity to convert each of the following binomialintervals to a normal distribution interval.

1. The probability of getting between 270 and 310 successes, inclusive

2. The probability of at least 158 successes

3. The probability of getting less than 63 successes

Solution

1. The discrete midpoint values are 270, 271, , 310. The correspondinginterval for the continuous normal distribution is

2. The discrete midpoint values are 158, 159, 160, . The correspondinginterval for the continuous normal distribution is

3. The discrete midpoint values are , 60, 61, 62. The corresponding intervalfor the continuous normal distribution is

� Try It Yourself 2Use a correction for continuity to convert each of the following binomialintervals to a normal distribution interval.

1. The probability of getting between 57 and 83 successes, inclusive2. The probability of getting at most 54 successes

a. List the midpoint values for the binomial probability.b. Use a correction for continuity to write the normal distribution interval.

Answer: Page A42

x 6 62.5.

Á

x 7 157.5.

Á

269.5 6 x 6 310.5.

Á

41831S4_0505.qxd 11/5/07 12:41 PM Page 287

� Approximating Binomial Probabilities


x

Number responding yes

μ = 33.15

39.5

21 27 33 39 4524 30 36 42

In a survey of U.S.adults, peoplewere asked if thelaw should allow

doctors to aid dying patientswho want to end their lives. Theresults of the survey are shownin the following pie chart. (Adaptedfrom Pew Research Center)

Assume that this survey is a true indication of the proportion of the populationwho believe in assisted deathfor terminally ill patients.If you sampled 50 adults at random, what is the probabilitythat between 21 and 25,inclusive, would believe inassisted death?

Do notbelieve in

assisted death54%

Believe inassisted death

46%

G U I D E L I N E SUsing the Normal Distribution to Approximate Binomial Probabilities

In Words In Symbols

1. Verify that the binomial distribution Specify and applies.

2. Determine if you can use the normal Is distribution to approximate the Is binomial variable.

3. Find the mean and standard deviationfor the distribution.

4. Apply the appropriate continuity Add or subtractcorrection. Shade the corresponding 0.5 from endpoints.area under the normal curve.

5. Find the corresponding -score(s).

6. Find the probability. Use the StandardNormal Table.

z =

x - m

sz

s = 1npqs

m = npm

nq Ú 5?x ,np Ú 5?

q .p ,n ,

G U I D E L I N E S

E X A M P L E 3

Approximating a Binomial ProbabilityFifty-one percent of adults in the United States whose New Year’s resolutionwas to exercise more achieved their resolution. You randomly select 65 adultsin the United States whose resolution was to exercise more and ask each if heor she achieved that resolution. What is the probability that fewer than 40 ofthem respond yes? (Source: Opinion Research Corporation)

Solution From Example 1, you know that you can use a normal distributionwith and to approximate the binomial distribution.Remember to apply the continuity correction for the value of In thebinomial distribution, the possible midpoint values for “fewer than 40” are

37, 38, 39.

To use the normal distribution, add 0.5 to the right-hand boundary 39 to getThe graph at the left shows a normal curve with andand a shaded area to the left of 39.5. The z-score that corresponds

to is

Using the Standard Normal Table,

Interpretation The probability that fewer than forty people respond yes isapproximately 0.9429, or about 94%.

P1z 6 1.582 = 0.9429.

L 1.58.

z =

39.5 - 33.15

265 # 0.51 # 0.49

x = 39.5s L 4.03

m = 33.15x = 39.5.

Á

x.s L 4.03m = 33.15

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Study TipIn a discrete distribution, there is a difference between and This is truebecause the probability that is exactly is notzero. In a continuous distribution, however,there is no differencebetween and

because the probability that is exactly

is zero.cx

P1x 7 c2P1x Ú c2

cx

P1x 7 c2.P1x Ú c2

� Try It Yourself 3Over the past 5 years, 80% of adults in the United States have made and kept1 or more New Year’s resolutions. You randomly select 70 adults in the UnitedStates who made a New Year’s resolution in the past 5 years and ask each if heor she kept at least 1 resolution. What is the probability that more than 50respond yes? (See Try it Yourself 1.) (Source: Opinion Research Corporation)

a. Determine whether you can use the normal distribution to approximate thebinomial variable (see part c of Try It Yourself 1).

b. Find the mean and the standard deviation for the distribution (see part dof Try It Yourself 1).

c. Apply the appropriate continuity correction and sketch a graph.d. Find the corresponding z-score.e. Use the Standard Normal Table to find the area to the left of z and calculate

the probability. Answer: Page A42

sm

E X A M P L E 4

Approximating a Binomial ProbabilityThirty-eight percent of people in the United States admit that they snoop inother people’s medicine cabinets.You randomly select 200 people in the UnitedStates and ask each if he or she snoops in other people’s medicine cabinets.What is the probability that at least 70 will say yes? (Source: USA TODAY)

Solution Because and thebinomial variable x is approximately normally distributed with

and

Using the correction for continuity, youcan rewrite the discrete probability

as the continuous probabilityThe graph shows a

normal curve with and and a shaded area to the right of 69.5.The z-score that corresponds to 69.5 is

So, the probability that at least 70 will say yes is

� Try It Yourself 4In Example 4, what is the probability that at most 85 people will say yes?

a. Determine whether you can use the normal distribution to approximate thebinomial variable (see Example 4).

b. Find the mean and the standard deviation for the distribution (seeExample 4).

c. Apply a continuity correction to rewrite and sketch a graph.d. Find the corresponding z-score.e. Use the Standard Normal Table to find the area to the left of z and calculate

the probability. Answer: Page A42

P1x … 852

sm

= 1 - 0.1711 = 0.8289. = 1 - P1z … -0.952

P1x Ú 69.52 = P1z Ú -0.952

L -0.95.z =

169.5 - 762

6.86

s = 6.86m = 76P1x Ú 69.52 .P1x Ú 702

s = 2200 # 0.38 # 0.62 L 6.86.m = np = 76

nq = 200 # 0.62 = 124,np = 200 # 0.38 = 76

55 60 65 70 75 8580 90 95 100x

69.5

μ = 76


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E X A M P L E 5

Approximating a Binomial ProbabilityA survey reports that 86% of Internet users use Windows® Internet Explorer®

as their browser. You randomly select 200 Internet users and ask each whetherhe or she uses Internet Explorer as his or her browser. What is the probabilitythat exactly 176 will say yes? (Source: OneStat.com)

Solution Because and thebinomial variable x is approximately normally distributed with

and

Using the correction for continuity, you can rewrite the discrete probabilityas the continuous probability The

following graph shows a normal curve with and and ashaded area between 175.5 and 176.5.

The z-scores that correspond to 175.5 and 176.5 are

and

So, the probability that exactly 176 Internet users will say they use InternetExplorer is

Interpretation The probability that exactly 176 of the Internet users will saythey use Internet Explorer is approximately 0.0601, or about 6%.

� Try It Yourself 5In Example 5, what is the probability that exactly 170 people will say yes?

a. Determine whether you can use the normal distribution to approximate thebinomial variable (see Example 5).

b. Find the mean and the standard deviation for the distribution (seeExample 5).

c. Apply a continuity correction to rewrite and sketch a graph.d. Find the corresponding z-scores.e. Use the Standard Normal Table to find the area to the left of each z-score

and calculate the probability. Answer: Page A42

P1x = 1702

sm

= 0.0601.

= 0.8212 - 0.7611

= P1z 6 0.922 - P1z 6 0.712

P1175.5 6 x 6 176.52 = P10.71 6 z 6 0.922

z2 =

176.5 - 172

2200 # 0.86 # 0.14.z1 =

175.5 - 172

2200 # 0.86 # 0.14

x

μ = 172


175.5 176.5

158 162 166 170 174 178 182 186

s = 4.91m = 172P1175.5 6 x 6 176.52.P1x = 1762

s = 2npq = 2200 # 0.86 # 0.14 L 4.91.m = np = 172

nq = 200 # 0.14 = 28,np = 200 # 0.86 = 172

The approximation in Example 5is only slightly less than the exact probability found using the binompdf( command on a TI-83/84.

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In Exercises 1–4, the sample size n, probability of success p, and probability offailure q are given for a binomial experiment. Decide whether you can use thenormal distribution to approximate the random variable x.

1. 2.

3. 4.

Approximating a Binomial Distribution In Exercises 5–8, a binomialexperiment is given. Decide whether you can use the normal distribution toapproximate the binomial distribution. If you can, find the mean and standarddeviation. If you cannot, explain why.

5. House Contract A survey of U.S. adults found that 85% read every word orat least enough to understand a contract for buying or selling a home beforesigning. You ask 10 adults selected at random if he or she reads every wordor at least enough to understand a contract for buying or selling a homebefore signing. (Source: FindLaw.com)

6. Organ Donors A survey of U.S. adults found that 63% would want theirorgans transplanted into a patient who needs them if they were killed in anaccident. You randomly select 20 adults and ask each if he or she would wanttheir organs transplanted into a patient who needs them if they were killedin an accident. (Source: USA TODAY)

7. Prostate Cancer In a recent year, the American Cancer Society said thatthe five-year survival rate for all men diagnosed with prostate cancer was99%. You randomly select 10 men who were diagnosed with prostate cancerand calculate their five-year survival rate. (Source: American Cancer Society)

8. Work Weeks A survey of workers in the United States found that 8.6%work fewer than 40 hours per week. You randomly select 30 workers in theUnited States and ask each if he or she works fewer than 40 hours per week.

In Exercises 9–12, match the binomial probability with the correct statement.

Probability Statement

9. (a) P(there are fewer than 65 successes)

10. (b) P(there are at most 65 successes)

11. (c) P(there are more than 65 successes)

12. (d) P(there are at least 65 successes)

In Exercises 13–16, use the correction for continuity and match the binomialprobability statement with the corresponding normal distribution statement.

Binomial Probability Normal Probability

13. (a)

14. (b)

15. (c)

16. (d) P1x Ú 108.52P1x 6 1092

P1x … 109.52P1x … 1092

P1x 6 108.52P1x Ú 1092

P1x 7 109.52P1x 7 1092

P1x 7 652

P1x 6 652

P1x … 652

P1x Ú 652

q = 0.35p = 0.65,n = 20,q = 0.10p = 0.90,n = 18,

q = 0.30p = 0.70,n = 15,q = 0.15p = 0.85,n = 24,

For Extra Help

5.5 EXERCISES

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Graphical Analysis In Exercises 17 and 18, write the binomial probability andthe normal probability for the shaded region of the graph. Find the value of eachprobability and compare the results.

17. 18.

Approximating Binomial Probabilities In Exercises 19–24, decidewhether you can use the normal distribution to approximate the binomialdistribution. If you can, use the normal distribution to approximate the indicatedprobabilities and sketch their graphs. If you cannot, explain why and use thebinomial distribution to find the indicated probabilities.

19. Blood Type O Seven percent of people in the United States have type blood. You randomly select 30 people in the United States and ask them iftheir blood type is (Source: American Association of Blood Banks)

(a) Find the probability that exactly 10 people say they have blood.

(b) Find the probability that at least 10 people say they have blood.

(c) Find the probability that fewer than 10 people say they have blood.

(d) A blood drive would like to get at least five donors with blood.There are 100 donors. What is the probability that there will not beenough blood donors?

20. Blood Type A Thirty-four percent of people in the United States havetype blood. You randomly select 32 people in the United States and askthem if their blood type is (Source: American Association of Blood Banks)

(a) Find the probability that exactly 12 people say they have blood.

(b) Find the probability that at least 12 people say they have blood.

(c) Find the probability that fewer than 12 people say they have blood.

(d) A blood drive would like to get at least 60 donors with blood. Thereare 150 donors. What is the probability that there will not be enough blood donors?

21. Public Transportation Five percent of workers in the United States usepublic transportation to get to work.You randomly select 250 workers and askthem if they use public transportation to get to work. (Source: U.S. CensusBureau)

(a) Find the probability that exactly 16 workers will say yes.

(b) Find the probability that at least 9 workers will say yes.

(c) Find the probability that fewer than 16 workers will say yes.

(d) A transit authority offers discount rates to companies that have at least30 employees who use public transportation to get to work. There are500 employees in a company. What is the probability that the companywill not get the discount?

A+

A+

A+

A+

A+

A+.A+

�

O-

O-

O-

O-

O-

O-.

O-�

0 6 8 10 122 4x

n = 12p = 0.5

0.04

0.08

0.12

0.16

0.20

0.24

P(x)

0 6 8 10 12 14 162 4x

n = 16p = 0.4

0.04

0.08

0.12

0.16

0.20

0.24

P(x)

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22. College Graduates Thirty-two percent of workers in the United States arecollege graduates. You randomly select 50 workers and ask each if he or sheis a college graduate. (Source: U.S. Bureau of Labor Statistics)

(a) Find the probability that exactly 12 workers are college graduates.

(b) Find the probability that at least 14 workers are college graduates.

(c) Find the probability that fewer than 18 workers are college graduates.

(d) A committee is looking for 30 working college graduates to volunteer ata career fair. The committee randomly selects 150 workers. What is theprobability that there will not be enough college graduates?

23. Favorite Cookie Fifty-two percent of adults say chocolate chip is theirfavorite cookie. You randomly select 40 adults and ask each if chocolate chipis his or her favorite cookie. (Source: Wearever)

(a) Find the probability that at most 23 people say chocolate chip is theirfavorite cookie.

(b) Find the probability that at least 18 people say chocolate chip is theirfavorite cookie.

(c) Find the probability that more than 20 people say chocolate chip is theirfavorite cookie.

(d) A community bake sale has prepared 350 chocolate chip cookies.The bake sale attracts 650 customers, and they each buy their favoritecookie. What is the probability there will not be enough chocolate chipcookies?

24. Long Work Weeks A survey of workers in the United States found that2.9% work more than 70 hours per week. You randomly select 10 workers inthe U.S. and ask each if he or she works more than 70 hours per week.

(a) Find the probability that at most 3 people say they work more than 70hours per week.

(b) Find the probability that at least 1 person says he or she works more than70 hours per week.

(c) Find the probability that more than 2 people say they work more than 70hours per week.

(d) A large company is concerned about overworked employees who workmore than 70 hours per week. The company randomly selects 50employees. What is the probability there will be no employee workingmore than 70 hours?

25. Bigger Home A survey of homeowners in the United States found that 24% feel their home is too small for their family. You randomly select 25 homeowners and ask them if they feel their home is too small for theirfamily.

(a) Verify that the normal distribution can be used to approximate thebinomial distribution.

(b) Find the probability that more than eight homeowners say their home istoo small for their family.

(c) Is it unusual for 8 out of 25 homeowners to say their home is too small?Why or why not?

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26. Driving to Work A survey of workers in the United States found that 80%rely on their own vehicle to get to work. You randomly select 40 workers andask them if they rely on their own vehicle to get to work.


(b) Find the probability that at most 26 workers say they rely on their ownvehicle to get to work.

(c) Is it unusual for 26 out of 40 workers to say they rely on their own vehicle to get to work? Why or why not?

� Extending Concepts

Getting Physical In Exercises 27 and 28, use the following information.The graph shows the results of a survey of adults in the United States ages 33 to 51who were asked if they participated in a sport. Seventy percent of adults said theyregularly participate in at least one sport, and they gave their favorite sport.

27. You randomly select 250 people in the United States ages 33 to 51 and askeach if he or she regularly participates in at least one sport. You find that60% say no. How likely is this result? Do you think the sample is a good one?Explain your reasoning.

28. You randomly select 300 people in the United States ages 33 to 51 and askeach if he or she regularly participates in at least one sport. Of the 200 whosay yes, 9% say they participate in hiking. How likely is this result? Is thesample a good one? Explain your reasoning.

Testing a Drug In Exercises 29 and 30, use the following information. A drugmanufacturer claims that a drug cures a rare skin disease 75% of the time. Theclaim is checked by testing the drug on 100 patients. If at least 70 patients are cured,the claim will be accepted.

29. Find the probability that the claim will be rejected assuming that themanufacturer’s claim is true.

30. Find the probability that the claim will be accepted assuming that the actual probability that the drug cures the skin disease is 65%.

Howadults get physical

16%

12%

10%

11%

9%

6%

4%

2%

Swimming

(tie) Bicycling, golf

Hiking

(tie) Softball, walking

Fishing

Tennis

(tie) Bowling, running

Aerobics

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UsesNormal Distributions Normal distributions can be used to describe many real-life situations and are widely used in the fields of science, business, andpsychology. They are the most important probability distributions in statistics and can be used to approximate other distributions, such as discrete binomialdistributions.

The most incredible application of the normal distribution lies in the CentralLimit Theorem. This theorem states that no matter what type of distribution apopulation may have, as long as the sample size is at least 30, the distribution ofsample means will be approximately normal. If the population is itself normal, thenthe distribution of sample means will be normal no matter how small the sample is.

The normal distribution is essential to sampling theory. Sampling theoryforms the basis of statistical inference, which you will begin to study in the nextchapter.

AbusesUnusual Events Suppose a population is normally distributed with a mean of 100and standard deviation of 15. It would not be unusual for an individual value takenfrom this population to be 115 or more. In fact, this will happen almost 16% of thetime. It would be, however, highly unusual to take random samples of 100 valuesfrom that population and obtain a sample with a mean of 115 or more. Because thepopulation is normally distributed, the mean of the sample distribution will be 100,and the standard deviation will be 1.5. A mean of 115 lies 10 standard deviationsabove the mean. This would be an extremely unusual event. When an event thisunusual occurs, it is a good idea to question the original claimed value of the mean.

Although normal distributions are common in many populations, people try tomake non-normal statistics fit the normal distribution. The statistics used fornormal distributions are often inappropriate when the distribution is obviouslynon-normal.

� EXERCISES1. Is It Unusual? A population is normally distributed with a mean of 100 and

a standard deviation of 15. Determine if the following event is unusual. Explainyour reasoning.

a. mean of a sample of 3 is 115 or more

b. mean of a sample of 20 is 105 or more

2. Find the Error The mean age of students at a high school is 16.5 with astandard deviation of 0.7. You use the Standard Normal Table to help youdetermine that the probability of selecting one student at random and findinghis or her age to be more than 17.5 years is about 8%. What is the error in this problem?

3. Give an example of a distribution that might be non-normal.

Uses & AbusesSta

tisti

cs i

n t

he R

eal

Wo

rld

295

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Section 5.1

� How to interpret graphs of normal probability distributions 1, 2 1, 2

� How to find and interpret z-scores

z =

x - m

s

3 3, 4

� How to find areas under the standard normal curve 4–6 5–16

Section 5.2

� How to find probabilities for normally distributed variables 1–3 17–24

Section 5.3

� How to find a z-score given the area under the normal curve 1, 2 25–30

� How to transform a z-score to an x-value

x = m + zs

3 31, 32

� How to find a specific data value of a normal distribution given the probability

4, 5 33–36

Section 5.4

� How to find sampling distributions and verify their properties 1 37, 38

� How to interpret the Central Limit Theorem

sx =

s

1nmx = m,

2, 3 39, 40

� How to apply the Central Limit Theorem to find the probability of a sample mean

4–6 41–46

Section 5.5

� How to decide when the normal distribution can approximate the binomialdistribution

s = 1npqm = np,

1 47, 48

� How to find the correction for continuity 2 49–52

� How to use the normal distribution to approximate binomial probabilities 3–5 53, 54

EXAMPLE(S)

5 CHAPTER SUMMARYREVIEW

EXERCISESWhat did you learn?

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REVIEW EXERCISES 297

Section 5.1

In Exercises 1 and 2, use the graph to estimate and

1. 2.

In Exercises 3 and 4, use the following information and standard scores toinvestigate observations about a normal population. A batch of 2500 resistors isnormally distributed, with a mean resistance of 1.5 ohms and a standard deviationof 0.08 ohm. Four resistors are randomly selected and tested.Their resistances weremeasured at 1.32, 1.54, 1.66, and 1.78 ohms.

3. How many standard deviations from the mean are these observations?

4. Are there any unusual observations?

In Exercises 5–16, use the Standard Normal Table to find the indicated area underthe standard normal curve. If convenient, use technology to find the area.

5. To the left of

6. To the left of

7. To the left of

8. To the left of

9. To the right of

10. To the right of

11. Between and the mean

12. Between and

13. Between and

14. Between and

15. To the left of and to the right of

16. To the left of and to the right of

Section 5.2

In Exercises 17–22, find the indicated probabilities. If convenient, use technologyto find the probability.

17. 18.

19. 20.

21. 22. P1z 6 0 or z 7 1.682P1z 6 -2.50 or z 7 2.502

P10.42 6 z 6 3.152P1-2.15 6 z 6 1.552

P1z 7 -0.742P1z 6 1.282

z = 2.16z = 0.64

z = 1.5z = -1.5

z = 1.96z = -1.96

z = 1.71z = 0.05

z = 1.04z = -1.55

z = -1.64

z = 0.12

z = 1.68

z = 1.72

z = -0.27

z = 2.55

z = 0.33

5101520 0 5 10 15x

− − − −5 10 15 20 25x

s.m

5 REVIEW EXERCISES

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In Exercises 23 and 24, find the indicated probabilities.

23. A study found that the mean migration distance of the green turtle was 2200kilometers and the standard deviation was 625 kilometers.Assuming that thedistances are normally distributed, find the probability that a randomlyselected green turtle migrates a distance of

(a) less than 1900 kilometers.

(b) between 2000 kilometers and 2500 kilometers.

(c) greater than 2450 kilometers.

(Adapted from Dorling Kindersley Visual Encyclopedia)

24. The world’s smallest mammal is the Kitti’s hog-nosed bat, with a meanweight of 1.5 grams and a standard deviation of 0.25 gram.Assuming that theweights are normally distributed, find the probability of randomly selecting abat that weighs

(a) between 1.0 gram and 2.0 grams.

(b) between 1.6 grams and 2.2 grams.

(c) more than 2.2 grams.

(Adapted from Dorling Kindersley Visual Encyclopedia)

Section 5.3

In Exercises 25–30, use the Standard Normal Table to find the z-score thatcorresponds to the given cumulative area or percentile. If the area is not in the table,use the entry closest to the area. If convenient, use technology to find the z-score.

25. 0.4721 26. 0.1 27. 0.8708

28. 29. 30.

In Exercises 31–36, use the following information. On a dry surface, the brakingdistance (in meters) of a Ford Expedition can be approximated by a normaldistribution, as shown in the graph. (Source: National Highway Traffic SafetyAdministration)

31. Find the braking distance of a Ford Expedition that corresponds toz = -2.4.

x

= 52 mμ= 2.5 mσ

Braking distance (in meters)

Braking Distance of aFord Expedition

45 47 49 51 53 55 57 59

P20P85P2

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REVIEW EXERCISES 299

32. Find the braking distance of a Ford Expedition that corresponds to

33. What braking distance of a Ford Expedition represents the 95th percentile?

34. What braking distance of a Ford Expedition represents the third quartile?

35. What is the shortest braking distance of a Ford Expedition that can be in thetop 10% of braking distances?

36. What is the longest braking distance of a Ford Expedition that can be in thebottom 5% of braking distances?

Section 5.4

In Exercises 37 and 38, use the given population to find the mean and standarddeviation of the population and the mean and standard deviation of the samplingdistribution. Compare the values.

37. A corporation has five executives. The number of minutes of overtime aweek reported by each is 90, 300, 120, 160, and 210. Draw three executives’names from this population, with replacement.

38. There are four residents sharing a house. The number of times each washeshis or her car each month is 1, 2, 0, and 3. Draw two names from thispopulation, with replacement.

In Exercises 39 and 40, use the Central Limit Theorem to find the mean andstandard error of the mean of the indicated sampling distribution. Then sketch agraph of the sampling distribution.

39. The consumption of processed fruits by people in the United States in arecent year was normally distributed, with a mean of 144.3 pounds and astandard deviation of 51.6 pounds. Random samples of size 35 are drawnfrom this population. (Adapted from U.S. Department of Agriculture)

40. The consumption of processed vegetables by people in the United States ina recent year was normally distributed, with a mean of 218.2 pounds and astandard deviation of 68.1 pounds. Random samples of size 40 are drawnfrom this population. (Adapted from U.S. Department of Agriculture)

In Exercises 41–46, find the probabilities for the sampling distributions.

41. Refer to Exercise 23.A sample of 12 green turtles is randomly selected. Find theprobability that the sample mean of the distance migrated is (a) less than 1900kilometers, (b) between 2000 kilometers and 2500 kilometers, and (c) greaterthan 2450 kilometers. Compare your answers with those in Exercise 23.

42. Refer to Exercise 24. A sample of seven Kitti’s hog-nosed bats is randomlyselected. Find the probability that the sample mean is (a) between 1.0 gramand 2.0 grams, (b) between 1.6 grams and 2.2 grams, and (c) more than 2.2 grams. Compare your answers with those in Exercise 24.

43. The mean annual salary for chauffeurs is $29,200. A random sample of size45 is drawn from this population. What is the probability that the mean annual salary is (a) less than $29,000 and (b) more than $31,000? Assume

(Source: Salary.com)

44. The mean value of land and buildings per acre for farms is $1300. A randomsample of size 36 is drawn. What is the probability that the mean value ofland and buildings per acre is (a) less than $1400 and (b) more than $1150?Assume s = $250.

s = $1500.

z = 1.2.

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45. The mean price of houses in a city is $1.5 million with a standard deviationof $500,000. The house prices are normally distributed. You randomly select15 houses in this city. What is the probability that the mean price will be lessthan $1.125 million?

46. Mean rent in a city is $500 per month with a standard deviation of $30. Therents are normally distributed. You randomly select 15 apartments in this city.What is the probability that the mean price will be more than $525?

Section 5.5

In Exercises 47 and 48, a binomial experiment is given. Decide whether you can usethe normal distribution to approximate the binomial distribution. If you can, findthe mean and standard deviation. If you cannot, explain why.

47. In a recent year, the American Cancer Society said that the five-year survivalrate for new cases of stage 1 kidney cancer is 95%. You randomly select 12men who were new stage 1 kidney cancer cases this year and calculate theirfive-year survival rate. (Source: American Cancer Society)

48. A survey indicates that 59% of men purchased perfume in the past year. Yourandomly select 15 men and ask them if they have purchased perfume in thepast year. (Source: USA TODAY)

In Exercises 49–52, write the binomial probability as a normal probability usingthe continuity correction.

Binomial Probability Normal Probability

49.

50.

51.

52.

In Exercises 53 and 54, decide whether you can use the normal distribution toapproximate the binomial distribution. If you can, use the normal distribution toapproximate the indicated probabilities and sketch their graphs. If you cannot,explain why and use the binomial distribution to find the indicated probabilities.

53. Seventy percent of children ages 12 to 17 keep at least part of their savingsin a savings account. You randomly select 45 children and ask each if he orshe keeps at least part of his or her savings in a savings account. Find theprobability that at most 20 children will say yes. (Source: InternationalCommunications Research for Merrill Lynch)

54. Thirty-three percent of adults graded public schools as excellent or good atpreparing students for college. You randomly select 12 adults and ask them ifthey think public schools are excellent or good at preparing students for college. Find the probability that more than five adults will say yes. (Source:Marist Institute for Public Opinion)

P1? 6 x 6 ?2P1x = 502

P1? 6 x 6 ?2P1x = 452

P1x 6 ?2P1x … 362

P1x 7 ?2P1x Ú 252

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CHAPTER QUIZ 301

Take this quiz as you would take a quiz in class. After you are done, check yourwork against the answers given in the back of the book.

1. Find each standard normal probability.

(a)

(b)

(c)

(d)

2. Find each normal probability for the given parameters.

(a)

(b)

(c)

In Exercises 3–10, use the following information. In a recent year, grade 8Minnesota State public school students taking a mathematics assessment test had amean score of 290 with a standard deviation of 37. Possible test scores could rangefrom 0 to 500. Assume that the scores are normally distributed. (Source: NationalCenter for Educational Statistics)

3. Find the probability that a student had a score higher than 320.

4. Find the probability that a student had a score between 250 and 300.

5. What percent of the students had a test score that is greater than 250?

6. If 2000 students are randomly selected, how many would be expected to havea test score that is less than 280?

7. What is the lowest score that would still place a student in the top 5% of thescores?

8. What is the highest score that would still place a student in the bottom 25%of the scores?

9. A random sample of 60 students is drawn from this population. What is theprobability that the mean test score is greater than 300?

10. Are you more likely to randomly select one student with a test score greaterthan 300 or are you more likely to select a sample of 15 students with a meantest score greater than 300? Explain.

In Exercises 11 and 12, use the following information. In a survey of adults, 75%strongly support using DNA research by scientists to find new ways to prevent ortreat diseases. You randomly select 24 adults and ask each if they strongly supportusing DNA research by scientists to find new ways to prevent or treat diseases.(Source: Harris Interactive)

11. Decide whether you can use the normal distribution to approximate thebinomial distribution. If you can, find the mean and standard deviation. Ifyou cannot, explain why.

12. Find the probability that at most 15 people say they strongly support usingDNA research by scientists to find new ways to prevent or treat diseases.

P1x 6 0 or x 7 372s = 9.25,m = 18.5,

P1-5.00 6 x 6 02s = 7.84,m = -8.2,

P15.36 6 x 6 5.642s = 0.08,m = 5.5,

P1z 6 -1.75 or z 7 -0.752

P1-2.33 6 z 6 2.332

P1z 6 3.222

P1z 7 -2.102

5 CHAPTER QUIZ

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REAL Statistics — Real Decisions

Putting It All Together

302

You work for a candy company as a statistical process analyst. Your job is to analyze processes and make sure they are in statistical control.In one process, a machine is supposed to drop 11.4 ounces of mints into a bag. (Assume this process can be approximated by a normaldistribution.) The acceptable range of weights for the bags of mints is 11.25ounces to 11.55 ounces, inclusive.

Because of an error with the release valve, the setting on the mintrelease machine “shifts” from 11.4 ounces. To check that the machine is placing the correct weight of mints into the bags, you select at random three samples of five bags of mints and find the mean weight (inounces) of each sample. A coworker asks why you take three samples of size 5 and find the mean instead of randomly choosing andmeasuring 15 bags of mints individually to check the machine’s settings.(Note: Both samples are chosen without replacement.)

� Exercises1. Sampling Individuals

You select one bag of mints and measure its weight. Assume themachine shifts and is filling the bags with a mean weight of 11.56ounces and a standard deviation of 0.05 ounces.

(a) What is the probability that you select a bag of mints that is notoutside the acceptable range (in other words, you do not detectthat the machine has shifted)? (See figure.)

(b) You randomly select 15 bags of mints. What is the probability thatyou select at least one bag that is not outside the acceptablerange?

2. Sampling Groups of Five

You select five bags of mints and find their mean weight. Assume themachine shifts and is filling the bags with a mean weight of 11.56ounces and a standard deviation of 0.05 ounces.

(a) What is the probability that you select a sample of five bags ofmints that has a mean that is not outside the acceptable range?(See figure.)

(b) You randomly select three samples of five bags of mints. What isthe probability that you select at least one sample of five bags ofmints that has a mean that is not outside the acceptable range?

(c) What is more sensitive to change an individual measure or themean?

3. Writing an Explanation

Write a paragraph to your coworker explaining why you take 3 samplesof size 5 and find the mean of each sample instead of randomly choosingand measuring 15 bags of mints individually to check the machine’ssettings.

—

Mean = 11.4 Mean = 11.56

Weight (in ounces)

Upper limitof acceptablerange

x

Originaldistributionof individualbags

Distribution whenmachine shifts

11.3 11.4 11.5 11.6 11.7

Mean = 11.4

Mean = 11.56

Weight (in ounces)

x

Upper limitof acceptablerange

Distribution whenmachine shifts

Originaldistributionof samplemeans,n = 5

11.3 11.4 11.5 11.6 11.7



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TECHNOLOGY 303

TECHNOLOGY MINITAB EXCEL T1-83/84

Extended solutions are given in the Technology Supplement. Technicalinstruction is provided for MINITAB, Excel, and the TI-83/84.

AGE DISTRIBUTION IN THE UNITED STATESOne of the jobs of the U.S. Census Bureau is to keep track of theage distribution in the country. The age distribution in 2005 isshown below.

Rel

ativ

e fr

eque

ncy

Age classes (in years)

7 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 972

1%

2%

3%

4%

5%

6%

7%

8%

9%

Age Distribution in the U.S.

www.census.gov

U.S. Census BureauClass

boundariesClass

midpointRelative

frequency

0–4 2 6.8%

5–9 7 6.6%

10–14 12 7.0%

15–19 17 7.1%

20–24 22 7.1%

25–29 27 6.8%

30–34 32 6.8%

35–39 37 7.1%

40–44 42 7.7%

45–49 47 7.6%

50–54 52 6.7%

55–59 57 5.9%

60–64 62 4.4%

65–69 67 3.4%

70–74 72 2.9%

75–79 77 2.5%

80–84 82 1.9%

85–89 87 1.1%

90–94 92 0.5%

95–99 97 0.1%

� EXERCISESWe used a technology tool to select random sampleswith from the age distribution of the UnitedStates. The means of the 36 samples were as follows.

28.14, 31.56, 36.86, 32.37, 36.12, 39.53,36.19, 39.02, 35.62, 36.30, 34.38, 32.98,36.41, 30.24, 34.19, 44.72, 38.84, 42.87,38.90, 34.71, 34.13, 38.25, 38.04, 34.07,39.74, 40.91, 42.63, 35.29, 35.91, 34.36,36.51, 36.47, 32.88, 37.33, 31.27, 35.80

1. Enter the age distribution of the United States intoa technology tool. Use the tool to find the mean agein the United States.

2. Enter the set of sample means into a technologytool. Find the mean of the set of sample means.How does it compare with the mean age in theUnited States? Does this agree with the resultpredicted by the Central Limit Theorem?

3. Are the ages of people in the United States normallydistributed? Explain your reasoning.

4. Sketch a relative frequency histogram for the 36sample means. Use nine classes. Is the histogramapproximately bell-shaped and symmetric? Doesthis agree with the result predicted by the CentralLimit Theorem?

5. Use a technology tool to find the standard deviation of the ages of people in the United States.

6. Use a technology tool to find the standard deviationof the set of 36 sample means. How does it comparewith the standard deviation of the ages? Does this agree with the result predicted by the CentralLimit Theorem?

n = 40

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C H A P T E R S 3 – 51. A survey of employees in the United States found that 56% do not use all of

their vacation time. You randomly select 30 employees and ask them if theyuse all of their vacation time. (Source: Rasmussen Reports)


(b) Find the probability that at most 14 employees say they do not use all oftheir vacation time.

(c) Is it unusual for 14 out of 30 employees to say they do not use all of theirvacation time? Why or why not?

In Exercises 2 and 3, use the probability distribution to find the (a) mean,(b) variance, (c) standard deviation, and (d) expected value of the probabilitydistribution, and (e) interpret the results.

2. The table shows the distribution of family household sizes in the UnitedStates for a recent year. (Source: U.S. Census Bureau)

3. The table shows the distribution of fouls per game for a player in a recentNBA season. (Source: NBA.com)

4. Use the probability distribution in Exercise 3 to find the probability ofrandomly selecting a game in which the player had (a) fewer than four fouls,(b) at least three fouls, and (c) between two and four fouls, inclusive.

5. From a pool of 16 candidates, 9 men and 7 women, the offices of president,vice president, secretary, and treasurer will be filled. (a) In how many differentways can the offices be filled? (b) What is the probability that all four of theoffices are filled by women?

In Exercises 6–11, use the Standard Normal Table to find the indicated area underthe standard normal curve.


8. To the right of 9. Between and

10. Between and 11. To the left of or to the right of

12. Seventy-eight percent of college graduates say they spent two years or less attheir first full-time job after graduating college. You randomly select 10college graduates and ask each how long they stayed at their first full-time jobafter graduating college. Find the probability that the number who say theyspent less than two years at their first full-time job after graduating college is(a) exactly six, (b) at least six, and (c) less than six. (Source: Experience.com)

z = 1.72z = -0.26z = 0.12z = -1.22

z = 3.09z = 0z = -0.84

z = -3.08z = 1.54

304

x 2 3 4 5 6 7

P1x2 0.421 0.233 0.202 0.093 0.033 0.017

x 0 1 2 3 4 5 6

P1x2 0.012 0.049 0.159 0.256 0.244 0.195 0.085

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13. An auto parts seller finds that 1 in every 200 parts sold is defective. Use thegeometric distribution to find the probability that (a) the first defective partis the tenth part sold, (b) the first defective part is the first, second, or thirdpart sold, and (c) none of the first ten parts sold are defective.

14. The table shows the results of a survey in which 3,188,100 public and 438,800private school teachers were asked about their full-time teaching experience.(Adapted from U.S. National Center for Education Statistics)

(a) Find the probability that a randomly selected private school teacher has10 to 20 years of full-time teaching experience.

(b) Given that a randomly selected teacher has 3 to 9 years of full-timeteaching experience, find the probability that the teacher is at a public school.

(c) Are the events “being a public school teacher” and “having 20 years or moreof full-time teaching experience” independent or dependent? Explain.

(d) Find the probability that a randomly selected teacher is either at a publicschool or has less than 3 years of full-time teaching experience.

(e) Find the probability that a randomly selected teacher has 3 to 9 years offull-time teaching experience or is at a private school.

15. The initial pressure for bicycle tires when first filled is normally distributed, witha mean of 70 pounds per square inch (psi) and a standard deviation of 1.2 psi.

(a) Random samples of size 40 are drawn from this population and the meanof each sample is determined. Use the Central Limit Theorem to find themean and standard error of the mean of the sampling distribution. Thensketch a graph of the sampling distribution of sample means.

(b) A random sample of 15 tires is drawn from this population. What is theprobability that the mean tire pressure of the sample, is less than 69 psi?

16. The life span of a car battery is normally distributed, with a mean of 44months and a standard deviation of 5 months.

(a) A car battery is selected at random. Find the probability that the lifespan of the battery is less than 36 months.

(b) A car battery is selected at random. Find the probability that the lifespan of the battery is between 42 and 60 months.

(c) What is the shortest life expectancy a car battery can have and still be inthe top 5% of life expectancies?

17. A florist has 12 different flowers from which floral arrangements can bemade. (a) If a centerpiece is to be made using four different flowers, howmany different centerpieces can be made? (b) What is the probability thatthe four flowers in the centerpiece are roses, gerbers, hydrangeas, and callas?

18. Forty-one percent of adults say they shop for a gift within a week of an event.You randomly select 20 adults and ask each how far in advance he or sheshops for a gift. Use the binomial formula to find the probability that thenumber who say they shop for a gift within a week of the event is (a) exactlyeight, (b) at least six, and (c) at most thirteen. (Source: Harris Interactive)

x,

CUMULATIVE REVIEW, PART I I 305

Public Private Total

388,500 106,600 495,100

1,048,000 144,900 1,192,900

833,350 98,300 931,800

918,100 89,000 1,007,100

3,188,100 438,800 3,626,900

Less than 3 years3 to 9 years10 to 20 years20 years or moreTotal

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