help on statistics

Question

A population of values has a normal distribution with  and . A random sample of size  is drawn. Find the probability that a sample of size  is randomly selected with a mean greater than 69.2. Round your answer to four decimal places.

 

Answer 1

M is N(66.3,40.2)

convert M to N(0,1) the standard normal distribution using the z-transformation for M = 69.2, mu = 66.3, sigma = 40.2 and n = 238:

z = (M - mu)/(sigma/sqrt(n)) = (69.2 - 66.3)/(40.2/sqrt(238)) = 1.11

so P(M < 69.2) = P(z > 1.11)

Because normal distribution is symmetric: P(z > 1.11) = P(z < -1.11) = 0.1335