Business Weekly conducted a survey of graduates from 30 top MBA programs. On the

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 141000 dollars. Assume the standard deviation is 36000 dollars. Suppose you take a simple random sample of 60 graduates. Find the probability that a single randomly selected salary exceeds 140000 dollars. P(X > 140000) = Find the probability that a sample of size n = 60 n=60 is randomly selected with a mean that exceeds 140000 dollars. P(M > 140000) = Enter your answers as numbers accurate to 4 decimal places.

Explanation / Answer

P(X>140000)=P[Z>(140000-141000)/36000] [Z=(X-mu)/sigma, where, mu is population mean and sigma is population standadrd deviation]

=P(Z>-0.03)=1-0.4880 (look into z table, the table gives area under standard normal curve to the left of z, thus to find area greater than a z value, subtract the area corresponding to the z value from 1)

=0.512

P(M>140000)=P[Z>(140000-141000)/(36000/sqrt 60)] [according to Central Limit Theorem, if repeated samples of size n are drawn from a population with mean mu and standard deviation sigma, then as n becomes large, the sampling distribution of sample means approach normality, with mean=mu and standadrd deviation, sigma/sqrt n.]

=P(Z>-0.22)=1-0.4129=0.5871

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