7.2 sampling distribution of the mean

Question

A normally distributed population has a mean of 550 and a standard deviation of 80. Determine the probability that a random sample of size 16 will have a sample mean greater than or equal to 609

Answer 1

E(Xbar) = µ = 550 as given; σ_{Xbar =} σ/√n = 80/√16 or 20. Then z = (Xbar - µ)/σ_Xbar or (609 - 550)/20 equal to 2.95.

Finally, P(X_bar ≥ 609|µ = 550, σ_Xbar=20) = P(z ≥ 2.95) or 1 - (0.5 + 0.4984) equivalent to 0.16%.