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Chapter 9: The Normal Probability Distribution
399
Figure 9-37:
Area associated with
z
0
such that
P
(
z
z
0
) = 0.3
Using the area of 0.2, we have from the body of the 0 to
-value table, a
corresponding
z
score of 0.52. Since, from
Figure 9-37
,
is to the right of
0, then
= 0.52. We can use this information to solve for
by using the
equation
or 0.52 = (
– 700)/15. Solving gives
= 707.8
708. That is, the minimum score the college will accept is approximately
708.
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