The statistic 100/Y is called the coefficient of variation. You can calculate it using

100 * sd(y)/mean(y) in R. For example, if the data are y=c(3,2,5,3,1,3) then the coefficient of variation is 100 * sd(y)/mean(y)=46.9% . If data are produced as iid from the normal distribution whose mean is =50 and whose standard deviation is =15, then the true coefficient of variation is 100(/)=100(15/50)=30%. But the estimator 100/Y is biased because (i) is itself biased, and (ii) the statistic is a nonlinear function of the mean and standard deviation.

A. Use R to estimate E(100/Y) whenn=5 observations Y1, Y2,..., Y5 iid N(50,152) Y1, Y2, ..., Y5 ~iid N(50, 152) are used to calculate and Y . Use the Law of Large Numbers to accomplish this. (You must generate many, many samples of n=5 observations each. Use the "apply" function.)

B. Draw the histogram of your estimates you found in A., and superimpose two vertical lines (i) the estimand, and (ii) your estimate of E(100/Y) that you got in A.

C. How does bias appear in your graph of B.?

D. In comparison with the variability in the estimator as shown in your graph of B., does the bias

appear problematic? Why or why not?