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Chapter 18: A. Basic Statistics and Applications 243
Continued
–3 0 2.15 3
Figure 18.35 Shaded area showing P(Z 2.15).
–3 –2.15
0
3
Figure 18.36 Shaded area showing P(Z –2.15).
So far in this section, we have considered problems of finding probabilities
of the standard normal variable Z, that is, a normal random variable with mean
m = 0 and standard deviation s = 1. Now we consider problems where m 0 and
s 1.
EXAMPLE 18.29
Let X be a random variable distributed normally with m = 6 and s = 4. Then determine
the following probabilities:
Solution:
(a) P(8.0 X 14.0) (b) P(2.0 X 10.0) (c) P(0 X 4.0)
a. In order to find the probability P(8.0 X 14.0), we first need to transform the
random variable X into the standard normal variable Z, which is done by
subtracting throughout the inequality, the mean m, and dividing by the standard
deviation s. Thus, as shown in Figures 18.37 and 18.38, we get
Part IV.A.5
P( 80 . X 140
. )= P
8 6 X 6 14
6
4 4 4
( )
= P 05 . Z 20 .
= P( 0 Z 2. 0)P 0Z
0.
50
( )
= 0. 4772
0.
1915= 0. 2857.
b. Proceeding in the same manner as in part (a) and using Figure 18.39, we have
Continued