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Advanced Algebra
The Normal Curve
Ernesto Diaz
Assistant Professor of Mathematics
Copyright © 2016 Brooks/Cole Cengage Learning
14.3
The Normal Curve
Copyright © Cengage Learning. All rights reserved.
Cumulative Distributions
3
Cumulative Distributions
A cumulative frequency is the sum of all preceding
frequencies in which some order has been established.
4
Example 1 – Find the mean, median, and mode
A judge ordered a survey to determine how many of the
offenders appearing in her court during the past year had
three or more previous appearances. The accumulated
data are shown in Table 14.9.
Number of Previous Appearances
Table 14.9
5
Example 1 – Find the mean, median, and mode
cont’d
Note the last column shows the cumulative percent. This is
usually called the cumulative relative frequency. Use the
cumulative relative frequency to find the percent who had
three or more previous appearances, and then find the
mean, median, and mode.
Solution:
From the cumulative relative frequency we see that 70%
had 2 or fewer court appearances; we see that since the
total is 100%, 30% must have had 3 or more previous
appearances.
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Example 1 – Solution
cont’d
The mean is found (using the idea of a weighted mean).
Note that the sum is 100%, or 1:
= 1.93
The median is the number of court appearances for which
the cumulative percent first exceeds 50%; we see that this
is 2 court appearances.
The mode is the number of court appearances that occurs
most frequently; we see that this is 1 court appearance.
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Bell-Shaped Curves
8
Bell-Shaped Curves
Suppose we survey the results of 20 children’s scores on
an IQ test.
The scores (rounded to the nearest 5 points) are 115, 90,
100, 95, 105, 95, 105, 105, 95, 125, 120, 110, 100, 100,
90, 110, 100, 115, 105, and 80.
We can find
= 103 and s  10.93.
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Bell-Shaped Curves
A frequency graph of these data is shown in part a of
Figure 14.34. If we consider 10,000 scores instead of only
20, we might obtain the frequency distribution shown in part
b of Figure 14.34.
a. IQs of 20 children
b. IQs of 10,000 children
Frequency distributions for IQ scores
Figure 14.34
10
The Normal Distribution:
as mathematical function (pdf)
f ( x) 
1
 2
Note constants:
=3.14159
e=2.71828
1 x 2
 (
)
2

e
This is a bell shaped curve
with different centers and
spreads depending on 
and 
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**The beauty of the normal curve:
No matter what  and  are, the area between - and
+ is about 68%; the area between -2 and +2 is
about 95%; and the area between -3 and +3 is
about 99.7%. Almost all values fall within 3 standard
deviations.
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68-95-99.7 Rule
68% of
the data
95% of the data
99.7% of the data
13
Bell-Shaped Curves
The data illustrated in Figure 14.34 approximate a
commonly used curve called a normal frequency curve, or
simply a normal curve. (See Figure 14.35.)
A normal curve
Figure 14.35
14
Bell-Shaped Curves
If we obtain the frequency distribution of a large number of
measurements (as with IQ), the corresponding graph tends
to look normal, or bell-shaped. The normal curve has
some interesting properties.
In it, the mean, the median, and the mode all have the
same value, and all occur exactly at the center of the
distribution; we denote this value by the Greek letter mu ().
The standard deviation for this distribution is  (sigma).
15
Bell-Shaped Curves
The normal distribution is a continuous (rather than a
discrete) distribution, and it extends indefinitely in both
directions, never touching the x-axis.
It is symmetric about a vertical line drawn through the
mean, . Graphs of this curve for several choices of  are
shown in Figure 14.36.
Variations of normal curves
Figure 14.36
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Example 2 – Find a normal distribution
Predict the distribution of IQ scores of 1,000 people if we
assume that IQ scores are normally distributed, with a
mean of 100 and a standard deviation of 15.
Solution:
First, find the breaking points around the mean.
For  = 100 and  = 15:
 +  = 100 + 15 = 115
 –  = 100 – 15 = 85
 + 2 = 100 + 2(15) = 130
 – 2 = 100 – 2(15) = 70
 + 3 = 100 + 3(15) = 145
 – 3 = 100 – 3(15) = 55
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Example 2 – Solution
cont’d
We use Figure 14.35 to find that 34.1% of the scores will
be between 100 and 115 (i.e.,between  and  + 1) :
0.341  1,000 = 341
A normal curve
Figure 14.35
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Example 2 – Solution
cont’d
About 13.6% will be between 115 and 130
(between  + 1 and  + 2):
0.136  1,000 = 136
About 2.2% will be between 130 and 145
(between  + 2 and  + 3):
0.022  1,000 = 22
About 0.1% will be above 145 (more than  + 3):
0.001  1,000 = 1
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Example 2 – Solution
cont’d
The distribution for intervals below the mean is identical,
since the normal curve is the same to the left and to the
right of the mean. The distribution is shown below.
20
z-Scores
21
z-Scores
First, we introduce some terminology. We use z-scores
(sometimes called standard scores) to determine how far,
in terms of standard deviations, a given score is from the
mean of the distribution.
We use the z-score to translate any normal curve into a
standard normal curve (the particular normal curve with a
mean of 0 and a standard deviation of 1) by using the
definition.
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z-Scores
23
Example 5 – Find probabilities in a normal distribution
The Eureka Lightbulb Company tested a new line of
lightbulbs and found their lifetimes to be normally
distributed, with a mean life of 98 hours and a standard
deviation of 13 hours.
a. What percentage of bulbs will last less than 72 hours?
b. What percentage of bulbs will last less than 100 hours?
c. What is the probability that a bulb selected at random will
last longer than 111 hours?
d. What is the probability that a bulb will last between 106
and 120 hours?
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Example 5 – Solution
cont’d
Draw a normal curve with mean 98 and standard deviation
13, as shown in Figure 14.38.
Lightbulb lifetimes are normally distributed
Figure 14.38
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Example 5 – Solution
cont’d
For this example, we are given  = 98 and  = 13 .
a. For x = 72, z =
= –2.
This is 2 standard deviations below the mean.
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Example 5 – Solution
cont’d
The percentage we seek is shown in blue in Figure 14.39a.
2 standard deviations below the mean
Using z -scores
Figure 14.39(a)
About 2.3% (2.2% + 0.1% = 2.3%) will last less than 72
hours. We can also use the z-score to find the percentage.
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Example 5 – Solution
cont’d
From Table 14.10, the area to the left of z = –2 is
0.5 – 0.4772 = 0.0228, so the probability is about 2.3%.
Standard Normal Distribution
Table 14.10
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Example 5 – Solution
cont’d
Standard Normal Distribution
Table 14.10
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Example 5 – Solution
b. For x = 100, z =
cont’d
 0.15.
This is 0.15 standard deviation above the mean. From
Table 14.10, we find 0.0596 (this is shown in green in
Figure 14.39b).
0.15 standard deviations above the mean
Using z-scores
Figure 14.39(b)
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Example 5 – Solution
cont’d
Since we want the percent of values less than 100, we
must add 50% for the numbers below the mean (shown
in blue). The percentage we seek is
0.5000 + 0.0596 = 0.5596 or about 56.0%.
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Example 5 – Solution
c. For x = 111, z =
cont’d
= 1.
This is one standard deviation above the mean, which is
the same as the z-score. The percentage we seek is
shown in blue (see Figure 14.39c).
1 standard deviations above the mean
Using z-scores
Figure 14.39(c)
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Example 5 – Solution
cont’d
We know that about 15.9% (13.6% + 2.2% + 0.1% = 15.9%)
of the bulbs will last longer than 111 hours, so
P(bulb life > 111 hours)  0.159
We can also use Table 14.10. For z = 1.00, the table entry
is 0.3413, and we are looking for the area to the right, so we
compute
0.5000 – 0.3413 = 0.1587
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Example 5 – Solution
cont’d
d. We first find the z-scores using  = 98 and  = 13. For
x = 106, z =
 0.62.
From Table 14.10, the area between the z-score and the
mean is 0.2324 (shown in green).
For x = 120, z =
 1.69.
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Example 5 – Solution
cont’d
From Table 14.10, the area between this z-score and the
mean is 0.4545. The desired answer (shown in yellow) is
approximately
0.4545 – 0.2324 = 0.2221
Since percent and probability are the same, we see the
probability that the life of the bulb is between 106 and
120 hours is about 22.2%.
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z-Scores
36
z-Scores
Sometimes data do not fall into a normal distribution, but
are skewed, which means their distribution has more tail
on one side or the other. For example, Figure 14.40a
shows that the 1941 scores on the SAT exam (when the
test was first used) were normally distributed.
1941 SAT scale
Distribution of SAT scores
Figure 14.40(a)
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z-Scores
However, by 1990, the scale had become skewed to the
left, as shown in Figure 14.40b.
1990 SAT scale
Distribution of SAT scores
Figure 14.40(b)
38
z-Scores
In a normal distribution, the mean, median, and mode all
have the same value, but if the distribution is skewed, the
relative positions of the mean, median, and mode would be
as shown in Figure 14.41.
a. Skewed to the right
(positive skew)
b. Normal distribution
c. Skewed to the left
(negative skew)
Comparison of three distributions
Figure 14.41
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