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 175000 dollars. Assume the standard deviation is 45000 dollars. Suppose you take a simple random sample of 65 graduates.

Find the probability that a single randomly selected policy has a mean value between 171651.1 and 190070.2 dollars.
P(171651.1 < X < 190070.2) = (Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a random sample of size n=65n=65 has a mean value between 171651.1 and 190070.2 dollars.
P(171651.1 < M < 190070.2) = (Enter your answers as numbers accurate to 4 decimal places.)

Explanation / Answer

mean is 175000 and s is 45000

a) z is given (x-mean)/s

P((171651.1-175000)/45000 < z <(190070.2-175000)/45000) =P(-0.07<z<0.33)=P(z<0.33)-(1-P(z<0.07))

0.6293-(1-0.5279)=0.1572

b) SE is s/sqrt(N)=45000/sqrt(65)=5581.56

P(171651.1 < M < 190070.2) =P((171651.1-175000)/5581.56<z<(190070.2-175000)/5581.56)=P(-0.6<z<2.7)

or P(z<2.7)-(1-P(z<0.6))=0.9965-(1-0.7257)=0.7222

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