Central Limit Theorem: Sample Proportion a Minority Representation Group Accuses a Major

Central limit theorem: Sample proportion A minority representation group accuses a major bank of racial discrimination in its recent hires for financial analysts. Exactly 10%of all applications were from minority members, and exactly 8%of the 900open positions were filled by members of the minority. a. Find the mean of p^, where p^is the proportion of minority member applications in a random sample of 900that is drawn from all applications. b. Find the standard deviation ofp^. c. Compute an approximation forP( p<=.08), which is the probability that there will be 8%or fewer minority member applications in a random sample of 900drawn from all applications. Round your answer to four decimal places.

Let x be the number of minority member applications in a random sample of 900 that is drawn from all applications. Then clearly x follows a Binomial distribution with parameters n = 900 and p = 0.10. Thus the mean and variance of x are np and np(1-p) respectively. Let p^ be the proportion of minority member applications in a random sample of 900that is drawn from all applications. Then p^ = x/n Now, the mean of p^ is given by The mean of p^ = E(p^)=E(x/n) =E(x)/n = np/n = p = 0.10 The variance op p^ is given by, Var(p^) =Var(x/n) = Var(x)/n2 = np(1-p)/n2 = p(1-p)/n Therefore, the standard deviation of p^ is given by, S.D(p^) =Sqrt[p(1-p)/n] = Sqrt[0.10*0.90/900] = Sqrt[0.0001] = 0.01 Since n is large by central limit theorem, p^ follows a Normal distribution with mean p=0.10 and standard deviation Sqrt[p(1-p)/n] = 0.01. Therefore, Z =(p^ - 0.10)/0.01 follows a Standard Normal distribution. Now the probability that there will be 8%or fewer minority member applications in a random sample of 900drawn from all applications is given by P[p^<=.08] = P[(p^ - 0.10)/0.01 <= (0.08 - 0.10)/0.01] = P[Z <= -2] = 0.0228

Note that if Z is a Standard Normal variable then P[Z<=-2] can be calculated using the Excel formula =NORMSDIST(-2)