Question 305160
Suppose the people have weights that are normally distibuted with a mean of 176 pounds and a standard deviation of 33 pounds. 
a.Find the probability that if a person is randomly selected, his weight will be greater than 169 pounds
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z(169) = (169-176)/33 = -0.2121
P(x > 169) = P(z> -0.2121) = 0.5840
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b. Find the probability that 12 randomly selected people will have a mean that is greater than 169 pounds
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t(169) = (169-176)/[33/sqrt(12)] = -0.7348
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P(x-bar > 169) = P(-0.7348 < t < 100 when df = 11) = 0.7611
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Cheers,
Stan H.