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Chapter 6
Probability Distributions
Section 6.2
Probabilities for Bell-Shaped Distributions
Copyright © 2013, 2009, and 2007, Pearson Education, Inc.
Normal Distribution
The normal distribution is symmetric, bell-shaped and
characterized by its mean  and standard deviation  .
 The normal distribution is the most important
distribution in statistics.
 Many distributions have an approximately normal
distribution.
 The normal distribution also can approximate
many discrete distributions well when there are a
large number of possible outcomes.
 Many statistical methods use it even when the
data are not bell shaped.
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Normal Distribution
Normal distributions are
 Bell shaped
 Symmetric around the mean
The mean ( ) and the standard deviation ( ) completely
describe the density curve.
 Increasing/decreasing  moves the curve along the
horizontal axis.
 Increasing/decreasing  controls the spread of the
curve.
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Normal Distribution
Within what interval do almost all of the men’s heights
fall? Women’s height?
Figure 6.4 Normal Distributions for Women’s Height and Men’s Height. For each different
combination of  and  values, there is a normal distribution with mean  and standard
deviation  . Question: Given that  = 70 and  = 4, within what interval do almost all of the
men’s heights fall?
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Normal Distribution: 68-95-99.7 Rule for
Any Normal Curve
≈ 68% of the observations fall within one standard deviation of the mean.
≈ 95% of the observations fall within two standard deviations of the mean.
≈ 99.7% of the observations fall within three standard deviations of the mean.
Figure 6.5 The Normal Distribution. The probability equals approximately 0.68 within
1 standard deviation of the mean, approximately 0.95 within 2 standard deviations,
and approximately 0.997 within 3 standard deviations. Question: How do these
probabilities relate to the empirical rule?
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Example : 68-95-99.7% Rule
Heights of adult women can be approximated by a normal
distribution,
  65
inches;
  3.5
inches
68-95-99.7 Rule for women’s heights:
68% are between 61.5 and 68.5 inches
 [     65  3.5]
95% are between 58 and 72 inches
 [   2  65  2(3.5)  65  7]
99.7% are between 54.5 and 75.5 inches
 [   3  65  3(3.5)  65  10.5]
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Z-Scores and the Standard Normal
Distribution
The z-score for a value x of a random variable is the
number of standard deviations that x falls from the mean.
z
x 

A negative (positive) z-score indicates that the value is
below (above) the mean.
Z-scores
can be used to calculate the probabilities of a

normal random variable using the normal tables in Table A
in the back of the book.
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Copyright © 2013, 2009, and 2007, Pearson Education, Inc.
Z-Scores and the Standard Normal
Distribution
A standard normal distribution has mean
 0
standard deviation   1 .
and
When a random variable has a normal distribution and
its values are converted to z-scores by subtracting the
mean and dividing by the standard deviation, the z-scores
follow the standard normal distribution.
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Table A: Standard Normal Probabilities
Table A enables us to find normal probabilities.
 It tabulates the normal cumulative probabilities
falling below the point   z .
To use the table:
 Find the corresponding z-score.
 Look up the closest standardized score (z) in the
table.
 First column gives z to the first decimal place.
 First row gives the second decimal place of z.
 The corresponding probability found in the body of
the table gives the probability of falling below the
z-score.
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Copyright © 2013, 2009, and 2007, Pearson Education, Inc.
Example: Using Table A
Find the probability that a normal random variable takes
a value less than 1.43 standard deviations above ;
P( z  1.43)  0.9236
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Copyright © 2013, 2009, and 2007, Pearson Education, Inc.
Example: Using Table A
Figure 6.7 The Normal Cumulative Probability, Less than z Standard Deviations
above the Mean. Table A lists a cumulative probability of 0.9236 for z  1.43, so
0.9236 is the probability less than 1.43 standard deviations above the mean of any
normal distribution (that is, below   1.43 ). The complement probability of 0.0764 is
the probability above   1.43 in the right tail.
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Example: Using Table A
Find the probability that a normal random variable
assumes a value within 1.43 standard deviations of  .
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
Probability below 1.43  0.9236

Probability below 1.43  0.0764

P(1.43  z  1.43)  0.9236  0.0764  0.8472
Copyright © 2013, 2009, and 2007, Pearson Education, Inc.
How Can We Find the Value of z for a
Certain Cumulative Probability?
To solve some of our problems, we will need to find the
value of z that corresponds to a certain normal cumulative
probability.
To do so, we use Table A in reverse.
 Rather than finding z using the first column (value
of z up to one decimal) and the first row (second
decimal of z).
 Find the probability in the body of the table.
 The z-score is given by the corresponding values
in the first column and row.
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Copyright © 2013, 2009, and 2007, Pearson Education, Inc.
How Can We Find the Value of z for a
Certain Cumulative Probability?
Example: Find the value of z for a cumulative probability of 0.025.
Look up the cumulative probability of 0.025 in the body of Table A.
A cumulative probability of 0.025
corresponds zto 1.96 .
Thus, the probability that a normal
random variable falls at least
1.96 standard deviations
below the mean is 0.025.
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SUMMARY: Using Z-Scores to Find Normal
Probabilities or Random Variable x Values
 If we’re given a value x and need to find a probability,
convert x to a z-score using z  ( x   ) /  , use a table of
normal probabilities (or software, or a calculator) to get a
cumulative probability and then convert it to the probability
of interest
 If we’re given a probability and need to find the value of
x , convert the probability to the related cumulative
probability, find the z-score using a normal table (or
software, or a calculator), and then evaluate x    z .
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Example: Comparing Test Scores That
Use Different Scales
Z-scores can be used to compare observations from
different normal distributions.
Picture the Scenario:
There are two primary standardized tests used by college
admissions, the SAT and the ACT.
You score 650 on the SAT which has   500 and   100
and 30 on the ACT which has   21.0 and   4.7.
How can we compare these scores to tell which score is
relatively higher?
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Using Z-scores to Compare Distributions

Compare z-scores:
SAT: z 
650  500
1.5
100
30  21
 1.91
ACT: z 

4.7
Since your z-score is greater for the ACT, you

performed
relatively better on this exam.
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