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 169000 dollars. Assume the standard deviation is 43000 dollars. Suppose you take a simple random sample of 77 graduates. Find the probability that a single randomly selected policy has a mean value between 168019.9 and 175860.4 dollars. P(168019.9 < X < 175860.4) = (Enter your answers as numbers accurate to 4 decimal places.) Find the probability that a random sample of size n = 77 n=77 has a mean value between 168019.9 and 175860.4 dollars. P(168019.9 < M < 175860.4) = (Enter your answers as numbers accurate to 4 decimal places.)

Answer 1

Solution :

Given that ,

mean = = 169000

standard deviation = = 43000

P(168019.9 < X < 175860.4)

= P((168019.9 - 169000) / 43000) < (x - ) /  < (175860.4 - 169000) / 43000) )

= P(-0.0228 < z < 0.1595)

= P(z < 0.1595) - P(z < -0.0228)

= 0.5634 - 0.4909

= 0.0725

Probability = 0.0725

n = 77

M = 169000

M = / n = 43000 / 77 = 4900.305

P(168019.9 < M < 175860.4)

= P((168019.9 - 169000) / 4900.305 <( - M) / M < (175860.4 - 169000) / 4900.305))

= P(-0.20 < Z < 1.40)

= P(Z < 1.40) - P(Z < -0.20)

= 0.9192 - 0.4207

= 0.4985

Probability = 0.4985