Solution:
Given in the question
Mean = 105
Standard deviation = 15
P(Xbar< 132)
z =(132-105)/15 = 27/15 = 1.8
So from z table we found p- value
P(Xbar<132) = 0.9641
So there is 96.41% probability that IQ is less than 132.
Assume that adults have scores that are normally distributed with a mean of μ-105 and a...
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