15.4 the normal distribution objectivesgarcia/4.4reading.pdf · 15.4 the normal distribution...

15.4 The Normal DistributionObjectives1. Understand the basic properties of the normal curve.2. Relate the area under a normal curve to z-scores.3. Make conversions between raw scores and z-scores.4. Use the normal distribution to solve applied problems.

In addition to shoes, what about the fit of movie seats, men’s ties, and headroom in cars?How do manufacturers decide how wide the seats at your favorite movie theater should be?Or, how long must a tie be so that it is neither too long nor too short for most men? How dowe design a car so that most people will not bump their heads on its roof? Or, how highshould a computer table be so that it does not put an unnecessary strain on your wrists and

If the shoe fits, wear it. —Well-known proverb

Copyright © 2010 Pearson Education, Inc.

μ68%

Inflection points

95%

99.7%

μ σ + μ σ − μ σ + 2μ σ − 2 μ σ + 3μ σ − 3

FIGURE 15.14 The 68-95-99.7 rule for anormal distribution.

KEY POINT

Many different types of datasets are normal distributions.

15.4 y The Normal Distribution 751

arms? In the relatively new science of ergonomics, scientists gather data to answer ques-tions such as these and to ensure most people fit comfortably into their environments.Much of the work of these scientists is based on a curve that you will study in this sectioncalled the normal curve, or normal distribution.*

The Normal DistributionThe normal distribution is the most common distribution in statistics and describes manyreal-life data sets. The histogram shown in Figure 15.12 will begin to give you an idea ofthe shape of a normal distribution.

Distributions of such diverse data sets as SAT scores, heights of people, the number ofmiles before an automobile tire wears out, the size of hamburgers at a fast-food restaurant,and the number of hours of use before a certain brand of DVD player breaks down are allexamples of normal distributions.

There are several patterns in Figure 15.12 that are common to all normal distribu-tions. First, if we were to take a wire and attach it to the tops of the bars in the histogramand then smooth out the wire to make a curve, the curve would have a bell-shapedappearance, as shown in Figure 15.13. This is why a normal distribution is often calleda bell-shaped curve.

Second, the mean, median, and mode of a normal distribution are the same. Third, thecurve is symmetric with respect to the mean. This means that if you see some pattern inthe graph on one side of the mean, then the mean acts as a mirror to give a reflection ofthat same pattern on the other side of the mean. Fourth, the area under a normal curveequals 1.

In Figure 15.14, we have marked the mean and two points on the normal curve calledinflection points. An inflection point is a point on the curve where the curve changes frombeing curved upward to being curved downward, or vice versa. For a normal curve,inflection points are located 1 standard deviation from the mean. Also, because a normalcurve is symmetric with respect to the mean, one-half, or 50%, of the area under thecurve is located on each side of the mean.

In normal distributions, approximately 68% of the data values occur within 1 standarddeviation of the mean, 95% of the data values lie within 2 standard deviations of the mean,and 99.7% of the data values lie within 3 standard deviations of the mean. We refer tothese facts as the 68-95-99.7 rule. Figure 15.14 illustrates the 68-95-99.7 rule. Indiscussing normal distributions, we usually assume that we are dealing with an entirepopulation rather than a sample, so in Figure 15.14 we represent the mean by m and thestandard deviation by s (rather than and s).

We summarize the properties of a normal distribution.xq

mean =median =

mode

FIGURE 15.13 Smoothing ahistogram of a normal distributioninto a normal curve.

FIGURE 15.12 A normaldistribution.

*Recall that a distribution is just another name for a set of data.


CHAPTER 15 y Descriptive Statistics752

mean450

68%

of scores

425 � onestandard deviationbelow mean

475 � onestandard deviationabove mean

FIGURE 15.15 Sixty-eight percent of the scoreslie within 1 standard deviation of the mean.

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

1. A normal curve is bell shaped.

2. The highest point on the curve is at the mean of the distribution.

3. The mean, median, and mode of the distribution are the same.

4. The curve is symmetric with respect to its mean.

5. The total area under the curve is 1.

6. Roughly 68% of the data values are within 1 standard deviation from the mean,95% of the data values are within 2 standard deviations from the mean, and99.7% of the data values are within 3 standard deviations from the mean.*

We can use the 68-95-99.7 rule to estimate how many values we expect to fall within 1,2, or 3 standard deviations of the mean of a normal distribution.

EXAMPLE 1 The Normal Distribution and Intelligence Tests

Suppose that the distribution of scores of 1,000 students who take a standardized intelli-gence test is a normal distribution. If the distribution’s mean is 450 and its standarddeviation is 25,

a) how many scores do we expect to fall between 425 and 475?

b) how many scores do we expect to fall above 500?

SOLUTION:

a) As you can see in Figure 15.15, the scores 425 and 475 are 1 standard deviation below andabove the mean, respectively.

*To keep this discussion simple, we are approximating these percentages. Shortly, we will use a table to slightlyimprove our estimate of the percentage of scores within 1 and 2 standard deviations from the mean.

From the 68-95-99.7 rule, we know that 68%, or about 0.68 of the scores, lie within1 standard deviation of the mean. Because we have 1,000 scores, we can expect thatabout 0.68 * 1,000 = 680 scores are in the range 425 to 475.

b) Recall in Figure 15.16, that 95% of the scores in a normal distribution fall between2 standard deviations below the mean and 2 standard deviations above the mean.


KEY POINT

Areas under a normal curverepresent percentages ofvalues in the distribution.

Quiz Yourself

Use the information given inExample 1 to determine thefollowing.

a) How many scores do weexpect to fall between 400 and500?

b) How many scores do weexpect to fall below 400?

11


This means that 5%, or 0.05, of the scores lie more than 2 standard deviations above orbelow the mean. Thus, we can expect to have 0.05 , 2 = 0.025 of the scores to be above 500. Again multiplying by 1,000, we can expect that 0.025 * 1,000 = 25 scores to be above 500.

Now try Exercises 5 to 16. ]

z-ScoresIn Example 1, we estimated how many values were within 1 standard deviation of the mean.It is natural to ask whether we can predict how many values lie other distances from themean. For example, knowing that the weights of women between ages 18 and 25 form anormal distribution, a clothing manufacturer may want to know what percentage of thefemale population is between 1.3 and 2.5 standard deviations above the mean weight forwomen. It is possible to find this by using Table 15.16, which is the table of areas under thestandard normal curve. The standard normal distribution has a mean of 0 and a standarddeviation of 1.

11

450mean

95%

of scores

Two standard deviationsbelow mean

400 500

Two standard deviationsabove mean

2.5 percentof scores here

2.5 percentof scores here

FIGURE 15.16 Five percent of the scores lie more than2 standard deviations from the mean.

Math in Your Life

People often argue about how much of your physical,intellectual, and psychological makeup is due to yourheredity and how much is due to your environment.Scientists have identified genes that affect characteristicssuch as height, body fat, blood pressure, and IQ.Surprisingly, even human anxiety has been associatedwith genetic makeup.

It has been estimated that there are as many as15 genes that can affect your anxiety level. If you haveall these genes, then you have a greater tendency† to be

anxious, and if you have none of them, then you tend tobe less anxious. If we were to plot a histogram of thenumber of anxiety genes present in a large number ofpeople, we would find very few people have almostnone of these genes, many people have a moderate num-ber of these genes, and very few people have almost allof them. If you were to plot the distribution of anxietygenes present in the population, it would look verymuch like the normal curves that you are studying inthis section.

How Worried Are You About Your Next Test?*

†It is believed that 45% of the variance in human anxiety is due to genetic factors. The other 55% is due to otherfactors.

*This note is based on a series of PowerPoint slides by Dr. Lee Bardwell, professor of genetics at the Universityof California at Irvine.



z A z A z A z A z A z A

.00 .000 .56 .212 1.12 .369 1.68 .454 2.24 .488 2.80 .497

.01 .004 .57 .216 1.13 .371 1.69 .455 2.25 .488 2.81 .498

.02 .008 .58 .219 1.14 .373 1.70 .455 2.26 .488 2.82 .498

.03 .012 .59 .222 1.15 .375 1.71 .456 2.27 .488 2.83 .498

.04 .016 .60 .226 1.16 .377 1.72 .457 2.28 .489 2.84 .498

.05 .020 .61 .229 1.17 .379 1.73 .458 2.29 .489 2.85 .498

.06 .024 .62 .232 1.18 .381 1.74 .459 2.30 .489 2.86 .498

.07 .028 .63 .236 1.19 .383 1.75 .460 2.31 .490 2.87 .498

.08 .032 .64 .239 1.20 .385 1.76 .461 2.32 .490 2.88 .498

.09 .036 .65 .242 1.21 .387 1.77 .462 2.33 .490 2.89 .498

.10 .040 .66 .245 1.22 .389 1.78 .463 2.34 .490 2.90 .498

.11 .044 .67 .249 1.23 .391 1.79 .463 2.35 .491 2.91 .498

.12 .048 .68 .252 1.24 .393 1.80 .464 2.36 .491 2.92 .498

.13 .052 .69 .255 1.25 .394 1.81 .465 2.37 .491 2.93 .498

.14 .056 .70 .258 1.26 .396 1.82 .466 2.38 .491 2.94 .498

.15 .060 .71 .261 1.27 .398 1.83 .466 2.39 .492 2.95 .498

.16 .064 .72 .264 1.28 .400 1.84 .467 2.40 .492 2.96 .499

.17 .068 .73 .267 1.29 .402 1.85 .468 2.41 .492 2.97 .499

.18 .071 .74 .270 1.30 .403 1.86 .469 2.42 .492 2.98 .499

.19 .075 .75 .273 1.31 .405 1.87 .469 2.43 .493 2.99 .499

.20 .079 .76 .276 1.32 .407 1.88 .470 2.44 .493 3.00 .499

.21 .083 .77 .279 1.33 .408 1.89 .471 2.45 .493 3.01 .499

.22 .087 .78 .282 1.34 .410 1.90 .471 2.46 .493 3.02 .499

.23 .091 .79 .285 1.35 .412 1.91 .472 2.47 .493 3.03 .499

.24 .095 .80 .288 1.36 .413 1.92 .473 2.48 .493 3.04 .499

.25 .099 .81 .291 1.37 .415 1.93 .473 2.49 .494 3.05 .499

.26 .103 .82 .294 1.38 .416 1.94 .474 2.50 .494 3.06 .499

.27 .106 .83 .297 1.39 .418 1.95 .474 2.51 .494 3.07 .499

.28 .110 .84 .300 1.40 .419 1.96 .475 2.52 .494 3.08 .499

.29 .114 .85 .302 1.41 .421 1.97 .476 2.53 .494 3.09 .499

.30 .118 .86 .305 1.42 .422 1.98 .476 2.54 .495 3.10 .499

.31 .122 .87 .308 1.43 .424 1.99 .477 2.55 .495 3.11 .499

.32 .126 .88 .311 1.44 .425 2.00 .477 2.56 .495 3.12 .499

.33 .129 .89 .313 1.45 .427 2.01 .478 2.57 .495 3.13 .499

.34 .133 .90 .316 1.46 .428 2.02 .478 2.58 .495 3.14 .499

.35 .137 .91 .319 1.47 .429 2.03 .479 2.59 .495 3.15 .499

.36 .141 .92 .321 1.48 .431 2.04 .479 2.60 .495 3.16 .499

.37 .144 .93 .324 1.49 .432 2.05 .480 2.61 .496 3.17 .499

.38 .148 .94 .326 1.50 .433 2.06 .480 2.62 .496 3.18 .499

.39 .152 .95 .329 1.51 .435 2.07 .481 2.63 .496 3.19 .499

.40 .155 .96 .332 1.52 .436 2.08 .481 2.64 .496 3.20 .499

.41 .159 .97 .334 1.53 .437 2.09 .482 2.65 .496 3.21 .499

.42 .163 .98 .337 1.54 .438 2.10 .482 2.66 .496 3.22 .499

.43 .166 .99 .339 1.55 .439 2.11 .483 2.67 .496 3.23 .499

.44 .170 1.00 .341 1.56 .441 2.12 .483 2.68 .496 3.24 .499

.45 .174 1.01 .344 1.57 .442 2.13 .483 2.69 .496 3.25 .499

.46 .177 1.02 .346 1.58 .443 2.14 .484 2.70 .497 3.26 .499

.47 .181 1.03 .349 1.59 .444 2.15 .484 2.71 .497 3.27 .500

.48 .184 1.04 .351 1.60 .445 2.16 .485 2.72 .497 3.28 .500

.49 .188 1.05 .353 1.61 .446 2.17 .485 2.73 .497 3.29 .500

.50 .192 1.06 .355 1.62 .447 2.18 .485 2.74 .497 3.30 .500

.51 .195 1.07 .358 1.63 .449 2.19 .486 2.75 .497 3.31 .500

.52 .199 1.08 .360 1.64 .450 2.20 .486 2.76 .497 3.32 .500

.53 .202 1.09 .362 1.65 .451 2.21 .487 2.77 .497 3.33 .500

.54 .205 1.10 .364 1.66 .452 2.22 .487 2.78 .497

.55 .209 1.11 .367 1.67 .453 2.23 .487 2.79 .497

TABLE 15.16 Standard normal distribution.



Table 15.16 gives the area under this curve between the mean and a number called a z-score. A z-score represents the number of standard deviations a data value is from themean. In Example 1, we found that for a normal distribution with a mean of 450 and astandard deviation of 25, the value 500 was 2 standard deviations above the mean.Another way of saying this is that the value 500 corresponds to a z-score of 2.

Notice that Table 15.16 gives areas only for positive z-scores—that is, for data that lieabove the mean. We use the symmetry of a normal curve to find areas corresponding tonegative z-scores.

Example 2 shows how to use Table 15.16. Because the total area under a normal curveis 1, we can interpret areas under the standard normal curve as percentages of the data val-ues in the distribution. Also note that because the mean is 0 and the standard deviation forthe standard normal curve is 1, the value of a z-score is also the same as the number ofstandard deviations that z-score is from the mean. After you have practiced working withthe standard normal distribution, we will show you how to use the techniques that you havelearned to solve problems involving real-life data.

PROBLEM SOLVING

The Three-Way PrincipleThe Three-Way Principle in Section 1.1 suggests that you can often understand a problemgraphically. This is certainly true in solving problems involving the normal curve. If youdraw a good picture of the situation, it often becomes more clear to you as to what to do toset up and solve the problem.

EXAMPLE 2 Finding Areas under the Standard Normal Curve

Use Table 15.16 to find the percentage of the data (area under the curve) that lie in thefollowing regions for a standard normal distribution:

a) between z = 0 and z = 1.3

b) between z = 1.5 and z = 2.1

c) between z = 0 and z = -1.83

SOLUTION:

a) The area under the curve between z = 0 and z = 1.3 is shown in Figure 15.17. We find thisarea by looking in Table 15.16 for the z-score 1.30. We see that A is 0.403 when z = 1.30.Thus, we expect 0.403, or 40.3%, of the data to fall between 0 and 1.3 standard deviationsabove the mean.

1.3z-score

0mean

The area ofthis region is0.403.

FIGURE 15.17 Area under thestandard normal curve between z = 0and z = 1.3.


›


0mean

1.5 2.1

A = 0.482 − 0.433A = 0.433

FIGURE 15.18 Finding the areaunder the standard normal curvebetween z = 1.5 and z = 2.1.

TI Screen showing area understandard normal curve between z = 1.5 and z = 2.1.

0mean

−1.83 1.83

A = 0.466 A = 0.466

FIGURE 15.19 Finding the areaunder the standard normal curvebetween z = -1.83 and z = 0.

c) Due to the symmetry of the normal distribution, the area between z = 0 and z = -1.83is the same as the area between z = 0 and z = 1.83 (Figure 15.19). From Table 15.16we see that when z = 1.83, A = 0.466. Therefore, 46.6% of the data values lie between0 and -1.83.


Some Good AdviceA common mistake in solving a problem such as Example 2(b) is to subtract 1.5 from 2.1 toget 0.6 and then wrongly use this for the z-score in Table 15.16. If you think about the graphof the standard normal curve, clearly the area between z = 0 and z = 0.6 is not the same as thearea between z = 1.5 and z = 2.1.

Converting Raw Scores to z-ScoresA real-life normal distribution, such as the set of all weights of women between ages 18 and25, may have a mean of 120 pounds and a standard deviation of 25 pounds. Such a distribu-tion will have the properties that we stated earlier for a normal distribution, but because thedistribution does not have a mean of 0 and a standard deviation of 1, we cannot useTable 15.16 directly, as we did in Example 2. We can use Table 15.16, however, if we firstconvert nonstandard values, called raw scores, to z-scores. The following formula shows thedesired relationship between values in a nonstandard normal distribution and z-scores.

F O R M U L A F O R C O N V E R T I N G R A W S C O R E S TO Z - S C O R E S Assume anormal distribution has a mean of m and a standard deviation of s. We use the equation

to convert a value x in the nonstandard distribution to a z-score.

z =x - ms

12

Quiz Yourself

Use Table 15.16 to find thefollowing areas under thestandard normal curve:

a) between z = 0 and z = 1.45

b) between z = 1.23 and z = 1.85

c) between z = 0 and z = -1.35

12

KEY POINT

We convert values in anonstandard normaldistribution to z-scores.

b) Figure 15.18 shows the area we want. Areas in Table 15.16 correspond to regions from z = 0 to the given z-score. Therefore, to find the area under the curve between z = 1.5 and z = 2.1, we must first find the area from z = 0 to z = 2.1 and then subtract thearea from z = 0 to z = 1.5. From Table 15.16, we see that when z = 2.1, A = 0.482, andwhen z = 1.5, A = 0.433. To finish this problem, we can set up our calculations as follows:

This means that in the standard normal distribution, the area under the curve between z = 1.5 and z = 2.1 is 0.049, or 4.9%.

subtract areas, not z-scores

area we want

larger area from z � 0 to z � 2.1 0.482� smaller area from z � 0 to z � 1.5 �0.433

0.049



EXAMPLE 3 Converting Raw Scores to z -ScoresSuppose the mean of a normal distribution is 20 and its standard deviation is 3.

a) Find the z-score corresponding to the raw score 25.

b) Find the z-score corresponding to the raw score 16.

SOLUTION:

a) Figure 15.20 gives us a picture of this situation. We will use the conversion formulawith the raw score x = 25, the mean m = 20, and the standard deviation z = x - m

s,

s = 3. Making the substitutions, we have, We can interpretthis result as telling us that in this distribution, 25 is 1.67 standard deviations abovethe mean.

z = 25 - 203 = 5

3 = 1.67.

� 20mean

�

� 3standard deviation

�x � 25raw score

x � 16raw score

FIGURE 15.20 Normal distribution with meanof 20 and standard deviation of 3.Quiz Yourself

Suppose a normal distributionhas a mean of 50 and a standarddeviation of 7. Convert each rawscore to a z-score.

a) 59 b) 50 c) 38

13

b) We will use Figure 15.20 and the conversion formula again, but this time with x = 16,m = 20, and s = 3. Therefore, the corresponding z-score is This tells us that 16 is 1.33 standard deviations below the mean.


The Three-Way Principle in Section 1.1suggests that you can remember the formula forconverting raw scores to z-scores by interpret-ing the formula geometrically. In Example 3,we had a normal distribution with a mean of 20and standard deviation 3. In order to make themean of 20 in the nonstandard distribution cor-respond to the mean of 0 in the standard normaldistribution, you can think of “sliding” the dis-tribution to the left 20 units, as we show inFigure 15.21(a). That is why we compute anexpression of the form x - 20 in the numeratorof the z-score formula. Because the standarddeviation was 3, we divided x - 20 by 3 to“narrow” the distribution so that it would havea standard deviation of 1, as you can see inFigure 15.21(b), and therefore allow us to useTable 15.16.

13

z = 16 - 203 = - 4

3 = -1.33.

�20 �10 0 10 20 30

0.08

0.06

0.04

0.02

Slide

FIGURE 15.21 (a) First slidethe distribution 20 units to the leftto get a mean of 0.

�10�15 �5 0 5 10 15

0.4

0.3

0.2

0.1SqueezeSqueeze

FIGURE 15.21 (b) Then squeezethe distribution to get a standarddeviation of 1.



ApplicationsNormal distributions have many practical applications.

EXAMPLE 4 Interpreting the Significance of an Exam ScoreSuppose that to qualify for a management training program offered by your employer youmust score in the top 10% of those employees who take a standardized test. Assume thatthe distribution of scores is normal and you received a score of 72 on the test, which had amean of 65 and a standard deviation of 4. What percentage of those who took this test hada score below yours?

SOLUTION: We will find the number of standard deviations your score is above the meanand then determine what percentage of scores fall below this number. First, we calculatethe z-score, which corresponds to your 72:

Now, looking in Table 15.16 for z = 1.75, we find that A = 0.460. Therefore, 46% of thescores fall between the mean and your score (Figure 15.22).

z =72 - 65

4=

7

4= 1.75

� 65mean

�

� 4standard deviation

�x � 75 (raw score)z � 1.75

A � 0.500

A � 0.460

FIGURE 15.22 Normal distribution with meanof 65 and standard deviation of 4.

However, we must not forget the scores that fall below the mean. Because a normal curveis symmetric, another 50% of the scores fall below the mean. So, there are 50% + 46% = 96%of the scores below yours. Congratulations! You qualify for the program. ]

We can use the ideas we have been discussing about the normal curve to compare data.

EXAMPLE 5 Using z -Scores to Compare DataConsider the following information:

Ty Cobb hit .420 in 1911.

Ted Williams hit .406 in 1941.

George Brett hit .390 in 1980.

In the 1910s, the mean batting average was .266 and the standard deviation was .0371.



Assume that the batting averages were normally distributed in each of these three decades.Use z-scores to determine which of the three batters was ranked the highest in relationshipto his contemporaries.



SOLUTION: We will convert each of the three batting averages to a corresponding z-scorefor the distribution of batting averages for the decade in which the athlete played.

In the 1910s, Ty Cobb’s average of .420 corresponded to a z-score of

In the 1940s, Ted Williams’s average of .406 corresponded to a z-score of

In the 1970s, George Brett’s average of .390 corresponded to a z-score of

We see that in using the standard deviation to compare each hitter with his contempo-raries, Ted Williams was ranked as the best hitter.

Now try Exercises 69 and 70. ]

We conclude our discussion of normal distributions by explaining how manufacturerscan use information regarding the life of their products as a basis for warranties.

EXAMPLE 6 Using the Normal Distribution to Write Warranties

To increase sales, a manufacturer plans to offer a warranty on a new model of the iPodTouch. In testing the iPod Touch, quality control engineers found that it has a mean time tofailure of 3,000 hours with a standard deviation of 500 hours. Assume that the typicalpurchaser will use the device for 4 hours per day. If the manufacturer does not want morethan 5% to be returned as defective within the warranty period, how long should thewarranty period be to guarantee this?

SOLUTION: In solving this problem, we first work with the standard normal distributionand then convert the answer to fit our nonstandard normal distribution. Figure 15.23 givesa picture of the situation for the standard normal curve.

.390 - .261

.0317=

.129

.0317= 4.0694.

.406 - .267

.0326=

.139

.0326= 4.2638.

.420 - .266

.0371=

.154

.0371= 4.1509.

z = −1.64 z = 1.64

45% A = 45%5%5%

0mean

= .45

FIGURE 15.23 Finding the area under thestandard normal curve for a negative z-score.

In Figure 15.23, we see that we need to find a z-score such that at least 95% of theentire area is beyond this point. Observe that this score is to the left of the mean and is neg-ative. Therefore, we use symmetry and find the z-score such that 95% of the entire area isbelow this score. However, Table 15.16 gives the areas only for the upper half of the distri-bution. However, this is not a problem because we know that 50% of the entire area lies



below the mean. Therefore, our problem reduces to finding a z-score greater than 0 suchthat 45% of the area lies between the mean and that z-score.

In Table 15.16, we see that if A = 0.450, the corresponding z-score is 1.64. This meansthat 95% of the area underneath the standard normal curve falls below z = 1.64. By sym-metry, we also can say that 95% of the values lie above -1.64. We must now interpret thisz-score in terms of original normal distribution of raw scores.

Recall the equation that relates values in a distribution to z-scores, namely,

(1)

Previously we used this equation to convert raw scores to z-scores. We now use the equa-tion in answering the reverse question. Specifically, what is x when m = 3,000, s = 500,and z = -1.64? Substituting these numbers in equation (1), we get

(2)

We solve for x in this equation by multiplying both sides by 500:

Simplifying the equation, we get

-820 = x - 3,000.

Last, we add 3,000 to both sides to obtain x = 2,180 hours. We expect 95% of the break-downs to occur beyond 2,180 hours.

Because we know that owners use the iPod Touch about 4 hours per day, we divide2,180 by 4 to get days. This is equivalent to months if we use31 days per month. Therefore, if the manufacturer wants to have no more than 5% of thebreakdowns occur within the warranty period, the warranty should be for roughly18 months. ] 14

54531 = 17.582,180

4 = 545

-1.64 # (500) =x - 3,000

500# (500)

-1.64 =x - 3,000

500.

z =x - ms

.

Quiz Yourself

Redo Example 6, but nowassume that the mean time tofailure is 4,200 hours, with astandard deviation of 600, andthe manufacturer wants no morethan 2% to be returned asdefective within the warrantyperiod.

14

Exercises 15.4Looking Back*These exercises follow the general outline of the topics presented inthis section and will give you a good overview of the material thatyou have just studied.

1. List six properties of the normal distribution.

2. What was the common error that we warned you against mak-ing in solving Example 2(b)?

3. What property of the normal distribution makes it unnecessaryto have negative z-scores in Table 15.16?

4. State the equation that allows you to convert raw scores to z-scores?

Sharpening Your SkillsAssume that all distributions in this exercise set are normal.Assume that the distribution in Exercises 5–10 has a mean of 10

and a standard deviation of 2. Use the 68-95-99.7 rule to find thepercentage of values in the distribution that we describe.

5. Between 10 and 12 6. Between 12 and 14

7. Above 14 8. Below 8


Assume that the distribution in Exercises 11–16 has a mean of 12and a standard deviation of 3. Use the 68-95-99.7 rule to find thepercentage of values in the distribution that we describe.

11. Below 9 12. Between 12 and 15


15. Between 15 and 18 16. Above 18

Use Table 15.16 to find the percentage of area under the standardnormal curve that is specified.

17. Between z = 0 and z = 1.23

18. Between z = 0 and z = 2.06

19. Between z = 0.89 and z = 1.43

*Before doing these exercises, you may find it useful to review the note How to Succeed at Mathematicson page xix.


15.4 y Exercises 761

20. Between z = 0.76 and z = 1.52

21. Between z = 0 and z = -0.75

22. Between z = 0 and z = -1.35

23. Between z = 1.25 and z = 1.95

24. Between z = 0.37 and z = 1.23

25. Between z = -0.38 and z = -0.76

26. Between z = -1.55 and z = -2.13

27. Above z = 1.45 28. Below z = -0.64

29. Below z = -1.40 30. Above z = 0.78

31. Below z = 1.33 32. Above z = -0.46

33. Above z = -0.84 34. Below z = 1.22

35. Find a z-score such that 10% of the area under the standardnormal curve is above that score.

36. Find a z-score such that 20% of the area under the standardnormal curve is above that score.

37. Find a z-score such that 12% of the area under the standardnormal curve is below that score.

38. Find a z-score such that 24% of the area under the standardnormal curve is below that score.

39. Find a z-score such that 60% of the area is below thatscore.

40. Find a z-score such that 90% of the area is below thatscore.

41. Find a z-score such that 75% of the area is above thatscore.

42. Find a z-score such that 60% of the area is above thatscore.

In Exercises 43–48, we give you a mean, a standard deviation, anda raw score. Find the corresponding z-score.

43. Mean 80, standard deviation 5, x = 87



46. Mean 52, standard deviation 7.5, x = 61

47. Mean 38, standard deviation 10.3, x = 48

48. Mean 8, standard deviation 2.4, x = 6.2

In Exercises 49–54, we give you a mean, a standard deviation, anda z-score. Find the corresponding raw score.

49. Mean 60, standard deviation 5, z = 0.84

50. Mean 20, standard deviation 6, z = 1.32

51. Mean 35, standard deviation 3, z = -0.45

52. Mean 62, standard deviation 7.5, z = -1.40

53. Mean 28, standard deviation 2.25, z = 1.64

54. Mean 8, standard deviation 3.5, z = -1.25

Applying What You’ve LearnedDue to random variations in the operation of an automatic coffeemachine, not every cup is filled with the same amount of coffee.Assume that the mean amount of coffee dispensed is 8 ounces andthe standard deviation is 0.5 ounce. Use Figure 15.14 to solveExercises 55 and 56.

55. Analyzing vending machines.

a. What percentage of the cups should have at least 8 ouncesof coffee?

b. What percentage of the cups should have less than 7.5 ouncesof coffee?


a. What percentage of the cups should have at least 8.5 ouncesof coffee?

b. What percentage of the cups should have less than 8 ouncesof coffee?

A machine fills bags of candy. Due toslight irregularities in the operationof the machine, not every bag getsexactly the same number of pieces.Assume that the number of piecesper bag has a mean of 200 with a standard deviation of 2. UseFigure 15.14 to solve Exercises 57 and 58.


a. How many of the bags do we expect to have between200 and 202 pieces in them?

b. How many of the bags do we expect to have at least202 pieces in them?


a. How many of the bags do we expect to have between198 and 200 pieces in them?

b. How many of the bags do we expect to have at least196 pieces in them?

59. Product reliability. Suppose that for a particular brand ofLCD television set, the distribution of failures has a mean of120 months with a standard deviation of 12 months. With respectto this distribution, 140 months corresponds to what z-score?

60. Product reliability. Redo Exercise 59 using 130 monthsinstead of 140.

61. Distribution of heights. Assume that in the NBA the distribu-tion of heights has a mean of 6 feet, 8 inches, and a standarddeviation of 3 inches. If there are 324 players in the league,how many players do you expect to be over 7 feet tall?

62. Distribution of heights. Redo Exercise 61, but now we wantto know how many players you would expect to be over 6 feet,6 inches.

63. Distribution of heart rates. Assume that the distribution of21-year-old women’s heart rates at rest is 68 beats per minutewith a standard deviation of 4 beats per minute. If 200 womenare examined, how many would you expect to have a heart rateof less than 70?

64. Distribution of heart rates. Redo Exercise 63, but now we askhow many you would expect to have a heart rate less than 75.

In Exercises 65 and 66, we assume that the mean length of a humanpregnancy is 268 days.

65. Birth statistics. If 95% of all human pregnancies last between250 and 286 days, what is the standard deviation of the distrib-ution of lengths of human pregnancies? What percentage ofhuman pregnancies do we expect to last at least 275 days?



66. Birth statistics. Babies born before the 37th week (that is,before day 252 of a pregnancy) are considered premature.What percentage of births do we expect to be premature? (SeeExercise 65.)

67. Analyzing Internet use. Comcast has analyzed the amount oftime that its customers are online per session. It found that thedistribution of connect times has a mean of 37 minutes and astandard deviation of 11 minutes. For this distribution, find theraw score that corresponds to a z-score of 1.5.

68. Analyzing customer service. The distribution of times thatcustomers spend waiting in line in a supermarket has a meanof 3.6 minutes and a standard deviation of 1.2 minutes. Forthis distribution, find the raw score that corresponds to a z-score of -1.3.

In Exercises 69 and 70, use the information and methods ofExample 5 to make the indicated comparisons.

69. Comparing athletes. In 1949, Jackie Robinson hit .342 forthe Brooklyn Dodgers; in 1973, Rod Carew hit .350 for theMinnesota Twins. Using z-scores to compare each batter withhis contemporaries, determine which batting average was moreimpressive.

70. Comparing athletes. In 1940, Joe DiMaggio hit .352 for theNew York Yankees; in 1975, Bill Madlock hit .354 for thePittsburgh Pirates. Using z-scores to compare each batter withhis contemporaries, determine which batting average was moreimpressive.

71. Writing a warranty. A manufacturer plans to provide a war-ranty on Wii Fit. In testing, the manufacturer has found that theset of failure times for Wii Fit is normally distributed with amean time of 2,000 hours and a standard deviation of 800 hours.Assume that the typical purchaser will use Wii Fit for 2 hoursper day. If the manufacturer does not want more than 4%returned as defective within the warranty period, how longshould the warranty period be to guarantee this? (Assume thatthere are 31 days in each month.)

72. Writing a warranty. Redo Exercise 71, but now assume thatthe mean of the distribution of failure times is 2,500 hours,the standard deviation is 600 hours, and the manufacturerwants no more than 2.5% returned during the warrantyperiod.

73. Analyzing investments. A certain mutual fund over the last15 years has a mean yearly return of 7.8% with a standarddeviation of 1.3%.

a. If you had invested in this fund over the past 15 years, inhow many of these years would you expect to earn at least9% on your investment?

b. In how many years would you expect to earn less than 6%?

74. Analyzing investments. A certain bond fund over the last12 years has a mean yearly return of 5.9% with a standarddeviation of 1.8%.

a. If you had invested in this fund over the past 12 years, inhow many of these years would you expect to earn at least8% on your investment?

b. In how many years would you expect to earn less than 4%?

75. Analyzing the SATs. Assume that the math SAT scores arenormally distributed with a mean of 500 and a standard devia-tion of 100. If you score 480 on this exam, what percentage ofthose taking the test scored below you?

76. Analyzing the SATs. Assume that the verbal SAT scores arenormally distributed with a mean of 500 and a standard devia-tion of 100. If you score 520 on this exam, what percentage ofthose taking the test scored below you?

Communicating Mathematics77. If a raw score corresponds to a z-score of 1.75, what does that

tell you about that score in relationship to the mean of thedistribution?

78. If a raw score corresponds to a z-score of -0.85, what does thattell you about that score in relationship to the mean of thedistribution?

79. Explain why, if you want to compute the area under a normalcurve between z = 1.3 and z = 2.0, it is not correct to subtract2.0 - 1.3 = 0.7 and then use the z-score of 0.7 to find the area.

80. Without looking at Table 15.16, can you determine whetherthe area under the standard normal curve between z = 0.5 andz = 1.0 is greater than or less than the area between z = 1.5 and

Explain.

81. Explain how you would estimate the mean and the standarddeviation of the two normal distributions in the given figure.

z = 2.0?

5 2015 25 30

Distribution 2

Distribution 1

82. Explain how you would estimate the mean and the standarddeviation of the two normal distributions in the given figure.

Using Technology to InvestigateMathematics83. You can find many interactive programs that illustrate various

properties of the normal distribution. Search the Internet for “nor-mal distribution applets,” run some, and report on your findings.

84. Search the Internet for some interesting applications of thenormal distribution. While searching, I found many applicationsin medicine and even some in music. Report on your findings.

Distribution 2

Distribution 1

3 6 9 12 15 18 21 24 27 30 36 4139


15.5 y Looking Deeper: Linear Correlation 763

For Extra Credit85. If you had a large collection of data, how might you go about

determining whether the distribution was normal?

86. Discuss whether you feel that the method we used to compareTy Cobb, Ted Williams, and George Brett would be a valid wayto compare the great African American athletes who played inthe Negro leagues with the great white athletes of the same era.

The nth percentile in a distribution is a number x such that n% ofthe values in that distribution fall below x. For example, 20% of thevalues in the distribution fall below the 20th percentile.

87. What z-score in the standard normal distribution is the 80thpercentile?

88. What z-score in the standard normal distribution is the 30thpercentile?

89. If a distribution has a mean of 40 and a standard deviation of 4,what is the 75th percentile?

90. If a distribution has a mean of 80 and a standard deviation of 6,what is the 35th percentile?

91. Finding percentiles. If you score a 75 on a commercial pilot’sexam that has a mean score of 68 and a standard deviation of 4,to what percentile does your score correspond?

92. Finding percentiles. If you know that the mean salary for yourprofession is $53,000 with a standard deviation of $2,500, towhat percentile does your salary of $57,000 correspond?


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