Posted by Anonymous on Tuesday, March 19, 2013 at 6:09pm.
A. Z = (score-mean)/SD
Z = (50-52)/17.2 = -2/17.2 = -.12
Z = (60-52)/17.2 = 8/17.2 = .47
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores.
.0478 +.1808 = .2286 = 23%
B. Similar process.
C. Correct
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