Question 1026360
The annual rainfall of a region has a terminal distribution with a mean rainfall of 100 inches and a standard deviation of 12 inches. What is the probability that the average mean rainfall during 49 randomly picked years will exceed 101.5 inches? 
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z(101.5) = (101.5-100)/[12/sqrt(49) = 1.5*7/12 = 0.875
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P(x-bar > 101.5) = P(z < 0.875) = 0.1908
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Cheers,
Stan H.