Essential Statistics

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Chapter 9: The Normal Probability Distribution 387
Figure 9-27: Area and probability value for P(-1.99  z  -0.66)
To use the -tables to solve again one can use the symmetric property of the
distribution. Because of the symmetry of the distribution, this is equivalent
to finding P(0.66   1.99) = P(0   1.99) – P(0  z  0.66) and solve
like in Example 9-5. From the standard normal tables at the end of the text,
for = 1.99, the corresponding value is 0.4767, and for = 0.66, the
corresponding value is 0.2454. Thus, the required area for P(0.66  
1.99) = P(0   1.99) – P(0   0.66) = 0.4767 – 0.2454 = 0.2313. This
is shown in Figure 9-28.