Business Week conducted a survey of graduates from 30 top MBA programs (Business

Question

Business Week conducted a survey of graduates from 30 top MBA programs (Business Week, September 22, 2003). The survey found that the average annual salary for male and female graduates 10 years after graduation was $168,000 and $117,000, respectively. Assume the population standard deviation for the male graduates is $35,000, and for the female graduates it is $22,000.

What is the probability that a simple random sample of 40 male graduates will provide a sample mean within $10,000 of the population mean, $168,000 (to 4 decimals)?

What is the probability that a simple random sample of 40 female graduates will provide a sample mean within $10,000 of the population mean, $117,000 (to 4 decimals)?

eBook Exercise 7.27 (Algorithmic Business Week conducted a survey of graduates from 30 top MBA programs (Business Week, September 22, 2003). The survey found that the average annual salary for male and female graduates 10 years after graduation was $168000 and $117,000, respectively. Assume the population standard deviation for the male graduates is $35000, and for the female graduates it is $22,000. a. What is the probability that a simple random sample of 40 male graduates will provide a sample mean within $10,000 of the population mean, $168,000 (to 4 decimals)? b. What is the probability that a simple random sample of 40 female graduates will provide a sample mean within $10,000 of the population mean, $117,000 (to 4 decimals)? G. In which of the preceding two cases, part (a) or part (b), do we have a higher probability of obtaining a sample estimate within $10,000 of the population mean? Part (b) because the population standard deviation for females is lower Hl d, what is the probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean (to 4 decimals)?

Explanation / Answer

a)here std error of mean =std deviaiton/(n)1/2 =35000/(40)1/2 =5533.986

hence probability=P(-10000/5533.986<Z<10000/5533.986)=P(-1.8070<Z<1.8070)=0.9646-0.0354 =0.9292

b)

std error of mean =std deviaiton/(n)1/2 =22000/(40)1/2 =3478.505

hence probability=P(-10000/3478.505<Z<10000/3478.505)=P(-2.8748<Z<2.8748)=0.9980-0.0020 =0.9960

d)

std error of mean =std deviaiton/(n)1/2 =35000/(100)1/2 =3500

hence probability=P(Z>-4000/3500)=P(Z>-1.1429)=0.8735

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