5.4.15 Suppose we have the situation describedN in Exercise 5.4.4, and we take a simple random sample of size n from H where |l'I|= (a) Prove that the mean of f X(0) is given by fX(0) (Hint: Note that we can write m0) = n" 221'. 1 1m} (Xma) and an (Had) ~ BemoulliUXm ) (b) Prove that the variance of f X(0) is given by 135(0) (1 - fX(0)) N n n N 1 ' (Hint: Use the hint in part (a), but note that the I{0} (X (3:1)) are not independent. Use Theorem 3.3 .403) and evaluate Cov (I{0} (X (a: I~)) , I{0} (X (a 0)) in terms of fX(0).) (c) Repeat the calculations in parts (a) and (b), but this time assume that you take a sample of n with replacement. (Hint: Use Exercise 5.4.4(c).) ((1) Explain why the factor (N n)/(N 1) in (5.4.1) is called the nite sample correction factor. (5.4.1) 5.4.4 Suppose we have a nite population H and a measurement X : H ) {0, 1} where |H| = N and H7: : X01?) = 0}| = a. (a) Determine fx(0) and f 35(1). Can you identify this population distribution? (b) For a simple random sample of size n, determine the probability that nf}(0) = x. (e) Under the assumption of i.i.d. sampling, determine the probability that 11]} (0) = x