393 / 762
Chapter 9: The Normal Probability Distribution
393
Solution:
For this example we cannot use the
-tables since they only
display probability values for two decimal places for the
-values. Thus we
will have to use the workbook. Thus we have to find
P
(-1.1234
2.0932). Using the
Normal Probability Distribution
workbook to solve, we get
the value of 0.8512. The result is shown in
Figure 9-33
.
Figure 9-33:
Area and probability value for
P
(-1.1234
z
2.0932)
Click here for the Normal Probability Distribution Workbook