P(159363.1 < X < 184576.5) =

Question

Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 170000 dollars. Assume the standard deviation is 39000 dollars. Suppose you take a simple random sample of 98 graduates.

Find the probability that a single randomly selected policy has a mean value between 159363.1 and 184576.5 dollars.

P(159363.1 < X < 184576.5) =

Answer 1

Zmin = (159363.1 - 170000)/39000 = -0.27274102564102564102564102564103

Zmax = (184576.5 - 170000)/39000 = 0.37375641025641025641025641025641

Prob( Zmin < z < Zmax) = Prob(z < Zmax) - Prob( z < Zmin)

= 0.645707224 - 0.392526158

= 0.253181