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 150000 dollars. Assume the standard deviation is 38000 dollars. Suppose you take a simple random sample of 98 graduates.

(a) Find the probability that a single randomly selected policy has a mean value between 139252 and 141171. 3 dollars.

P(139252 < X < 141171. 3) =?

(Enter your answers as numbers accurate to 4 decimal places. )

(b) Find the probability that a random sample of size n=98 has a mean value between 139252 and 141171. 3 dollars.

P(139252 < M < 141171. 3) =?

(Enter your answers as numbers accurate to 4 decimal places. )

Answer 1

Based on the mean annual salary and the standard deviation, the following are true:

• Probability of P(139252 < X < 141171. 3) = 0.0193.
• Probability that random sample has P(139252 < M < 141171. 3) = 0.0081.

What is the Probability of P(139252 < X < 141171. 3)?

= P ( (139,252 - 150,000) / 38,000) < Z score < P ( (141,171.30 - 150,000) / 38,000)

= P(-0.28 < Z < -0.23)

Using Z table:

= (1 - 0.5910) - ( 1 - 0.6103)

= 0.0193

What is the Probability that random sample has P(139252 < M < 141171. 3) ?

= P ( (139,252 - 150,000) / (38,000 / √98)) < Z score < P ( (141,171.30 - 150,000) /(38,000 / √98))

= P(-2.80 < Z < -2.30)

Using Z table:

= (1 - 0.9893) - (1 - 0.9974)

= 0.0107 - 0.0026

= 0.0081

Find out more on the z table at https://brainly.com/question/6096474.