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In this chapter we cover...
Density curves
Describing density curves
Normal distributions
The 68–95–99.7 rule
The standard Normal
distribution
Finding Normal
proportions
Using the standard Normal
table
Finding a value given a
proportion
Stone/Getty Images
CHAPTER
3
The Normal Distributions
We now have a kit of graphical and numerical tools for describing distributions.
What is more, we have a clear strategy for exploring data on a single quantitative
variable.
EXPLORING A DISTRIBUTION
1. Always plot your data: make a graph, usually a histogram or a stemplot.
2. Look for the overall pattern (shape, center, spread) and for striking
deviations such as outliers.
3. Calculate a numerical summary to briefly describe center and spread.
In this chapter, we add one more step to this strategy:
4. Sometimes the overall pattern of a large number of observations is so regular
that we can describe it by a smooth curve.
Density curves
64
Figure 3.1 is a histogram of the scores of all 947 seventh-grade students in Gary,
Indiana, on the vocabulary part of the Iowa Test of Basic Skills.1 Scores of many
students on this national test have a quite regular distribution. The histogram is
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Density curves
2
4
6
8
10
12
Iowa Test vocabulary score
F I G U R E 3 . 1 Histogram of the vocabulary scores of all seventh-grade students in
Gary, Indiana. The smooth curve shows the overall shape of the distribution.
symmetric, and both tails fall off smoothly from a single center peak. There are
no large gaps or obvious outliers. The smooth curve drawn through the tops of the
histogram bars in Figure 3.1 is a good description of the overall pattern of the data.
EXAMPLE 3.1
From histogram to density curve
Our eyes respond to the areas of the bars in a histogram. The bar areas represent proportions of the observations. Figure 3.2(a) is a copy of Figure 3.1 with the leftmost bars
shaded. The area of the shaded bars in Figure 3.2(a) represents the students with vocabulary scores 6.0 or lower. There are 287 such students, who make up the proportion
287/947 = 0.303 of all Gary seventh-graders.
Now look at the curve drawn through the bars. In Figure 3.2(b), the area under the
curve to the left of 6.0 is shaded. We can draw histogram bars taller or shorter by adjusting
the vertical scale. In moving from histogram bars to a smooth curve, we make a specific
choice: adjust the scale of the graph so that the total area under the curve is exactly 1. The
total area represents the proportion 1, that is, all the observations. We can then interpret
areas under the curve as proportions of the observations. The curve is now a density curve.
The shaded area under the density curve in Figure 3.2(b) represents the proportion of
students with score 6.0 or lower. This area is 0.293, only 0.010 away from the actual
proportion 0.303. Areas under the density curve give quite good approximations to the
actual distribution of the 947 test scores.
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C H A P T E R 3 • The Normal Distributions
2
4
6
8
10
12
Iowa Test vocabulary score
F I G U R E 3 . 2 ( a ) The proportion of scores less than
or equal to 6.0 from the histogram is 0.303.
2
4
6
8
10
12
Iowa Test vocabulary score
F I G U R E 3 . 2 ( b ) The proportion of scores less than
or equal to 6.0 from the density curve is 0.293.
DENSITY CURVE
A density curve is a curve that
• is always on or above the horizontal axis, and
• has area exactly 1 underneath it.
A density curve describes the overall pattern of a distribution. The area
under the curve and above any range of values is the proportion of all
observations that fall in that range.
CAUTION
UTION
Density curves, like distributions, come in many shapes. Figure 3.3 shows a
strongly skewed distribution, the survival times of guinea pigs from Exercise 2.34
(page 59). The histogram and density curve were both created from the data by
software. Both show the overall shape and the “bumps” in the long right tail. The
density curve shows a higher single peak as a main feature of the distribution. The
histogram divides the observations near the peak between two bars, thus reducing
the height of the peak. A density curve is often a good description of the overall
pattern of a distribution. Outliers, which are deviations from the overall pattern,
are not described by the curve. Of course, no set of real data is exactly described by
a density curve. The curve is an idealized description that is easy to use and accurate
enough for practical use.
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Describing density curves
0
100
200
300
400
500
600
Survival time (days)
F I G U R E 3 . 3 A right-skewed distribution pictured by both a histogram and a
density curve.
APPLY YOUR KNOWLEDGE
3.1
Sketch density curves. Sketch density curves that describe distributions with
the following shapes:
(a) Symmetric, but with two peaks (that is, two strong clusters of observations).
(b) Single peak and skewed to the left.
Describing density curves
Our measures of center and spread apply to density curves as well as to actual sets
of observations. The median and quartiles are easy. Areas under a density curve
represent proportions of the total number of observations. The median is the point
with half the observations on either side. So the median of a density curve is the
equal-areas point, the point with half the area under the curve to its left and the
remaining half of the area to its right. The quartiles divide the area under the
curve into quarters. One-fourth of the area under the curve is to the left of the first
quartile, and three-fourths of the area is to the left of the third quartile. You can
roughly locate the median and quartiles of any density curve by eye by dividing
the area under the curve into four equal parts.
Because density curves are idealized patterns, a symmetric density curve is exactly symmetric. The median of a symmetric density curve is therefore at its center.
Figure 3.4(a) shows a symmetric density curve with the median marked. It isn’t so
easy to spot the equal-areas point on a skewed curve. There are mathematical ways
of finding the median for any density curve. That’s how we marked the median on
the skewed curve in Figure 3.4(b).
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C H A P T E R 3 • The Normal Distributions
The long right tail pulls
the mean to the right.
Mean
Median
Median and mean
F I G U R E 3 . 4 ( a ) The median and mean of
a symmetric density curve both lie at the center
of symmetry.
F I G U R E 3 . 4 ( b ) The median and mean of a
right-skewed density curve. The mean is pulled away
from the median toward the long tail.
What about the mean? The mean of a set of observations is their arithmetic
average. If we think of the observations as weights strung out along a thin rod, the
mean is the point at which the rod would balance. This fact is also true of density
curves. The mean is the point at which the curve would balance if made of solid material.
Figure 3.5 illustrates this fact about the mean. A symmetric curve balances at its
center because the two sides are identical. The mean and median of a symmetric
density curve are equal, as in Figure 3.4(a). We know that the mean of a skewed
distribution is pulled toward the long tail. Figure 3.4(b) shows how the mean of a
skewed density curve is pulled toward the long tail more than is the median. It’s
hard to locate the balance point by eye on a skewed curve. There are mathematical
ways of calculating the mean for any density curve, so we are able to mark the mean
as well as the median in Figure 3.4(b).
MEDIAN AND MEAN OF A DENSITY CURVE
The median of a density curve is the equal-areas point, the point that
divides the area under the curve in half.
The mean of a density curve is the balance point, at which the curve would
balance if made of solid material.
The median and mean are the same for a symmetric density curve. They
both lie at the center of the curve. The mean of a skewed curve is pulled
away from the median in the direction of the long tail.
F I G U R E 3 . 5 The mean is the balance point of a density curve.
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Describing density curves
We can roughly locate the mean, median, and quartiles of any density curve by
eye. This is not true of the standard deviation. When necessary, we can once again
call on more advanced mathematics to learn the value of the standard deviation.
The study of mathematical methods for doing calculations with density curves is
part of theoretical statistics. Though we are concentrating on statistical practice,
we often make use of the results of mathematical study.
Because a density curve is an idealized description of a distribution of data, we
need to distinguish between the mean and standard deviation of the density curve
and the mean x and standard deviation s computed from the actual observations.
The usual notation for the mean of a density curve is μ (the Greek letter mu).
We write the standard deviation of a density curve as σ (the Greek letter sigma).
69
mean μ
standard deviation σ
APPLY YOUR KNOWLEDGE
3.2
A uniform distribution. Figure 3.6 displays the density curve of a uniform
distribution. The curve takes the constant value 1 over the interval from 0 to 1
and is zero outside that range of values. This means that data described by this
distribution take values that are uniformly spread between 0 and 1. Use areas
under this density curve to answer the following questions.
(a) Why is the total area under this curve equal to 1?
(b) What percent of the observations lie above 0.8?
(c) What percent of the observations lie below 0.6?
(d) What percent of the observations lie between 0.25 and 0.75?
height = 1
0
3.3
3.4
F I G U R E 3 . 6 The density
curve of a uniform
distribution, for Exercises 3.2
and 3.3.
1
Mean and median. What is the mean μ of the density curve pictured in Figure
3.6? What is the median?
Mean and median. Figure 3.7 displays three density curves, each with three
points marked on them. At which of these points on each curve do the mean and
the median fall?
A
A BC
(a)
B
(b)
AB C
C
(c)
F I G U R E 3 . 7 Three density
curves, for Exercise 3.4.
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C H A P T E R 3 • The Normal Distributions
Normal distributions
One particularly important class of density curves has already appeared in Figures
3.1 and 3.2. These density curves are symmetric, single-peaked, and bell-shaped.
They are called Normal curves, and they describe Normal distributions. Normal distributions play a large role in statistics, but they are rather special and not at all
“normal” in the sense of being usual or average. We capitalize Normal to remind
you that these curves are special. All Normal distributions have the same overall
shape. The exact density curve for a particular Normal distribution is described by giving
its mean μ and its standard deviation σ . The mean is located at the center of the
symmetric curve and is the same as the median. Changing μ without changing
σ moves the Normal curve along the horizontal axis without changing its spread.
The standard deviation σ controls the spread of a Normal curve. Figure 3.8 shows
two Normal curves with different values of σ . The curve with the larger standard
deviation is more spread out.
σ
σ
μ
μ
F I G U R E 3 . 8 Two Normal curves, showing the mean μ and standard deviation σ .
The standard deviation σ is the natural measure of spread for Normal distributions. Not only do μ and σ completely determine the shape of a Normal curve,
but we can locate σ by eye on the curve. Here’s how. Imagine that you are skiing
down a mountain that has the shape of a Normal curve. At first, you descend at
an ever-steeper angle as you go out from the peak:
Fortunately, before you find yourself going straight down, the slope begins to
grow flatter rather than steeper as you go out and down:
The points at which this change of curvature takes place are located at distance σ on either
side of the mean μ. You can feel the change as you run a pencil along a Normal
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The 68–95–99.7 rule
curve, and so find the standard deviation. Remember that μ and σ alone do not
specify the shape of most distributions, and that the shape of density curves in general
does not reveal σ . These are special properties of Normal distributions.
NORMAL DISTRIBUTIONS
A Normal distribution is described by a Normal density curve. Any
particular Normal distribution is completely specified by two numbers, its
mean and standard deviation.
The mean of a Normal distribution is at the center of the symmetric
Normal curve. The standard deviation is the distance from the center to
the change-of-curvature points on either side.
Why are the Normal distributions important in statistics? Here are three reasons. First, Normal distributions are good descriptions for some distributions of real
data. Distributions that are often close to Normal include scores on tests taken by
many people (such as Iowa Tests and SAT exams), repeated careful measurements
of the same quantity, and characteristics of biological populations (such as lengths
of crickets and yields of corn). Second, Normal distributions are good approximations to the results of many kinds of chance outcomes, such as the proportion of
heads in many tosses of a coin. Third, we will see that many statistical inference
procedures based on Normal distributions work well for other roughly symmetric distributions. However, many sets of data do not follow a Normal distribution.
Most income distributions, for example, are skewed to the right and so are not
Normal. Non-Normal data, like nonnormal people, not only are common but are
sometimes more interesting than their Normal counterparts.
The 68–95–99.7 rule
Although there are many Normal curves, they all have common properties. In
particular, all Normal distributions obey the following rule.
THE 68–95–99.7 RULE
In the Normal distribution with mean μ and standard deviation σ :
• Approximately 68% of the observations fall within σ of the mean μ.
• Approximately 95% of the observations fall within 2σ of μ.
• Approximately 99.7% of the observations fall within 3σ of μ.
Figure 3.9 illustrates the 68–95–99.7 rule. By remembering these three numbers, you can think about Normal distributions without constantly making detailed
calculations.
CAUTION
UTION
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C H A P T E R 3 • The Normal Distributions
F I G U R E 3 . 9 The
68–95–99.7 rule for Normal
distributions.
68% of data
95% of data
99.7% of data
−3
−2
−1
0
1
2
3
Standard deviations
EXAMPLE 3.2
Iowa Test scores
Figures 3.1 and 3.2 show that the distribution of Iowa Test vocabulary scores for seventhgrade students in Gary, Indiana, is close to Normal. Suppose that the distribution is
exactly Normal with mean μ = 6.84 and standard deviation σ = 1.55. (These are the
mean and standard deviation of the 947 actual scores.)
Figure 3.10 applies the 68–95–99.7 rule to Iowa Test scores. The 95 part of the rule
says that 95% of all scores are between
μ − 2σ = 6.84 − (2)(1.55) = 6.84 − 3.10 = 3.74
and
μ + 2σ = 6.84 + (2)(1.55) = 6.84 + 3.10 = 9.94
The other 5% of scores are outside this range. Because Normal distributions are symmetric, half these scores are lower than 3.74 and half are higher than 9.94. That is, 2.5% of
the scores are below 3.74 and 2.5% are above 9.94.
CAUTION
UTION
The 68–95–99.7 rule describes distributions that are exactly Normal. Real data such
as the actual Gary scores are never exactly Normal. For one thing, Iowa Test scores
are reported only to the nearest tenth. A score can be 9.9 or 10.0, but not 9.94.
We use a Normal distribution because it’s a good approximation, and because we
think the knowledge that the test measures is continuous rather than stopping at
tenths.
How well does our work in Example 3.2 describe the actual Iowa Test scores?
Well, 900 of the 947 scores are between 3.74 and 9.94. That’s 95.04%, very accurate indeed. Of the remaining 47 scores, 20 are below 3.74 and 27 are above 9.94.
The tails of the actual data are not quite equal, as they would be in an exactly
Normal distribution. Normal distributions often describe real data better in the
center of the distribution than in the extreme high and low tails.
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The 68–95–99.7 rule
One standard deviation
is 1.55.
68% of data
95% of data
2.5% of scores
are below 3.74.
99.7% of data
2.19
3.74
5.29
6.84
8.39
9.94
11.49
Iowa Test score
F I G U R E 3 . 1 0 The 68–95–99.7 rule applied to the distribution of Iowa Test scores in
Gary, Indiana, with μ = 6.84 and σ = 1.55.
EXAMPLE 3.3
Iowa Test scores
Look again at Figure 3.10. A score of 5.29 is one standard deviation below the mean.
What percent of scores are higher than 5.29? Find the answer by adding areas in the
figure. Here is the calculation in pictures:
=
+
68%
5.29
8.39
percent between 5.29 and 8.39 +
68%
+
16%
84%
8.39
percent above 8.39
16%
5.29
=
=
percent above 5.29
84%
Be sure you see where the 16% came from: 32% of scores are outside the range 5.29 to
8.39, and half of these are above 8.39.
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C H A P T E R 3 • The Normal Distributions
Because we will mention Normal distributions often, a short notation is helpful. We abbreviate the Normal distribution with mean μ and standard deviation
σ as N(μ, σ ). For example, the distribution of Gary Iowa Test scores is approximately N(6.84, 1.55).
APPLY YOUR KNOWLEDGE
Jim McGuire/Index Stock Imagery/
Picture Quest
3.5
Heights of young women. The distribution of heights of women aged 20 to 29
is approximately Normal with mean 64 inches and standard deviation 2.7
inches.2 Draw a Normal curve on which this mean and standard deviation are
correctly located. (Hint: Draw the curve first, locate the points where the
curvature changes, then mark the horizontal axis.)
3.6
Heights of young women. The distribution of heights of women aged 20 to 29
is approximately Normal with mean 64 inches and standard deviation 2.7 inches.
Use the 68–95–99.7 rule to answer the following questions. (Start by making a
sketch like Figure 3.10.)
(a) Between what heights do the middle 95% of young women fall?
(b) What percent of young women are taller than 61.3 inches?
3.7
Length of pregnancies. The length of human pregnancies from conception to
birth varies according to a distribution that is approximately Normal with mean
266 days and standard deviation 16 days. Use the 68–95–99.7 rule to answer the
following questions.
(a) Between what values do the lengths of almost all (99.7%) pregnancies fall?
(b) How short are the shortest 2.5% of all pregnancies?
The standard Normal distribution
As the 68–95–99.7 rule suggests, all Normal distributions share many common
properties. In fact, all Normal distributions are the same if we measure in units of
size σ about the mean μ as center. Changing to these units is called standardizing.
To standardize a value, subtract the mean of the distribution and then divide by
the standard deviation.
STANDARDIZING AND z-SCORES
If x is an observation from a distribution that has mean μ and standard
deviation σ , the standardized value of x is
x −μ
z=
σ
A standardized value is often called a z-score.
A z-score tells us how many standard deviations the original observation
falls away from the mean, and in which direction. Observations larger than the
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The standard Normal distribution
75
mean are positive when standardized, and observations smaller than the mean are
negative.
EXAMPLE 3.4
Standardizing women’s heights
The heights of young women are approximately Normal with μ = 64 inches and σ =
2.7 inches. The standardized height is
height − 64
z=
2.7
A woman’s standardized height is the number of standard deviations by which her height
differs from the mean height of all young women. A woman 70 inches tall, for example,
has standardized height
70 − 64
z=
= 2.22
2.7
or 2.22 standard deviations above the mean. Similarly, a woman 5 feet (60 inches) tall
has standardized height
60 − 64
z=
= −1.48
2.7
or 1.48 standard deviations less than the mean height.
We often standardize observations from symmetric distributions to express
them in a common scale. We might, for example, compare the heights of two
children of different ages by calculating their z-scores. The standardized heights
tell us where each child stands in the distribution for his or her age group.
If the variable we standardize has a Normal distribution, standardizing does
more than give a common scale. It makes all Normal distributions into a single
distribution, and this distribution is still Normal. Standardizing a variable that
has any Normal distribution produces a new variable that has the standard Normal
distribution.
STANDARD NORMAL DISTRIBUTION
The standard Normal distribution is the Normal distribution N(0, 1) with
mean 0 and standard deviation 1.
If a variable x has any Normal distribution N(μ, σ ) with mean μ and
standard deviation σ , then the standardized variable
x −μ
z=
σ
has the standard Normal distribution.
APPLY YOUR KNOWLEDGE
3.8
SAT versus ACT. Eleanor scores 680 on the mathematics part of the SAT. The
distribution of SAT math scores in recent years has been Normal with mean 518
He said, she said.
The height and weight distributions
in this chapter come from actual
measurements by a government
survey. Good thing that is. When
asked their weight, almost all
women say they weigh less than
they really do. Heavier men also
underreport their weight—but
lighter men claim to weigh more
than the scale shows. We leave you
to ponder the psychology of the two
sexes. Just remember that “say so” is
no substitute for measuring.
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C H A P T E R 3 • The Normal Distributions
and standard deviation 114. Gerald takes the ACT Assessment mathematics test
and scores 27. ACT math scores are Normally distributed with mean 20.7 and
standard deviation 5.0. Find the standardized scores for both students. Assuming
that both tests measure the same kind of ability, who has the higher score?
3.9
Men’s and women’s heights. The heights of women aged 20 to 29 are
approximately Normal with mean 64 inches and standard deviation 2.7 inches.
Men the same age have mean height 69.3 inches with standard deviation 2.8
inches. What are the z-scores for a woman 6 feet tall and a man 6 feet tall? Say
in simple language what information the z-scores give that the actual heights do
not.
Finding Normal proportions
Spencer Grant/PhotoEdit
Areas under a Normal curve represent proportions of observations from that Normal distribution. There is no formula for areas under a Normal curve. Calculations
use either software that calculates areas or a table of areas. The table and most software calculate one kind of area, cumulative proportions.
CUMULATIVE PROPORTIONS
The cumulative proportion for a value x in a distribution is the proportion
of observations in the distribution that lie at or below x.
Cumulative
proportion
x
The key to calculating Normal proportions is to match the area you want with
areas that represent cumulative proportions. If you make a sketch of the area you
want, you will almost never go wrong. Find areas for cumulative proportions either
from software or (with an extra step) from a table. The following example shows
the method in a picture.
EXAMPLE 3.5
Who qualifies for college sports?
The National Collegiate Athletic Association (NCAA) requires Division I athletes to
score at least 820 on the combined mathematics and verbal parts of the SAT exam
in order to compete in their first college year. (Higher scores are required for students
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Finding Normal proportions
with poor high school grades.) The scores of the millions of high school seniors taking the SATs in recent years are approximately Normal with mean 1026 and standard deviation 209. What percent of high school seniors qualify for Division I college
sports?
Here is the calculation in a picture: the proportion of scores above 820 is the area
under the curve to the right of 820. That’s the total area under the curve (which is
always 1) minus the cumulative proportion up to 820.
−
=
820
820
area right of 820
=
=
total area
1
−
−
area left of 820
0.1622
= 0.8378
About 84% of all high school seniors meet the NCAA requirement to compete in Division I college sports.
There is no area under a smooth curve and exactly over the point 820. Consequently, the area to the right of 820 (the proportion of scores > 820) is the same as
the area at or to the right of this point (the proportion of scores ≥ 820). The actual
data may contain a student who scored exactly 820 on the SAT. That the proportion of scores exactly equal to 820 is 0 for a Normal distribution is a consequence
of the idealized smoothing of Normal distributions for data.
To find the numerical value 0.1622 of the cumulative proportion in Example
3.5 using software, plug in mean 1026 and standard deviation 209 and ask for
the cumulative proportion for 820. Software often uses terms such as “cumulative
distribution” or “cumulative probability.” We will learn in Chapter 10 why the
language of probability fits. Here, for example, is Minitab’s output:
Cumulative Distribution Function
Normal with mean = 1026 and standard deviation = 209
x
820
P ( X <= x )
0.162153
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C H A P T E R 3 • The Normal Distributions
APPLET
The P in the output stands for “probability,”but we can read it as “proportion of the
observations.” CrunchIt! and the Normal Curve applet are even handier because
they draw pictures as well as finding areas. If you are not using software, you can
find cumulative proportions for Normal curves from a table. That requires an extra
step.
Using the standard Normal table*
The extra step in finding cumulative proportions from a table is that we must
first standardize to express the problem in the standard scale of z-scores. This allows us to get by with just one table, a table of standard Normal cumulative proportions. Table A in the back of the book gives cumulative proportions for the
standard Normal distribution. The pictures at the top of the table remind us that
the entries are cumulative proportions, areas under the curve to the left of a
value z.
EXAMPLE 3.6
The standard Normal table
What proportion of observations on a standard Normal variable z take values less than
1.47?
Solution: To find the area to the left of 1.47, locate 1.4 in the left-hand column of
Table A, then locate the remaining digit 7 as .07 in the top row. The entry opposite 1.4
and under .07 is 0.9292. This is the cumulative proportion we seek. Figure 3.11 illustrates
this area.
Table entry:
area = 0.9292
F I G U R E 3 . 1 1 The area
under a standard Normal
curve to the left of the point
z = 1.47 is 0.9292. Table A
gives areas under the
standard Normal curve.
z = 1.47
Now that you see how Table A works, let’s redo Example 3.5 using the table.
We can break Normal calculations using the table into three steps.
∗
This section is unnecessary if you will always use software for Normal distribution calculations.
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Using the standard Normal table
EXAMPLE 3.7
Who qualifies for college sports?
Scores of high school seniors on the SAT exam follow the Normal distribution with
mean μ = 1026 and standard deviation σ = 209. What percent of seniors score at least
820?
1. Draw a picture. The picture is exactly as in Example 3.5.
2. Standardize. Call the SAT score x. Subtract the mean, then divide by the standard
deviation, to transform the problem about x into a problem about a standard
Normal z:
x ≥ 820
820 − 1026
x − 1026
≥
209
209
z ≥ −0.99
3. Use the table. The picture says that we want the cumulative proportion for
x = 820. Step 2 says this is the same as the cumulative proportion for z = −0.99.
The Table A entry for z = −0.99 says that this cumulative proportion is 0.1611.
The area to the right of −0.99 is therefore 1 − 0.1611 = 0.8389.
The area from the table in Example 3.7 (0.8389) is slightly less accurate than
the area from software in Example 3.5 (0.8378) because we must round z to two
decimal places when we use Table A. The difference is rarely important in practice.
Here’s the method in outline form.
USING TABLE A TO FIND NORMAL PROPORTIONS
1. State the problem in terms of the observed variable x. Draw a picture
that shows the proportion you want in terms of cumulative proportions.
2. Standardize x to restate the problem in terms of a standard Normal
variable z.
3. Use Table A and the fact that the total area under the curve is 1 to find
the required area under the standard Normal curve.
EXAMPLE 3.8
Who qualifies for an athletic scholarship?
The NCAA considers a student a “partial qualifier”if the combined SAT score is at least
720. Partial qualifiers can receive athletic scholarships and practice with the team, but
they can’t compete during their first college year. What proportion of all students who
take the SAT would be partial qualifiers?
1. State the problem and draw a picture. Call the SAT score x. The variable x has the
N(1026, 209) distribution. What proportion of SAT scores fall between 720 and
820? See the following picture.
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C H A P T E R 3 • The Normal Distributions
−
=
720
820
820
z = −0.99
720
z = −1.46
2. Standardize. Subtract the mean, then divide by the standard deviation, to turn x
into a standard Normal z:
720
≤
x
<
820
720 − 1026
x − 1026
820 − 1026
≤
<
209
209
209
−1.46
≤
z
<
−0.99
3. Use the table. Follow the picture (we added the z-scores to the picture label to help
you):
area between −1.46 and −0.99 = (area left of −0.99) − (area left of −1.46)
= 0.1611 − 0.0721 = 0.0890
About 9% of high school seniors would be partial qualifiers.
Sometimes we encounter a value of z more extreme than those appearing in
Table A. For example, the area to the left of z = −4 is not given directly in the
table. The z-values in Table A leave only area 0.0002 in each tail unaccounted
for. For practical purposes, we can act as if there is zero area outside the range of
Table A.
APPLY YOUR KNOWLEDGE
CORBIS
3.10 Use the Normal table. Use Table A to find the proportion of observations from
a standard Normal distribution that satisfies each of the following statements. In
each case, sketch a standard Normal curve and shade the area under the curve
that is the answer to the question.
(a) z < 2.85
(b) z > 2.85
(c) z > −1.66
(d) −1.66 < z < 2.85
3.11 How hard do locomotives pull? An important measure of the performance of a
locomotive is its “adhesion,” which is the locomotive’s pulling force as a multiple
of its weight. The adhesion of one 4400-horsepower diesel locomotive model
varies in actual use according to a Normal distribution with mean μ = 0.37 and
standard deviation σ = 0.04.
(a) What proportion of adhesions measured in use are higher than 0.40?
(b) What proportion of adhesions are between 0.40 and 0.50?
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Finding a value given a proportion
3.12 A better locomotive. Improvements in the locomotive’s computer controls
change the distribution of adhesion to a Normal distribution with mean μ = 0.41
and standard deviation σ = 0.02. Find the proportions in (a) and (b) of the
previous exercise after this improvement.
Finding a value given a proportion
Examples 3.5 to 3.8 illustrate the use of software or Table A to find what proportion of the observations satisfies some condition, such as “SAT score above 820.”
We may instead want to find the observed value with a given proportion of the
observations above or below it. Statistical software will do this directly.
EXAMPLE 3.9
Find the top 10% using software
Scores on the SAT verbal test in recent years follow approximately the N(504, 111)
distribution. How high must a student score in order to place in the top 10% of all
students taking the SAT?
We want to find the SAT score x with area 0.1 to its right under the Normal curve
with mean μ = 504 and standard deviation σ = 111. That’s the same as finding the
SAT score x with area 0.9 to its left. Figure 3.12 poses the question in graphical form.
Most software will tell you x when you plug in mean 504, standard deviation 111, and
cumulative proportion 0.9. Here is Minitab’s output:
Inverse Cumulative Distribution Function
Normal with mean = 504 and standard deviation = 111
P( X <= x )
x
0.9
646.252
Minitab gives x = 646.252. So scores above 647 are in the top 10%. (Round up
because SAT scores can only be whole numbers.)
Area = 0.90
x = 504
z=0
Area = 0.10
x=?
z = 1.28
F I G U R E 3 . 1 2 Locating the point on a Normal curve with area 0.10 to its right.
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C H A P T E R 3 • The Normal Distributions
Without software, use Table A backward. Find the given proportion in the
body of the table and then read the corresponding z from the left column and top
row. There are again three steps.
EXAMPLE 3.10
Find the top 10% using Table A
Scores on the SAT verbal test in recent years follow approximately the N(504, 111)
distribution. How high must a student score in order to place in the top 10% of all
students taking the SAT?
1. State the problem and draw a picture. This step is exactly as in Example 3.9. The
picture is Figure 3.12.
2. Use the table. Look in the body of Table A for the entry closest to 0.9. It is 0.8997.
This is the entry corresponding to z = 1.28. So z = 1.28 is the standardized value
with area 0.9 to its left.
3. Unstandardize to transform z back to the original x scale. We know that the
standardized value of the unknown x is z = 1.28. So x itself satisfies
x − 504
= 1.28
111
Solving this equation for x gives
x = 504 + (1.28)(111) = 646.1
This equation should make sense: it says that x lies 1.28 standard deviations above
the mean on this particular Normal curve. That is the “unstandardized” meaning
of z = 1.28. A student must score at least 647 to place in the highest 10%.
Here is the general formula for unstandardizing a z-score. To find the value x
from the Normal distribution with mean μ and standard deviation σ corresponding to a given standard Normal value z, use
x = μ + zσ
EXAMPLE 3.11
Find the third quartile
High levels of cholesterol in the blood increase the risk of heart disease. For 14-year-old
boys, the distribution of blood cholesterol is approximately Normal with mean μ = 170
milligrams of cholesterol per deciliter of blood (mg/dl) and the standard deviation σ =
30 mg/dl.3 What is the third quartile of the distribution of blood cholesterol?
1. State the problem and draw a picture. Call the cholesterol level x. The variable x has
the N(170, 30) distribution. The third quartile is the value with 75% of the
distribution to its left. Figure 3.13 is the picture.
2. Use the table. Look in the body of Table A for the entry closest to 0.75. It is 0.7486.
This is the entry corresponding to z = 0.67. So z = 0.67 is the standardized value
with area 0.75 to its left.
3. Unstandardize. The cholesterol level corresponding to z = 0.67 is
x = μ + zσ
= 170 + (0.67)(30) = 190.1
The third quartile of blood cholesterol levels in 14-year-old boys is about 190 mg/dl.
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Chapter 3 Summary
83
F I G U R E 3 . 1 3 Locating the
third quartile of a Normal curve.
The third quartile
always has area
0.75 to its left.
Area = 0.75
x=?
z = 0.67
APPLY YOUR KNOWLEDGE
3.13 Table A. Use Table A to find the value z of a standard Normal variable that
satisfies each of the following conditions. (Use the value of z from Table A that
comes closest to satisfying the condition.) In each case, sketch a standard Normal
curve with your value of z marked on the axis.
(a) The point z with 25% of the observations falling below it.
(b) The point z with 40% of the observations falling above it.
3.14 IQ test scores. Scores on the Wechsler Adult Intelligence Scale are
approximately Normally distributed with μ = 100 and σ = 15.
(a) What scores fall in the lowest 25% of the distribution?
(b) How high a score is needed to be in the highest 5%?
The bell curve?
C H A P T E R 3 SUMMARY
We can sometimes describe the overall pattern of a distribution by a density
curve. A density curve has total area 1 underneath it. An area under a density
curve gives the proportion of observations that fall in a range of values.
A density curve is an idealized description of the overall pattern of a distribution
that smooths out the irregularities in the actual data. We write the mean of a
density curve as μ and the standard deviation of a density curve as σ to
distinguish them from the mean x and standard deviation s of the actual data.
The mean, the median, and the quartiles of a density curve can be located by eye.
The mean μ is the balance point of the curve. The median divides the area
under the curve in half. The quartiles and the median divide the area under the
curve into quarters. The standard deviation σ cannot be located by eye on most
density curves.
Does the distribution of human
intelligence follow the “bell curve”
of a Normal distribution? Scores on
IQ tests do roughly follow a Normal
distribution. That is because a test
score is calculated from a person’s
answers in a way that is designed to
produce a Normal distribution. To
conclude that intelligence follows a
bell curve, we must agree that the
test scores directly measure
intelligence. Many psychologists
don’t think there is one human
characteristic that we can call
“intelligence” and can measure by a
single test score.
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C H A P T E R 3 • The Normal Distributions
The mean and median are equal for symmetric density curves. The mean of a
skewed curve is located farther toward the long tail than is the median.
The Normal distributions are described by a special family of bell-shaped,
symmetric density curves, called Normal curves. The mean μ and standard
deviation σ completely specify a Normal distribution N(μ, σ ). The mean is the
center of the curve, and σ is the distance from μ to the change-of-curvature
points on either side.
To standardize any observation x, subtract the mean of the distribution and then
divide by the standard deviation. The resulting z-score
x −μ
σ
says how many standard deviations x lies from the distribution mean.
All Normal distributions are the same when measurements are transformed to
the standardized scale. In particular, all Normal distributions satisfy the
68–95–99.7 rule, which describes what percent of observations lie within one,
two, and three standard deviations of the mean.
If x has the N(μ, σ ) distribution, then the standardized variable z = (x − μ)/σ
has the standard Normal distribution N(0, 1) with mean 0 and standard
deviation 1. Table A gives the cumulative proportions of standard Normal
observations that are less than z for many values of z. By standardizing, we can
use Table A for any Normal distribution.
z=
CHECK YOUR SKILLS
3.15 Which of these variables is least likely to have a Normal distribution?
(a) Income per person for 150 different countries
(b) Lengths of 50 newly hatched pythons
(c) Heights of 100 white pine trees in a forest
3.16 To completely specify the shape of a Normal distribution, you must give
(a) the mean and the standard deviation.
(b) the five-number summary.
(c) the mean and the median.
3.17 Figure 3.14 shows a Normal curve. The mean of this distribution is
(a) 0.
(b) 2.
(c) 3.
3.18 The standard deviation of the Normal distribution in Figure 3.14 is
(a) 2.
(b) 3.
(c) 5.
3.19 The length of human pregnancies from conception to birth varies according to a
distribution that is approximately Normal with mean 266 days and standard
deviation 16 days. 95% of all pregnancies last between
(a) 250 and 282 days.
(b) 234 and 298 days.
(c) 218 and 314 days.
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Chapter 3 Exercises
−8 −6 −4 −2 0
2
4
6
8 10 12
F I G U R E 3 . 1 4 A Normal curve, for Exercises 3.17 and 3.18.
3.20 The scale of scores on an IQ test is approximately Normal with mean 100 and
standard deviation 15. The organization MENSA, which calls itself “the high-IQ
society,” requires an IQ score of 130 or higher for membership. What percent of
adults would qualify for membership?
(a) 95%
(b) 5%
(c) 2.5%
3.21 The scores of adults on an IQ test are approximately Normal with mean 100 and
standard deviation 15. Corinne scores 118 on such a test. Her z-score is about
(a) 1.2.
(b) 7.87.
(c) 18.
3.22 The proportion of observations from a standard Normal distribution that take
values less than 1.15 is about
(a) 0.1251.
(b) 0.8531.
(c) 0.8749.
3.23 The proportion of observations from a standard Normal distribution that take
values larger than −0.75 is about
(a) 0.2266.
(b) 0.7734.
(c) 0.8023.
3.24 The scores of adults on an IQ test are approximately Normal with mean 100 and
standard deviation 15. Corinne scores 118 on such a test. She scores higher than
what percent of all adults?
(a) About 12%
(b) About 88%
(b) About 98%
C H A P T E R 3 EXERCISES
3.25 Understanding density curves. Remember that it is areas under a density curve,
not the height of the curve, that give proportions in a distribution. To illustrate
this, sketch a density curve that has its peak at 0 on the horizontal axis but has
greater area within 0.25 on either side of 1 than within 0.25 on either side of 0.
3.26 Are the data Normal? Soil penetrability. Table 2.3 (page 61) gives data on the
penetrability of soil at each of three levels of compression. We might expect the
penetrability of specimens of the same soil at the same level of compression to
follow a Normal distribution. Make stemplots of the data for loose and for
intermediate compression. Does either sample seem roughly Normal? Does either
appear distinctly non-Normal? If so, what kind of departure from Normality does
your stemplot show?
3.27 IQ test scores. The Wechsler Adult Intelligence Scale (WAIS) is the most
common “IQ test.” The scale of scores is set separately for each age group and is
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C H A P T E R 3 • The Normal Distributions
approximately Normal with mean 100 and standard deviation 15. According to
the 68–95–99.7 rule, about what percent of people have WAIS scores
(a) above 100?
(b) above 145?
(c) below 85?
3.28 Low IQ test scores. Scores on the Wechsler Adult Intelligence Scale (WAIS)
are approximately Normal with mean 100 and standard deviation 15. People with
WAIS scores below 70 are considered mentally retarded when, for example,
applying for Social Security disability benefits. According to the 68–95–99.7 rule,
about what percent of adults are retarded by this criterion?
3.29 Actual IQ test scores. Here are the IQ test scores of 31 seventh-grade girls in a
Midwest school district:4
114
108
111
100
130
103
104
120
74
89
132
112
102
111
107
91
128
103
114
118
98
114
119
96
103
86
112
105
72
112
93
(a) We expect IQ scores to be approximately Normal. Make a stemplot to check
that there are no major departures from Normality.
(b) Nonetheless, proportions calculated from a Normal distribution are not
always very accurate for small numbers of observations. Find the mean x and
standard deviation s for these IQ scores. What proportions of the scores are
within one standard deviation and within two standard deviations of the mean?
What would these proportions be in an exactly Normal distribution?
3.30 Standard Normal drill. Use Table A to find the proportion of observations from
a standard Normal distribution that falls in each of the following regions. In each
case, sketch a standard Normal curve and shade the area representing the region.
(a) z ≤ −2.25
(b) z ≥ −2.25
(c) z > 1.77
(d) −2.25 < z < 1.77
3.31 Standard Normal drill.
(a) Find the number z such that the proportion of observations that are less than
z in a standard Normal distribution is 0.8.
(b) Find the number z such that 35% of all observations from a standard Normal
distribution are greater than z.
ACT versus SAT. There are two major tests of readiness for college: the ACT and
the SAT. ACT scores are reported on a scale from 1 to 36. The distribution of ACT
scores in recent years has been roughly Normal with mean μ = 20.9 and standard
deviation σ = 4.8. SAT scores are reported on a scale from 400 to 1600. SAT scores
have been roughly Normal with mean μ = 1026 and standard deviation σ = 209.
Exercises 3.32 to 3.43 are based on this information.
3.32 Tonya scores 1318 on the SAT. Jermaine scores 27 on the ACT. Assuming that
both tests measure the same thing, who has the higher score?
3.33 Jacob scores 16 on the ACT. Emily scores 670 on the SAT. Assuming that both
tests measure the same thing, who has the higher score?
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Chapter 3 Exercises
3.34 José scores 1287 on the SAT. Assuming that both tests measure the same thing,
what score on the ACT is equivalent to José’s SAT score?
3.35 Maria scores 28 on the ACT. Assuming that both tests measure the same thing,
what score on the SAT is equivalent to Maria’s ACT score?
3.36 Reports on a student’s ACT or SAT usually give the percentile as well as the
actual score. The percentile is just the cumulative proportion stated as a percent:
the percent of all scores that were lower than this one. Tonya scores 1318 on the
SAT. What is her percentile?
3.37 Reports on a student’s ACT or SAT usually give the percentile as well as the
actual score. The percentile is just the cumulative proportion stated as a percent:
the percent of all scores that were lower than this one. Jacob scores 16 on the
ACT. What is his percentile?
3.38 It is possible to score higher than 1600 on the SAT, but scores 1600 and above are
reported as 1600. What proportion of SAT scores are reported as 1600?
3.39 It is possible to score higher than 36 on the ACT, but scores 36 and above are
reported as 36. What proportion of ACT scores are reported as 36?
3.40 What SAT scores make up the top 10% of all scores?
3.41 How well must Abigail do on the ACT in order to place in the top 20% of all
students?
3.42 The quartiles of any distribution are the values with cumulative proportions 0.25
and 0.75. What are the quartiles of the distribution of ACT scores?
3.43 The quintiles of any distribution are the values with cumulative proportions 0.20,
0.40, 0.60, and 0.80. What are the quintiles of the distribution of SAT scores?
3.44 Heights of men and women. The heights of women aged 20 to 29 follow
approximately the N(64, 2.7) distribution. Men the same age have heights
distributed as N(69.3, 2.8). What percent of young women are taller than the
mean height of young men?
3.45 Heights of men and women. The heights of women aged 20 to 29 follow
approximately the N(64, 2.7) distribution. Men the same age have heights
distributed as N(69.3, 2.8). What percent of young men are shorter than the
mean height of young women?
3.46 A surprising calculation. Changing the mean of a Normal distribution by a
moderate amount can greatly change the percent of observations in the tails.
Suppose that a college is looking for applicants with SAT math scores 750 and
above.
(a) In 2004, the scores of men on the math SAT followed the N(537, 116)
distribution. What percent of men scored 750 or better?
(b) Women’s SAT math scores that year had the N(501, 110) distribution. What
percent of women scored 750 or better? You see that the percent of men above
750 is almost three times the percent of women with such high scores. Why this is
true is controversial.
3.47 Grading managers. Many companies “grade on a bell curve” to compare the
performance of their managers and professional workers. This forces the use of
some low performance ratings so that not all workers are listed as “above
average.” Ford Motor Company’s “performance management process” for a
time assigned 10% A grades, 80% B grades, and 10% C grades to the company’s
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C H A P T E R 3 • The Normal Distributions
18,000 managers. Suppose that Ford’s performance scores really are Normally
distributed. This year, managers with scores less than 25 received C’s and those
with scores above 475 received A’s. What are the mean and standard deviation of
the scores?
APPLET
3.48 Osteoporosis. Osteoporosis is a condition in which the bones become brittle
due to loss of minerals. To diagnose osteoporosis, an elaborate apparatus measures
bone mineral density (BMD). BMD is usually reported in standardized form. The
standardization is based on a population of healthy young adults. The World
Health Organization (WHO) criterion for osteoporosis is a BMD 2.5 standard
deviations below the mean for young adults. BMD measurements in a population
of people similar in age and sex roughly follow a Normal distribution.
(a) What percent of healthy young adults have osteoporosis by the WHO
criterion?
(b) Women aged 70 to 79 are of course not young adults. The mean BMD in this
age is about −2 on the standard scale for young adults. Suppose that the standard
deviation is the same as for young adults. What percent of this older population
has osteoporosis?
3.49 Are the data Normal? ACT scores. Scores on the ACT test for the 2004 high
school graduating class had mean 20.9 and standard deviation 4.8. In all,
1,171,460 students in this class took the test, and 1,052,490 of them had scores of
27 or lower.5 If the distribution of scores were Normal, what percent of scores
would be 27 or lower? What percent of the actual scores were 27 or lower? Does
the Normal distribution describe the actual data well?
3.50 Are the data Normal? Student loans. A government report looked at the
amount borrowed for college by students who graduated in 2000 and had taken
out student loans.6 The mean amount was x = $17,776 and the standard
deviation was s = $12,034. The quartiles were Q1 = $9900, M = $15,532, and
Q3 = $22,500.
(a) Compare the mean x and the median M. Also compare the distances of Q1
and Q3 from the median. Explain why both comparisons suggest that the
distribution is right-skewed.
(b) The right skew pulls the standard deviation up. So a Normal distribution
with the same mean and standard deviation would have a third quartile larger
than the actual Q3 . Find the third quartile of the Normal distribution with
μ = $17,776 and σ = $12,034 and compare it with Q3 = $22,500.
The Normal Curve applet allows you to do Normal calculations quickly. It is somewhat
limited by the number of pixels available for use, so that it can’t hit every value exactly.
In the exercises below, use the closest available values. In each case, make a sketch of the
curve from the applet marked with the values you used to answer the questions.
3.51 How accurate is 68–95–99.7? The 68–95–99.7 rule for Normal distributions is
a useful approximation. To see how accurate the rule is, drag one flag across the
other so that the applet shows the area under the curve between the two flags.
(a) Place the flags one standard deviation on either side of the mean. What
is the area between these two values? What does the 68–95–99.7 rule say this
area is?
(b) Repeat for locations two and three standard deviations on either side of the
mean. Again compare the 68–95–99.7 rule with the area given by the applet.
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Chapter 3 Exercises
3.52 Where are the quartiles? How many standard deviations above and below the
mean do the quartiles of any Normal distribution lie? (Use the standard Normal
distribution to answer this question.)
3.53 Grading managers. In Exercise 3.47, we saw that Ford Motor Company grades
its managers in such a way that the top 10% receive an A grade, the bottom 10%
a C, and the middle 80% a B. Let’s suppose that performance scores follow a
Normal distribution. How many standard deviations above and below the mean
do the A/B and B/C cutoffs lie? (Use the standard Normal distribution to answer
this question.)
APPLET
APPLET
89