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242 Part IV: Quality Assurance
Continued
b. The probability P(Z 0.70) is shown by the shaded area in Figure 18.33. This area
is equal to the sum of the area to the left of z = 0 and the area between z = 0 and
z = 0.7, which implies that
P(Z 0.70) = P(Z 0) + P(0 Z 0.7) = 0.5 + .2580 = 0.7580.
c. By using the same argument as in part (b) and Figure 18.34, we get
P(Z –1.0) = P(–1.0 Z 0) + P(Z 0)
= P(0 Z 1.0) + P(Z 0)
= .3413 + .5 = 0.8413.
–3 0 0.7
3
Figure 18.33 Shaded area showing P(Z 0.70).
–3 –1 0
3
Figure 18.34 Shaded area showing P(Z –1.0).
Part IV.A.5
EXAMPLE 18.28
Use the table in Appendix D to find the following probabilities:
(a) P(Z 2.15) (b) P(Z –2.15)
Solution:
a. The desired probability P(Z 2.15) is equal to the shaded area under the normal
curve to the right of z = 2.15, shown in Figure 18.35. This area is equal to the area
to the right of z = 0 minus the area between z = 0 and z = 2.15. Since the area to the
right of z = 0 is 0.5. Thus, we have
P(Z 2.15) = 0.5 – P(0 Z 2.15) = 0.5 – 0.4842 = 0.0158
b. Using the symmetric property of the normal distribution (see Figure 18.36) and
using part (a), we have
P(Z –2.15) = P(Z 2.15) = 0.0158
Continued