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```Applications of the
Normal Distribution
Section 6.4
Find the probabilities for a normally
distributed variable by transforming it into
a standard normal variable
 Find specific data values given the
percentages, using the standard normal
distribution

Objectives
This section presents methods for working
with normal distributions that are not
standard (NON-STANDARD). That is the
mean, m, is not 0 or the standard
deviation, s, is not 1 or both.
 The key concept is that we transform the
original variable, x, to a standard normal
distribution by using the following
formula:

Key Concept
original x-value  mean
z
standard deviation
xm

s
Always round z-scores to 2 decimal places
Conversion Formula
Converting to Standard Normal
Distribution
z=
x-m
s
P
P
(a)
m
x
(b)
0
z

Choose the correct (left/right) of the
graph
 Negative z-score implies it is located to the left
of the mean
 Positive z-score implies it is located to the right
of the mean
 Area less than 50% is to the left, while area
more than 50% is to the right

Areas (or probabilities) are positive or
zero values, but they are never negative
Cautions!!!!

Don’t confuse z-scores and areas
 Z-scores are distances on the horizontal
scale. Table E lists z-scores in the left
column and across the top row
 Areas (probabilities or percentages) are
regions UNDER the normal curve. Table E
lists areas in the body of the table
Cautions!!!!
Draw a normal distribution curve labeling
the mean and x
 Convert x to standard normal distribution
using z-score formula
 Use procedures on page 287-288

Finding Areas given a specified
variable, x

According to the American College Test
(ACT), results from the 2004 ACT testing
score of 21.3 with a standard deviation of
6.0. Assuming that the scores are normally
distributed:
◦ Find the probability that a randomly selected
student has a reading ACT score less than 20
◦ Find the probability that a randomly selected
student has a reading ACT score between 18 and
24
◦ Find the probability that a randomly selected
student has a reading ACT score greater than 30
Example

Women’s heights are normally distributed
with a mean 63.6 inches and standard
deviation 2.5 inches. The US Army
requires women’s heights to be between
58 inches and 80 inches. Find the
percentage of women meeting that height
requirement. Are many women being
denied the opportunity to join the Army
because they are too short or too tall?
Example
Draw a normal distribution labeling the
given probability (percentage) under the
curve
 Using Table E, find the closest area
(probability) to the given and then
identify the corresponding z-score
 Substitute z, mean, and standard
deviation in z-score formula and solve for
x.

Finding variable x given a specific
probability

According to the American College Test
(ACT), results from the 2004 ACT testing