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Copyright © 2009 Pearson Education, Inc.
Chapter 13 Section 7 – Slide 1
Chapter 13
Statistics
Copyright © 2009 Pearson Education, Inc.
Chapter 13 Section 7 – Slide 2
WHAT YOU WILL LEARN
•
•
•
•
Sampling techniques
Misuses of statistics
Frequency distributions
Histograms, frequency polygons,
stem-and-leaf displays
• Mode, median, mean, and
midrange
• Percentiles and quartiles
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Chapter 13 Section 7 – Slide 3
WHAT YOU WILL LEARN
• Range and standard deviation
• z-scores and the normal distribution
• Correlation and regression
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Chapter 13 Section 7 – Slide 4
Section 7
The Normal Curve
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Chapter 13 Section 7 – Slide 5
Types of Distributions
Rectangular Distribution

J-shaped distribution
Rectangular Distribution
Frequency

Values
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Chapter 13 Section 7 – Slide 6
Types of Distributions (continued)

Bimodal
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
Skewed to right
Chapter 13 Section 7 – Slide 7
Types of Distributions (continued)

Skewed to left
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
Normal
Chapter 13 Section 7 – Slide 8
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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Chapter 13 Section 7 – Slide 9
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).
Copyright © 2009 Pearson Education, Inc.
Chapter 13 Section 7 – Slide 10
z-Scores

z-scores determine how far, in terms of
standard deviations, a given score is from the
mean of the distribution.
value of piece of data  mean x  
z

standard deviation

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Chapter 13 Section 7 – Slide 11
Example: z-scores



A normal distribution has a mean of 50 and a
standard deviation of 5. Find z-scores for the
following values.
a) 55 b) 60 c) 43
value of piece of data  mean
a)
z
standard deviation
55  50 5
z55 
 1
5
5
A score of 55 is one standard deviation
above the mean.
Copyright © 2009 Pearson Education, Inc.
Chapter 13 Section 7 – Slide 12
Example: z-scores (continued)

60  50 10
b) z60 

2
5
5
A score of 60 is 2 standard deviations above the
mean.

43  50 7
c) z43 

 1.4
5
5
A score of 43 is 1.4 standard deviations below
the mean.
Copyright © 2009 Pearson Education, Inc.
Chapter 13 Section 7 – Slide 13
To Find the Percent of Data Between
any Two Values
1.
2.
3.
Draw a diagram of the normal curve,
indicating the area or percent to be
determined.
Use the formula to convert the given
values to z-scores. Indicate these zscores on the diagram.
Look up the percent that corresponds to
each z-score in Table 13.7.
Copyright © 2009 Pearson Education, Inc.
Chapter 13 Section 7 – Slide 14
To Find the Percent of Data Between
any Two Values (continued)
4.
a) When finding the percent of data to the left of
a negative z-score, use Table 13.7(a).
b) When finding the percent of data to the left of
a positive z-score, use Table 13.7(b).
c) When finding the percent of data to the right
of a z-score, subtract the percent of data to
the left of that z-score from 100%.
d) When finding the percent of data between
two z-scores, subtract the smaller percent
from the larger percent.
Copyright © 2009 Pearson Education, Inc.
Chapter 13 Section 7 – Slide 15
Example
Assume that the waiting times for customers at a
popular restaurant before being seated for lunch
are normally distributed with a mean of 12
minutes and a standard deviation of 3 min.
a) Find the percent of customers who wait for at
least 12 minutes before being seated.
b) Find the percent of customers who wait between
9 and 18 minutes before being seated.
c) Find the percent of customers who wait at least
17 minutes before being seated.
d) Find the percent of customers who wait less than
8 minutes before being seated.
Copyright © 2009 Pearson Education, Inc.
Chapter 13 Section 7 – Slide 16
Solution
a. wait for at least 12
minutes
Since 12 minutes is the
mean, half, or 50% of
customers wait at least
12 min before being
seated.
b. between 9 and 18
minutes
9  12
z9 
 1.00
3
18  12
z18 
 2.00
3
Use table 13.7 on pages
889-89 in the 8th edition.
97.7% - 15.9% = 81.8%
Copyright © 2009 Pearson Education, Inc.
Chapter 13 Section 7 – Slide 17
Solution (continued)
c. at least 17 min
17  12
z17 
 1.67
3
Use table 13.7(b) page
889.
100% - 95.3% = 4.7%
Thus, 4.7% of
customers wait at least
17 minutes.
Copyright © 2009 Pearson Education, Inc.
d. less than 8 min
8  12
z8 
 1.33
3
Use table 13.7(a) page
889.
Thus, 9.2% of
customers wait less
than 8 minutes.
Chapter 13 Section 7 – Slide 18