Download Section 5-2 Properties of Normal Distribution

5.2 Properties of the Normal
Distribution
LEARNING GOAL
Know how to interpret the normal distribution in terms of
the 68-95-99.7 rule, standard scores, and percentiles.
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Consider a Consumer Reports survey in which
participants were asked how long they owned their
last TV set before they replaced it. The variable of
interest in this survey is replacement time for
television sets.
Based on the survey, the distribution of replacement
times has a mean of about 8.2 years, which we denote
as m (the Greek letter mu).
The standard deviation of the distribution is about 1.1
years, which we denote as s (the Greek letter sigma).
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Slide 5.2- 2
Making the reasonable assumption that the
distribution of TV replacement times is approximately
normal, we can picture it as shown in Figure 5.16.
Figure 5.16 Normal distribution for replacement times for TV sets with a
mean of m 8.2 years and a standard deviation of s 1.1 years.
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Slide 5.2- 3
TIME OUT TO THINK
Apply the four criteria for a normal distribution (see
Section 5.1) to explain why the distribution of TV
replacement times should be approximately normal.
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Slide 5.2- 4
A simple rule, called the 68-95-99.7 rule, gives
precise guidelines for the percentage of data values
that lie within 1, 2, and 3 standard deviations of the
mean for any normal distribution.
Figure 5.17 Normal distribution illustrating the 68-95-99.7 rule.
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Slide 5.2- 5
The 68-95-99.7 Rule for a Normal Distribution
• About 68% (more precisely, 68.3%), or just over twothirds, of the data points fall within 1 standard deviation
of the mean.
• About 95% (more precisely, 95.4%) of the data points
fall within 2 standard deviations of the mean.
• About 99.7% of the data points fall within 3 standard
deviations of the mean.
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Slide 5.2- 6
EXAMPLE 1 SAT Scores
The tests that make up the verbal (critical reading) and
mathematics SAT (and the GRE, LSAT, and GMAT) are
designed so that their scores are normally distributed with a
mean of m = 500 and a standard deviation of s = 100. Interpret
this statement.
Solution: From the 68-95-99.7 rule, about 68% of students
have scores within 1 standard deviation (100 points) of the
mean of 500 points; that is, about 68% of students score
between 400 and 600.
About 95% of students score within 2 standard deviations (200
points) of the mean, or between 300 and 700.
And about 99.7% of students score within 3 standard deviations
(300 points) of the mean, or between 200 and 800.
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Slide 5.2- 7
EXAMPLE 1 SAT Scores
Solution: (cont.)
Figure 5.18 shows this interpretation graphically; note
that the horizontal axis shows both actual scores and
distance from the mean in standard deviations.
Figure 5.18 Normal distribution for SAT scores, showing the percentages
associated with 1, 2, and 3 standard deviations.
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Slide 5.2- 8
EXAMPLE 2 Detecting Counterfeits
Vending machines can be adjusted to reject coins above and
below certain weights. The weights of legal U.S. quarters have a
normal distribution with a mean of 5.67 grams and a standard
deviation of 0.0700 gram. If a vending machine is adjusted to
reject quarters that weigh more than 5.81 grams and less than 5.53
grams, what percentage of legal quarters will be rejected by the
machine?
Solution: A weight of 5.81 is 0.14 gram, or 2 standard deviations,
above the mean. A weight of 5.53 is 0.14 gram, or 2 standard
deviations, below the mean. Therefore, by accepting
only quarters within the weight range 5.53 to 5.81 grams, the
machine accepts quarters that are within 2 standard deviations of
the mean and rejects those that are more than 2 standard
deviations from the mean. By the 68-95-99.7 rule, 95% of legal
quarters will be accepted and 5% of legal quarters will be rejected.
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Slide 5.2- 9
Applying the 68-95-99.7 Rule
We can apply the 68-95-99.7 rule to determine when
data values lie 1, 2, or 3 standard deviations from the
mean.
For example, suppose that 1,000 students take an
exam and the scores are normally distributed with a
mean of m = 75 and a standard deviation of s = 7.
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Slide 5.2- 10
Figure 5.19 A normal distribution of test scores with a mean of 75 and a
standard deviation of 7. (a) 68% of the scores lie within 1 standard
deviation of the mean. (b) 95% of the scores lie within 2 standard deviations
of the mean.
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Slide 5.2- 11
Identifying Unusual Results
In statistics, we often need to distinguish values that are
typical, or “usual,” from values that are “unusual.” By
applying the 68-95-99.7 rule, we find that about 95% of all
values from a normal distribution lie within 2 standard
deviations of the mean.
This implies that, among all values, 5% lie more than 2
standard deviations away from the mean. We can use
this property to identify values that are relatively
“unusual”:
Unusual values are values that are more than 2 standard
deviations away from the mean.
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Slide 5.2- 12
EXAMPLE 4 Normal Heart Rate
You measure your resting heart rate at noon every day for a year
and record the data. You discover that the data have a normal
distribution with a mean of 66 and a standard deviation of 4. On
how many days was your heart rate below 58 beats per minute?
Solution: A heart rate of 58 is 8 (or 2 standard deviations) below
the mean. According to the 68-95-99.7 rule, about 95% of the data
points are within 2 standard deviations of the mean.
Therefore, 2.5% of the data points are more than 2 standard
deviations below the mean, and 2.5% of the data points are more
than 2 standard deviations above the mean. On 2.5% of 365 days,
or about 9 days, your measured heart rate was below 58 beats per
minute.
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Slide 5.2- 13
TIME OUT TO THINK
As Example 4 suggests, many measurements of the
resting heart rate of a single individual are normally
distributed. Would you expect the average resting heart
rates of many individuals to be normally distributed?
Which distribution would you expect to have the larger
standard deviation? Why?
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Slide 5.2- 14
EXAMPLE 5 Finding a Percentile
On a visit to the doctor’s office, your fourth-grade daughter is told
that her height is 1 standard deviation above the mean for her age
and sex. What is her percentile for height? Assume that heights of
fourth-grade girls are normally distributed.
Solution: Recall that a data value lies in the nth percentile of a
distribution if n% of the data values are less than or equal to it
(see Section 4.3). According to the 68-95-99.7 rule, 68% of the
heights are within 1 standard deviation of the mean. Therefore,
34% of the heights (half of 68%) are between 0 and 1 standard
deviation above the mean. We also know that, because the
distribution is symmetric, 50% of all heights are below the mean.
Therefore, 50% + 34% = 84% of all heights are less than 1
standard deviation above the mean (Figure 5.21). Your daughter is
in the 84th percentile for heights among fourth-grade girls.
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Slide 5.2- 15
Figure 5.21 Normal distribution curve showing 84% of scores less than 1
standard deviation above the mean.
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Slide 5.2- 16
Standard Scores
Computing Standard Scores
The number of standard deviations a data value lies
above or below the mean is called its standard score (or
z-score), defined by
data value – mean
z = standard score =
standard deviation
The standard score is positive for data values above the
mean and negative for data values below the mean.
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Slide 5.2- 17
EXAMPLE 6 Finding Standard Scores
The Stanford-Binet IQ test is scaled so that scores have a mean of
100 and a standard deviation of 16. Find the standard scores for
IQs of 85, 100, and 125.
Solution: We calculate the standard scores for these IQs by using
the standard score formula with a mean of 100 and standard
deviation of 16.
standard score for 125: z =
85 – 100
= -0.94
16
100 – 100
standard score for 100: z =
= 0.00
16
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Slide 5.2- 18
EXAMPLE 6 Finding Standard Scores
Solution: (cont.)
125 – 100
standard score for 125: z =
= 1.56
16
We can interpret these standard scores as follows: 85 is 0.94
standard deviation below the mean, 100 is equal to the mean, and
125 is 1.56 standard deviations above the mean.
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Slide 5.2- 19
Figure 5.22 shows the values on the distribution of IQ
scores from Example 6.
Figure 5.22 Standard scores for IQ scores of 85, 100, and 125.
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Slide 5.2- 20
Standard Scores and Percentiles
Once we know the standard score of a data value, the
properties of the normal distribution allow us to find
its percentile in the distribution. This is usually done
with a standard score table, such as Table 5.1 (next
slide).
(Appendix A has a more detailed standard score
table.)
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Slide 5.2- 21
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Slide 5.2- 22
EXAMPLE 7 Cholesterol Levels
Cholesterol levels in men 18 to 24 years of age are normally
distributed with a mean of 178 and a standard deviation of 41.
a. What is the percentile for a 20-year-old man with a cholesterol
level of 190?
Solution:
a.The standard score for a cholesterol level of 190 is
data value – mean
190 – 178
z = standard score =
=
≈ 0.29
standard deviation
41
Table 5.1 shows that a standard score of 0.29 corresponds to about
the 61st percentile.
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Slide 5.2- 23
EXAMPLE 7 Cholesterol Levels
Cholesterol levels in men 18 to 24 years of age are normally
distributed with a mean of 178 and a standard deviation of 41.
b. What cholesterol level corresponds to the 90th percentile, the
level at which treatment may be necessary?
Solution:
b. Table 5.1 shows that 90.32% of all data values have a standard
score less than 1.3. Thus, the 90th percentile is about 1.3 standard
deviations above the mean. Given the mean cholesterol level of
178 and the standard deviation of 41, a cholesterol level 1.3
standard deviations above the mean is
A cholesterol level of about 231 corresponds to the 90th percentile.
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Slide 5.2- 24
Toward Probability
Suppose you pick a baby at random and ask whether the baby
was born more than 15 days prior to his or her due date. Because
births are normally distributed around the due date with a
standard deviation of 15 days, we know that 16% of all births
occur more than 15 days prior to the due date (see Example 3).
For an individual baby chosen at random, we can therefore say
that there’s a 0.16 chance (about 1 in 6) that the baby was born
more than 15 days early.
In other words, the properties of the normal distribution allow us
to make a probability statement about an individual. In this case,
our statement is that the probability of a birth occurring more
than 15 days early is 0.16.
This example shows that the properties of the normal distribution
can be restated in terms of ideas of probability.
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The End
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