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Chapter 18: A. Basic Statistics and Applications 241
Continued
c. By using the same argument as in part (b) and using the table in Appendix D
(also see Figure 18.31), we get
P(–2.2 Z –1.0) = P(1.0 Z 2.2)
= P(0 Z 2.2) – P(0 Z 1.0)
= .4861 – .3413 = 0.1448.
–3 –2.2 –1 1 2.2 3
Figure 18.31 Two shaded areas showing P(–2.2 Z –1.0) = P(1.0 Z 2.2).
EXAMPLE 18.27
Use the table in Appendix D to determine the following probabilities:
(a) P(–1.50 Z 0.80) (b) P(Z 0.70) (c) P(Z –1.0)
Solution:
a. Since the standard normal distribution table (Appendix D) gives the probabilities
of z-values starting from zero to positive z-values, we have to break the interval
–1.5 to 0.8 into two parts, that is –1.5 to zero plus zero to 0.8 (see Figure 18.32), so
that we get
Thus, we have
P(–1.50 Z 0.80) = P(–1.50 Z 0) + P(0 Z 0.80).
P(–1.50 Z 0.80) = P(–1.50 Z 0) + P(0 Z 0.80)
= P(0 Z 1.50) + P(0 Z 0.80)
= .4332 + .2881 = 0.7213.
Part IV.A.5
–3 –1.5 0 0.8
3
Figure 18.32 Showing P(–1.50 Z .80) = P(–1.50 Z 0) +P(0 Z 0.80).
Continued