Download 15.4 The Normal Distribution

15.4
The Normal Distribution
Objectives
1.
2.
3.
4.
Understand the basic properties of the normal curve.
Relate the area under a normal curve to z-scores.
Make conversions between raw scores and z-scores.
Use the normal distribution to solve applied problems.
If the shoe fits, wear it.
—Well-known proverb
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 do
we design a car so that most people will not bump their heads on its roof? Or, how high
should a computer table be so that it does not put an unnecessary strain on your wrists and
Copyright © 2010 Pearson Education, Inc.
15.4 y The Normal Distribution
751
arms? In the relatively new science of ergonomics, scientists gather data to answer questions 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 section
called the normal curve, or normal distribution.*
KEY POINT
Many different types of data
sets are normal distributions.
FIGURE 15.12 A normal
distribution.
The Normal Distribution
The normal distribution is the most common distribution in statistics and describes many
real-life data sets. The histogram shown in Figure 15.12 will begin to give you an idea of
the shape of a normal distribution.
Distributions of such diverse data sets as SAT scores, heights of people, the number of
miles 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 all
examples of normal distributions.
There are several patterns in Figure 15.12 that are common to all normal distributions. First, if we were to take a wire and attach it to the tops of the bars in the histogram
and then smooth out the wire to make a curve, the curve would have a bell-shaped
appearance, as shown in Figure 15.13. This is why a normal distribution is often called
a bell-shaped curve.
Second, the mean, median, and mode of a normal distribution are the same. Third, the
curve is symmetric with respect to the mean. This means that if you see some pattern in
the graph on one side of the mean, then the mean acts as a mirror to give a reflection of
that same pattern on the other side of the mean. Fourth, the area under a normal curve
equals 1.
In Figure 15.14, we have marked the mean and two points on the normal curve called
inflection points. An inflection point is a point on the curve where the curve changes from
being 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 normal
curve is symmetric with respect to the mean, one-half, or 50%, of the area under the
curve is located on each side of the mean.
In normal distributions, approximately 68% of the data values occur within 1 standard
deviation 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 to
these facts as the 68-95-99.7 rule. Figure 15.14 illustrates the 68-95-99.7 rule. In
discussing normal distributions, we usually assume that we are dealing with an entire
population rather than a sample, so in Figure 15.14 we represent the mean by m and the
standard deviation by s (rather than qx and s).
We summarize the properties of a normal distribution.
mean =
median =
mode
Inflection points
FIGURE 15.13 Smoothing a
histogram of a normal distribution
into a normal curve.
μ − 3σ μ − 2σ μ − σ
μ
μ + σ μ + 2σ μ + 3σ
68%
95%
99.7%
FIGURE 15.14 The 68-95-99.7 rule for a
normal distribution.
*Recall that a distribution is just another name for a set of data.
Copyright © 2010 Pearson Education, Inc.
752
CHAPTER 15 y Descriptive Statistics
PROPERTIES OF A NORMAL DISTRIBUTION
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, and
99.7% 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 intelligence test is a normal distribution. If the distribution’s mean is 450 and its standard
deviation 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 and
above the mean, respectively.
68%
of scores
mean
450
425 one
standard deviation
below mean
475 one
standard deviation
above mean
FIGURE 15.15 Sixty-eight percent of the scores
lie within 1 standard deviation of the mean.
From the 68-95-99.7 rule, we know that 68%, or about 0.68 of the scores, lie within
1 standard deviation of the mean. Because we have 1,000 scores, we can expect that
about 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 between
2 standard deviations below the mean and 2 standard deviations above the mean.
*To keep this discussion simple, we are approximating these percentages. Shortly, we will use a table to slightly
improve our estimate of the percentage of scores within 1 and 2 standard deviations from the mean.
Copyright © 2010 Pearson Education, Inc.
15.4 y The Normal Distribution
95%
of scores
2.5 percent
of scores here
Quiz Yourself
400
11
Use the information given in
Example 1 to determine the
following.
a) How many scores do we
expect to fall between 400 and
500?
b) How many scores do we
expect to fall below 400?
Areas under a normal curve
represent percentages of
values in the distribution.
2.5 percent
of scores here
500
Two standard
deviations
above mean
FIGURE 15.16 Five percent of the scores lie more than
2 standard deviations from the mean.
This means that 5%, or 0.05, of the scores lie more than 2 standard deviations above or
below 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. ]
KEY POINT
450
mean
Two standard
deviations
below mean
753
11
z-Scores
In 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 the
mean. For example, knowing that the weights of women between ages 18 and 25 form a
normal distribution, a clothing manufacturer may want to know what percentage of the
female population is between 1.3 and 2.5 standard deviations above the mean weight for
women. It is possible to find this by using Table 15.16, which is the table of areas under the
standard normal curve. The standard normal distribution has a mean of 0 and a standard
deviation of 1.
Math in Your Life
How Worried Are You About Your Next Test?*
People often argue about how much of your physical,
intellectual, and psychological makeup is due to your
heredity and how much is due to your environment.
Scientists have identified genes that affect characteristics
such as height, body fat, blood pressure, and IQ.
Surprisingly, even human anxiety has been associated
with genetic makeup.
It has been estimated that there are as many as
15 genes that can affect your anxiety level. If you have
all these genes, then you have a greater tendency† to be
anxious, and if you have none of them, then you tend to
be less anxious. If we were to plot a histogram of the
number of anxiety genes present in a large number of
people, we would find very few people have almost
none of these genes, many people have a moderate number of these genes, and very few people have almost all
of them. If you were to plot the distribution of anxiety
genes present in the population, it would look very
much like the normal curves that you are studying in
this section.
*This note is based on a series of PowerPoint slides by Dr. Lee Bardwell, professor of genetics at the University
of California at Irvine.
†It is believed that 45% of the variance in human anxiety is due to genetic factors. The other 55% is due to other
factors.
Copyright © 2010 Pearson Education, Inc.
754
CHAPTER 15 y Descriptive Statistics
z
A
z
A
z
A
z
A
z
A
z
A
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.10
.11
.12
.13
.14
.15
.16
.17
.18
.19
.20
.21
.22
.23
.24
.25
.26
.27
.28
.29
.30
.31
.32
.33
.34
.35
.36
.37
.38
.39
.40
.41
.42
.43
.44
.45
.46
.47
.48
.49
.50
.51
.52
.53
.54
.55
.000
.004
.008
.012
.016
.020
.024
.028
.032
.036
.040
.044
.048
.052
.056
.060
.064
.068
.071
.075
.079
.083
.087
.091
.095
.099
.103
.106
.110
.114
.118
.122
.126
.129
.133
.137
.141
.144
.148
.152
.155
.159
.163
.166
.170
.174
.177
.181
.184
.188
.192
.195
.199
.202
.205
.209
.56
.57
.58
.59
.60
.61
.62
.63
.64
.65
.66
.67
.68
.69
.70
.71
.72
.73
.74
.75
.76
.77
.78
.79
.80
.81
.82
.83
.84
.85
.86
.87
.88
.89
.90
.91
.92
.93
.94
.95
.96
.97
.98
.99
1.00
1.01
1.02
1.03
1.04
1.05
1.06
1.07
1.08
1.09
1.10
1.11
.212
.216
.219
.222
.226
.229
.232
.236
.239
.242
.245
.249
.252
.255
.258
.261
.264
.267
.270
.273
.276
.279
.282
.285
.288
.291
.294
.297
.300
.302
.305
.308
.311
.313
.316
.319
.321
.324
.326
.329
.332
.334
.337
.339
.341
.344
.346
.349
.351
.353
.355
.358
.360
.362
.364
.367
1.12
1.13
1.14
1.15
1.16
1.17
1.18
1.19
1.20
1.21
1.22
1.23
1.24
1.25
1.26
1.27
1.28
1.29
1.30
1.31
1.32
1.33
1.34
1.35
1.36
1.37
1.38
1.39
1.40
1.41
1.42
1.43
1.44
1.45
1.46
1.47
1.48
1.49
1.50
1.51
1.52
1.53
1.54
1.55
1.56
1.57
1.58
1.59
1.60
1.61
1.62
1.63
1.64
1.65
1.66
1.67
.369
.371
.373
.375
.377
.379
.381
.383
.385
.387
.389
.391
.393
.394
.396
.398
.400
.402
.403
.405
.407
.408
.410
.412
.413
.415
.416
.418
.419
.421
.422
.424
.425
.427
.428
.429
.431
.432
.433
.435
.436
.437
.438
.439
.441
.442
.443
.444
.445
.446
.447
.449
.450
.451
.452
.453
1.68
1.69
1.70
1.71
1.72
1.73
1.74
1.75
1.76
1.77
1.78
1.79
1.80
1.81
1.82
1.83
1.84
1.85
1.86
1.87
1.88
1.89
1.90
1.91
1.92
1.93
1.94
1.95
1.96
1.97
1.98
1.99
2.00
2.01
2.02
2.03
2.04
2.05
2.06
2.07
2.08
2.09
2.10
2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2.19
2.20
2.21
2.22
2.23
.454
.455
.455
.456
.457
.458
.459
.460
.461
.462
.463
.463
.464
.465
.466
.466
.467
.468
.469
.469
.470
.471
.471
.472
.473
.473
.474
.474
.475
.476
.476
.477
.477
.478
.478
.479
.479
.480
.480
.481
.481
.482
.482
.483
.483
.483
.484
.484
.485
.485
.485
.486
.486
.487
.487
.487
2.24
2.25
2.26
2.27
2.28
2.29
2.30
2.31
2.32
2.33
2.34
2.35
2.36
2.37
2.38
2.39
2.40
2.41
2.42
2.43
2.44
2.45
2.46
2.47
2.48
2.49
2.50
2.51
2.52
2.53
2.54
2.55
2.56
2.57
2.58
2.59
2.60
2.61
2.62
2.63
2.64
2.65
2.66
2.67
2.68
2.69
2.70
2.71
2.72
2.73
2.74
2.75
2.76
2.77
2.78
2.79
.488
.488
.488
.488
.489
.489
.489
.490
.490
.490
.490
.491
.491
.491
.491
.492
.492
.492
.492
.493
.493
.493
.493
.493
.493
.494
.494
.494
.494
.494
.495
.495
.495
.495
.495
.495
.495
.496
.496
.496
.496
.496
.496
.496
.496
.496
.497
.497
.497
.497
.497
.497
.497
.497
.497
.497
2.80
2.81
2.82
2.83
2.84
2.85
2.86
2.87
2.88
2.89
2.90
2.91
2.92
2.93
2.94
2.95
2.96
2.97
2.98
2.99
3.00
3.01
3.02
3.03
3.04
3.05
3.06
3.07
3.08
3.09
3.10
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
3.20
3.21
3.22
3.23
3.24
3.25
3.26
3.27
3.28
3.29
3.30
3.31
3.32
3.33
.497
.498
.498
.498
.498
.498
.498
.498
.498
.498
.498
.498
.498
.498
.498
.498
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.499
.500
.500
.500
.500
.500
.500
.500
TABLE 15.16 Standard normal distribution.
Copyright © 2010 Pearson Education, Inc.
15.4 y The Normal Distribution
755
Table 15.16 gives the area under this curve between the mean and a number called a zscore. A z-score represents the number of standard deviations a data value is from the
mean. In Example 1, we found that for a normal distribution with a mean of 450 and a
standard 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 lie
above the mean. We use the symmetry of a normal curve to find areas corresponding to
negative z-scores.
Example 2 shows how to use Table 15.16. Because the total area under a normal curve
is 1, we can interpret areas under the standard normal curve as percentages of the data values in the distribution. Also note that because the mean is 0 and the standard deviation for
the standard normal curve is 1, the value of a z-score is also the same as the number of
standard deviations that z-score is from the mean. After you have practiced working with
the standard normal distribution, we will show you how to use the techniques that you have
learned to solve problems involving real-life data.
PROBLEM SOLVING
The Three-Way Principle
The Three-Way Principle in Section 1.1 suggests that you can often understand a problem
graphically. This is certainly true in solving problems involving the normal curve. If you
draw a good picture of the situation, it often becomes more clear to you as to what to do to
set 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 the
following 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 this
area 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 deviations
above the mean.
The area of
this region is
0.403.
0
mean
1.3
z-score
FIGURE 15.17 Area under the
standard normal curve between z = 0
and z = 1.3.
Copyright © 2010 Pearson Education, Inc.
CHAPTER 15 y Descriptive Statistics
756
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 the
area from z = 0 to z = 1.5. From Table 15.16, we see that when z = 2.1, A = 0.482, and
when z = 1.5, A = 0.433. To finish this problem, we can set up our calculations as follows:
TI Screen showing area under
standard normal curve between
z = 1.5 and z = 2.1.
larger area from z 0 to z 2.1 0.482
smaller area from z 0 to z 1.5 0.433
0.049
area we want
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%.
A = 0.482 − 0.433
A = 0.433
A = 0.466
subtract areas, not z-scores
A = 0.466
0
mean
1.5 2.1
FIGURE 15.18 Finding the area
under the standard normal curve
between z = 1.5 and z = 2.1.
−1.83
0
mean
1.83
FIGURE 15.19 Finding the area
under the standard normal curve
between z = -1.83 and z = 0.
Quiz Yourself
12
Use Table 15.16 to find the
following areas under the
standard 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
KEY POINT
We convert values in a
nonstandard normal
distribution to z-scores.
c) Due to the symmetry of the normal distribution, the area between z = 0 and z = -1.83
is the same as the area between z = 0 and z = 1.83 (Figure 15.19). From Table 15.16
we see that when z = 1.83, A = 0.466. Therefore, 46.6% of the data values lie between
0 and -1.83.
Now try Exercises 17 to 34. ]
12
› Some Good Advice
A common mistake in solving a problem such as Example 2(b) is to subtract 1.5 from 2.1 to
get 0.6 and then wrongly use this for the z-score in Table 15.16. If you think about the graph
of the standard normal curve, clearly the area between z = 0 and z = 0.6 is not the same as the
area between z = 1.5 and z = 2.1.
Converting Raw Scores to z-Scores
A real-life normal distribution, such as the set of all weights of women between ages 18 and
25, may have a mean of 120 pounds and a standard deviation of 25 pounds. Such a distribution will have the properties that we stated earlier for a normal distribution, but because the
distribution does not have a mean of 0 and a standard deviation of 1, we cannot use
Table 15.16 directly, as we did in Example 2. We can use Table 15.16, however, if we first
convert nonstandard values, called raw scores, to z-scores. The following formula shows the
desired 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 T O Z - S C O R E S Assume a
normal distribution has a mean of m and a standard deviation of s. We use the equation
z=
x-m
s
to convert a value x in the nonstandard distribution to a z-score.
Copyright © 2010 Pearson Education, Inc.
15.4 y The Normal Distribution
EXAMPLE 3
757
Converting Raw Scores to z -Scores
Suppose 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 formula
m
z = x -s , with the raw score x = 25, the mean m = 20, and the standard deviation
s = 3. Making the substitutions, we have, z = 25 -3 20 = 53 = 1.67. We can interpret
this result as telling us that in this distribution, 25 is 1.67 standard deviations above
the mean.
20
mean
3
standard deviation
x 16
raw score
Quiz Yourself
FIGURE 15.20 Normal distribution with mean
of 20 and standard deviation of 3.
13
Suppose a normal distribution
has a mean of 50 and a standard
deviation of 7. Convert each raw
score to a z-score.
a) 59
b) 50
c) 38
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 z = 16 -3 20 = -34 = -1.33.
This tells us that 16 is 1.33 standard deviations below the mean.
Now try Exercises 43 to 48. ]
13
0.4
0.08
Slide
0.06
0.3
0.04
0.2
Squeeze
0
Squeeze
0.1
0.02
20 10
10
20
x 25
raw score
30
FIGURE 15.21 (a) First slide
the distribution 20 units to the left
to get a mean of 0.
15 10 5
0
5
10
15
FIGURE 15.21 (b) Then squeeze
the distribution to get a standard
deviation of 1.
Copyright © 2010 Pearson Education, Inc.
The Three-Way Principle in Section 1.1
suggests that you can remember the formula for
converting raw scores to z-scores by interpreting the formula geometrically. In Example 3,
we had a normal distribution with a mean of 20
and standard deviation 3. In order to make the
mean of 20 in the nonstandard distribution correspond to the mean of 0 in the standard normal
distribution, you can think of “sliding” the distribution to the left 20 units, as we show in
Figure 15.21(a). That is why we compute an
expression of the form x - 20 in the numerator
of the z-score formula. Because the standard
deviation was 3, we divided x - 20 by 3 to
“narrow” the distribution so that it would have
a standard deviation of 1, as you can see in
Figure 15.21(b), and therefore allow us to use
Table 15.16.
758
CHAPTER 15 y Descriptive Statistics
Applications
Normal distributions have many practical applications.
EXAMPLE 4
Interpreting the Significance of an Exam Score
Suppose that to qualify for a management training program offered by your employer you
must score in the top 10% of those employees who take a standardized test. Assume that
the distribution of scores is normal and you received a score of 72 on the test, which had a
mean of 65 and a standard deviation of 4. What percentage of those who took this test had
a score below yours?
SOLUTION: We will find the number of standard deviations your score is above the mean
and then determine what percentage of scores fall below this number. First, we calculate
the z-score, which corresponds to your 72:
z=
72 - 65 7
= = 1.75
4
4
Now, looking in Table 15.16 for z = 1.75, we find that A = 0.460. Therefore, 46% of the
scores fall between the mean and your score (Figure 15.22).
A
0.500
A
0.460
65
mean
4
standard deviation
x 75 (raw score)
z 1.75
FIGURE 15.22 Normal distribution with mean
of 65 and standard deviation of 4.
However, we must not forget the scores that fall below the mean. Because a normal curve
is 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 Data
Consider 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.
In the 1940s, the mean batting average was .267 and the standard deviation was .0326.
In the 1970s, the mean batting average was .261 and the standard deviation was .0317.
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 relationship
to his contemporaries.
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15.4 y The Normal Distribution
759
SOLUTION: We will convert each of the three batting averages to a corresponding z-score
for 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
.420 - .266 .154
=
= 4.1509.
.0371
.0371
In the 1940s, Ted Williams’s average of .406 corresponded to a z-score of
.406 - .267 .139
=
= 4.2638.
.0326
.0326
In the 1970s, George Brett’s average of .390 corresponded to a z-score of
.390 - .261 .129
=
= 4.0694.
.0317
.0317
We see that in using the standard deviation to compare each hitter with his contemporaries, Ted Williams was ranked as the best hitter.
Now try Exercises 69 and 70. ]
We conclude our discussion of normal distributions by explaining how manufacturers
can 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 iPod
Touch. In testing the iPod Touch, quality control engineers found that it has a mean time to
failure of 3,000 hours with a standard deviation of 500 hours. Assume that the typical
purchaser will use the device for 4 hours per day. If the manufacturer does not want more
than 5% to be returned as defective within the warranty period, how long should the
warranty period be to guarantee this?
SOLUTION: In solving this problem, we first work with the standard normal distribution
and then convert the answer to fit our nonstandard normal distribution. Figure 15.23 gives
a picture of the situation for the standard normal curve.
5%
45%
z = −1.64
A = 45%
= .45
0
mean
5%
z = 1.64
FIGURE 15.23 Finding the area under the
standard 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 the
entire area is beyond this point. Observe that this score is to the left of the mean and is negative. Therefore, we use symmetry and find the z-score such that 95% of the entire area is
below this score. However, Table 15.16 gives the areas only for the upper half of the distribution. However, this is not a problem because we know that 50% of the entire area lies
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760
CHAPTER 15 y Descriptive Statistics
below the mean. Therefore, our problem reduces to finding a z-score greater than 0 such
that 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 means
that 95% of the area underneath the standard normal curve falls below z = 1.64. By symmetry, we also can say that 95% of the values lie above -1.64. We must now interpret this
z-score in terms of original normal distribution of raw scores.
Recall the equation that relates values in a distribution to z-scores, namely,
z=
x-m
.
s
(1)
Previously we used this equation to convert raw scores to z-scores. We now use the equation 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
-1.64 =
x - 3,000
.
500
(2)
We solve for x in this equation by multiplying both sides by 500:
-1.64 # (500) =
Quiz Yourself
14
Simplifying the equation, we get
Redo Example 6, but now
assume that the mean time to
failure is 4,200 hours, with a
standard deviation of 600, and
the manufacturer wants no more
than 2% to be returned as
defective within the warranty
period.
Exercises
x - 3,000 #
(500)
500
-820 = x - 3,000.
Last, we add 3,000 to both sides to obtain x = 2,180 hours. We expect 95% of the breakdowns to occur beyond 2,180 hours.
Because we know that owners use the iPod Touch about 4 hours per day, we divide
545
2,180 by 4 to get 2,180
4 = 545 days. This is equivalent to 31 = 17.58 months if we use
31 days per month. Therefore, if the manufacturer wants to have no more than 5% of the
breakdowns occur within the warranty period, the warranty should be for roughly
18 months. ] 14
15.4
Looking Back*
These exercises follow the general outline of the topics presented in
this section and will give you a good overview of the material that
you have just studied.
1. List six properties of the normal distribution.
2. What was the common error that we warned you against making in solving Example 2(b)?
3. What property of the normal distribution makes it unnecessary
to have negative z-scores in Table 15.16?
4. State the equation that allows you to convert raw scores to
z-scores?
and a standard deviation of 2. Use the 68-95-99.7 rule to find the
percentage of values in the distribution that we describe.
5. Between 10 and 12
6. Between 12 and 14
7. Above 14
8. Below 8
9. Above 12
10. Below 10
Assume that the distribution in Exercises 11–16 has a mean of 12
and a standard deviation of 3. Use the 68-95-99.7 rule to find the
percentage of values in the distribution that we describe.
11. Below 9
12. Between 12 and 15
13. Above 6
14. Below 12
15. Between 15 and 18
16. Above 18
Use Table 15.16 to find the percentage of area under the standard
normal curve that is specified.
17. Between z = 0 and z = 1.23
Sharpening Your Skills
Assume that all distributions in this exercise set are normal.
Assume that the distribution in Exercises 5–10 has a mean of 10
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 Mathematics
on page xix.
Copyright © 2010 Pearson Education, Inc.
15.4 y Exercises
20. Between z = 0.76 and z = 1.52
761
55. Analyzing vending machines.
21. Between z = 0 and z = -0.75
a. What percentage of the cups should have at least 8 ounces
of coffee?
22. Between z = 0 and z = -1.35
b. What percentage of the cups should have less than 7.5 ounces
of coffee?
23. Between z = 1.25 and z = 1.95
24. Between z = 0.37 and z = 1.23
56. Analyzing vending machines.
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 standard
normal curve is above that score.
36. Find a z-score such that 20% of the area under the standard
normal curve is above that score.
37. Find a z-score such that 12% of the area under the standard
normal curve is below that score.
a. What percentage of the cups should have at least 8.5 ounces
of coffee?
b. What percentage of the cups should have less than 8 ounces
of coffee?
A machine fills bags of candy. Due to
slight irregularities in the operation
of the machine, not every bag gets
exactly the same number of pieces.
Assume that the number of pieces
per bag has a mean of 200 with a standard deviation of 2. Use
Figure 15.14 to solve Exercises 57 and 58.
57. Analyzing vending machines.
38. Find a z-score such that 24% of the area under the standard
normal curve is below that score.
a. How many of the bags do we expect to have between
200 and 202 pieces in them?
39. Find a z-score such that 60% of the area is below that
score.
40. Find a z-score such that 90% of the area is below that
score.
41. Find a z-score such that 75% of the area is above that
score.
b. How many of the bags do we expect to have at least
202 pieces in them?
42. Find a z-score such that 60% of the area is above that
score.
In Exercises 43–48, we give you a mean, a standard deviation, and
a raw score. Find the corresponding z-score.
43. Mean 80, standard deviation 5, x = 87
44. Mean 100, standard deviation 15, x = 117
58. Analyzing vending machines.
a. How many of the bags do we expect to have between
198 and 200 pieces in them?
b. How many of the bags do we expect to have at least
196 pieces in them?
59. Product reliability. Suppose that for a particular brand of
LCD television set, the distribution of failures has a mean of
120 months with a standard deviation of 12 months. With respect
to this distribution, 140 months corresponds to what z-score?
60. Product reliability. Redo Exercise 59 using 130 months
instead of 140.
45. Mean 21, standard deviation 4, x = 14
61. Distribution of heights. Assume that in the NBA the distribution of heights has a mean of 6 feet, 8 inches, and a standard
deviation of 3 inches. If there are 324 players in the league,
how many players do you expect to be over 7 feet tall?
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, and
a 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
62. Distribution of heights. Redo Exercise 61, but now we want
to know how many players you would expect to be over 6 feet,
6 inches.
63. Distribution of heart rates. Assume that the distribution of
21-year-old women’s heart rates at rest is 68 beats per minute
with a standard deviation of 4 beats per minute. If 200 women
are examined, how many would you expect to have a heart rate
of less than 70?
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
64. Distribution of heart rates. Redo Exercise 63, but now we ask
how many you would expect to have a heart rate less than 75.
Applying What You’ve Learned
Due to random variations in the operation of an automatic coffee
machine, not every cup is filled with the same amount of coffee.
Assume that the mean amount of coffee dispensed is 8 ounces and
the standard deviation is 0.5 ounce. Use Figure 15.14 to solve
Exercises 55 and 56.
In Exercises 65 and 66, we assume that the mean length of a human
pregnancy is 268 days.
65. Birth statistics. If 95% of all human pregnancies last between
250 and 286 days, what is the standard deviation of the distribution of lengths of human pregnancies? What percentage of
human pregnancies do we expect to last at least 275 days?
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762
CHAPTER 15 y Descriptive Statistics
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? (See
Exercise 65.)
75. Analyzing the SATs. Assume that the math SAT scores are
normally distributed with a mean of 500 and a standard deviation of 100. If you score 480 on this exam, what percentage of
those taking the test scored below you?
67. Analyzing Internet use. Comcast has analyzed the amount of
time that its customers are online per session. It found that the
distribution of connect times has a mean of 37 minutes and a
standard deviation of 11 minutes. For this distribution, find the
raw score that corresponds to a z-score of 1.5.
76. Analyzing the SATs. Assume that the verbal SAT scores are
normally distributed with a mean of 500 and a standard deviation of 100. If you score 520 on this exam, what percentage of
those taking the test scored below you?
68. Analyzing customer service. The distribution of times that
customers spend waiting in line in a supermarket has a mean
of 3.6 minutes and a standard deviation of 1.2 minutes. For
this distribution, find the raw score that corresponds to a
z-score of -1.3.
In Exercises 69 and 70, use the information and methods of
Example 5 to make the indicated comparisons.
69. Comparing athletes. In 1949, Jackie Robinson hit .342 for
the Brooklyn Dodgers; in 1973, Rod Carew hit .350 for the
Minnesota Twins. Using z-scores to compare each batter with
his contemporaries, determine which batting average was more
impressive.
70. Comparing athletes. In 1940, Joe DiMaggio hit .352 for the
New York Yankees; in 1975, Bill Madlock hit .354 for the
Pittsburgh Pirates. Using z-scores to compare each batter with
his contemporaries, determine which batting average was more
impressive.
71. Writing a warranty. A manufacturer plans to provide a warranty on Wii Fit. In testing, the manufacturer has found that the
set of failure times for Wii Fit is normally distributed with a
mean time of 2,000 hours and a standard deviation of 800 hours.
Assume that the typical purchaser will use Wii Fit for 2 hours
per day. If the manufacturer does not want more than 4%
returned as defective within the warranty period, how long
should the warranty period be to guarantee this? (Assume that
there are 31 days in each month.)
72. Writing a warranty. Redo Exercise 71, but now assume that
the mean of the distribution of failure times is 2,500 hours,
the standard deviation is 600 hours, and the manufacturer
wants no more than 2.5% returned during the warranty
period.
73. Analyzing investments. A certain mutual fund over the last
15 years has a mean yearly return of 7.8% with a standard
deviation of 1.3%.
a. If you had invested in this fund over the past 15 years, in
how many of these years would you expect to earn at least
9% 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 last
12 years has a mean yearly return of 5.9% with a standard
deviation of 1.8%.
a. If you had invested in this fund over the past 12 years, in
how many of these years would you expect to earn at least
8% on your investment?
b. In how many years would you expect to earn less than 4%?
Communicating Mathematics
77. 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 the
distribution?
78. If a raw score corresponds to a z-score of -0.85, what does that
tell you about that score in relationship to the mean of the
distribution?
79. Explain why, if you want to compute the area under a normal
curve between z = 1.3 and z = 2.0, it is not correct to subtract
2.0 - 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 whether
the area under the standard normal curve between z = 0.5 and
z = 1.0 is greater than or less than the area between z = 1.5 and
z = 2.0? Explain.
81. Explain how you would estimate the mean and the standard
deviation of the two normal distributions in the given figure.
Distribution 1
Distribution 2
5
15
20
25
30
82. Explain how you would estimate the mean and the standard
deviation of the two normal distributions in the given figure.
Distribution 1
Distribution 2
3
6
9
12 15 18 21 24 27 30 36 39 41
Using Technology to Investigate
Mathematics
83. You can find many interactive programs that illustrate various
properties of the normal distribution. Search the Internet for “normal distribution applets,” run some, and report on your findings.
84. Search the Internet for some interesting applications of the
normal distribution. While searching, I found many applications
in medicine and even some in music. Report on your findings.
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15.5 y Looking Deeper: Linear Correlation
For Extra Credit
85. 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 compare
Ty Cobb, Ted Williams, and George Brett would be a valid way
to compare the great African American athletes who played in
the Negro leagues with the great white athletes of the same era.
The nth percentile in a distribution is a number x such that n% of
the values in that distribution fall below x. For example, 20% of the
values in the distribution fall below the 20th percentile.
87. What z-score in the standard normal distribution is the 80th
percentile?
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88. What z-score in the standard normal distribution is the 30th
percentile?
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’s
exam 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 your
profession is $53,000 with a standard deviation of $2,500, to
what percentile does your salary of $57,000 correspond?
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