Question

The Economic Policy Institute periodically issues reports on worker’s wages. The institute reported that mean wages for male college graduates were $37.39 per hour and for female college graduates were $27.83 per hour in 2017. Assume the standard deviation for male graduates is $4.60, and for female graduates it is $4.10. What is the probability that a sample of 50 male graduates will provide a sample mean within $1.00 of the population mean, $37.39?

Answer 1

Here, μ = 37.39, σ = 0.6505, x1 = 36.39 and x2 = 38.39. We need to compute P(36.39<= X <= 38.39). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (36.39 - 37.39)/0.6505 = -1.54
z2 = (38.39 - 37.39)/0.6505 = 1.54

Therefore, we get
P(36.39 <= X <= 38.39) = P((38.39 - 37.39)/0.6505) <= z <= (38.39 - 37.39)/0.6505)
= P(-1.54 <= z <= 1.54) = P(z <= 1.54) - P(z <= -1.54)
= 0.9382 - 0.0618
= 0.8764