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Section 13.6
The Normal
Curve
Copyright 2013, 2010, 2007, Pearson, Education, Inc.
What You Will Learn
Rectangular Distribution
J-shaped Distribution
Bimodal Distribution
Skewed Distribution
Normal Distribution
z-Scores
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Rectangular Distribution
All the observed values occur with the
same frequency.
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J-shaped Distribution
The frequency is either constantly
increasing or constantly decreasing.
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Bimodal Distribution
Two nonadjacent values occur more
frequently than any other values in a
set of data.
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Skewed Distribution
Has more of a “tail” on one side than
the other.
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Skewed Distribution
Smoothing the histograms of the
skewed distributions to form curves.
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Skewed Distribution
The relationship between the mean,
median, and mode for curves that are
skewed to the right and left.
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Normal Distribution
The most important distribution is the
normal distribution.
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Properties of a Normal Distribution
The graph of a normal distribution is
called the normal curve.
The normal curve is bell shaped and
symmetric about the mean.
In a normal distribution, the mean,
median, and mode all have the same
value and all occur at the center of the
distribution.
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Empirical Rule
Approximately 68% of all the data lie
within one standard deviation of the
mean (in both directions).
Approximately 95% of all the data lie
within two standard deviations of the
mean (in both directions).
Approximately 99.7% of all the data lie
within three standard deviations of the
mean (in both directions).
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z-Scores
z-scores (or standard scores)
determine how far, in terms of
standard deviations, a given score is
from the mean of the distribution.
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z-Scores
The formula for finding z-scores (or
standard scores) is
value of piece of data − mean
z=
standard deviation
x−µ
=
σ
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Example 2: Finding z-scores
A normal distribution has a mean of 80
and a standard deviation of 10.
Find z-scores for the following values.
a) 90 b) 95 c) 80 d) 64
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Example 2: Finding z-scores
Solution
a) 90
value of piece of data − mean
z=
standard deviation
90 − 80
10
z90 =
=
=1
10
10
A value of 90 is 1 standard deviation
above the mean.
13.6-15
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Example 2: Finding z-scores
Solution
b) 95
value of piece of data − mean
z=
standard deviation
95 − 80
15
z95 =
=
= 1.5
10
10
A value of 90 is 1.5 standard
deviations above the mean.
13.6-16
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Example 2: Finding z-scores
Solution
c) 80
value of piece of data − mean
z=
standard deviation
80 − 80
0
z80 =
=
=0
10
10
The mean always has a z-score of 0.
13.6-17
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Example 2: Finding z-scores
Solution
d) 64
value of piece of data − mean
z=
standard deviation
z64
64 − 80
−16
=
=
= −1.6
10
10
A value of 64 is 1.6 standard
deviations below the mean.
13.6-18
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To Determine the Percent of Data
Between any Two Values
1. Draw a diagram of the normal curve
indicating the area or percent to be
determined.
2. Use the formula to convert the given
values to z-scores. Indicate these
z-scores on the diagram.
3. Look up the percent that corresponds
to each z-score in Table 13.7.
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To Determine the Percent of Data
Between any Two Values
a) When finding the percent of data to
the left of a negative z-score, use
Table 13.7(a).
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To Determine the Percent of Data
Between any Two Values
b) When finding the percent of data to
the left of a positive z-score, use
Table 13.7(b).
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To Determine the Percent of Data
Between any Two Values
c) When finding the percent of data to
the right of a z-score, subtract the
percent of data to the left of that zscore from 100%.
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To Determine the Percent of Data
Between any Two Values
c) Or use the symmetry of a normal
distribution.
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To Determine the Percent of Data
Between any Two Values
d) When finding the percent of data
between two z-scores, subtract the
smaller percent from the larger
percent.
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To Determine the Percent of Data
Between any Two Values
4. Change the areas you found in Step
3 to percents as explained earlier.
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Example 5: Horseback Rides
Assume that the length of time for a
horseback ride on the trail at Triple R
Ranch is normally distributed with a
mean of 3.2 hours and a standard
deviation of 0.4 hour.
a) What percent of horseback rides
last at least 3.2 hours?
Solution
In a normal distribution, half the data
are above the mean. Since 3.2 hours
is the mean, 50%, of the horseback
rides last at least 3.2 hours.
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Example 5: Horseback Rides
b) What percent of horseback rides last
less than 2.8 hours?
Solution
Convert 2.8 to a z-score.
2.8 − 3.2
z2.8 =
= −1.00
0.4
The area to the left of –1.00 is 0.1587.
The percent of horseback rides that
last less than 2.8 hours is 15.87%.
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Example 5: Horseback Rides
c) What percent of horseback rides are
at least 3.7 hours?
Solution
Convert 3.7 to a z-score.
3.7 − 3.2
z3.7 =
= 1.25
0.4
Area to left of 1.25 is .8944 = 89.44%.
% above 1.25: 1 – 89.44% = 10.56%.
Thus, 10.56% of horseback rides last
at least 3.7 hours.
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Example 5: Horseback Rides
d) What percent of horseback rides are
between 2.8 hours and 4.0 hours?
Solution
Convert 4.0 to a z-score.
4.0 − 3.2
z4.0 =
= 2.00
0.4
Area to left of 2.00 is .9722 = 97.22%.
Percent below 2.8 is 15.87%.
The percent of data between –1.00 and
2.00 is 97.22% – 15.87% =
81.58%.
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Example 5: Horseback Rides
Solution
Thus, the percent of horseback rides
that last between 2.8 hours and 4.0
hours is 81.85%.
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Example 5: Horseback Rides
e) In a random sample of 500
horseback rides at Triple R Ranch,
how many are at least 3.7 hours?
Solution
In part (c), we determined that
10.56% of all horseback rides last
at least 3.7 hours.
Thus, 0.1056 × 500 = 52.8, or
approximately 53, horseback rides
last at least 3.7 hours.
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