(BusinessWe lation TmCaln. 27. Business Week conducted a survey of graduates fro
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Question
(BusinessWe lation TmCaln. 27. Business Week conducted a survey of graduates from 30 top MBA programs or male September 22, 2003). On the basis of the survey, assume that the mean annual salary f and female graduates 10 years after graduation is $168,000 and S117,000, respectivel Assume the standard deviation for the male graduates is $40,000, and for the female gradi a. What is the probability that a simple random sample of 40 male graduates will provide b. What is the probability that a simple random sample of 40 female graduates will c. In which of the preceding two cases, part (a) or part (b), do we have a higher pro ates it is $25,000. a sample mean within $10,000 of the population mean, $168,000? provide a sample mean within $10,000 of the population mean, $117,000? bility of obtaining a sample estimate within $10,000 of the population mean? Why d. What is the probability that a simple random sample of 100 male graduates will po vide a sample mean more than $4000 below the population mean?Explanation / Answer
a)as we know that z score =(X-mean)/std error
also std error=std deviation/sqrt(n)
P(sample mean within 10000 of population mean)=P(168000-10000<Xbar<168000+10000)
=P(-10000/(40000/sqrt(40)<Z<10000/(40000/sqrt(40))=P(-1.58<Z<1.58)=0.9429-0.0571=0.8858
b)
P(sample mean within 10000 of population mean)=P(-10000/(25000/sqrt(40))<Z<10000/(25000/sqrt(40)))
=P(-2.53<Z<2.53)=0.9943-0.0057=0.9886
c)
in part b we have higher probability of being close to mean due to low std deviation and therefore sample mean will be closer to population mean
d)
P(sample mean is more than 4000 below the populaiton mean)=P(Z<-4000/(168000/sqrt(100)))
=P(Z<-0.24)=0.4052
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