Please help explain to me how to solve this problem. I am very confused

5. Let f (m) = 6(1 + 03:) be a family of functions dened on the interval 1 S m S 1. The parameter 9 is in the interval 1 S 9 S 1. (a) Show that for any 0, f (16') is a valid probability density only if c = 1/2. We need the density to integrate to I. So 1 = 111 0(1 + 93:)dzs = [cm + 9332/2]1_1 = 20. Thus (2 = 1 / 2 (b) Dene X to be a continuous random variable with pdf f (3:) Let's investigate the mean of X: p: = E(X). Show that E(X) = 9/ 3. E(X) = [ll 93' - g1 + 92:)(13: = [1/43:2 + 1/29323/3]1_1 = 9/3 (c) For a random sample sample X1, . . . ,Xn from the distribution f (:13), we will estimate 0 using a\": 3?. Does this seem like a sensible choice of an estimator for 9? Explain your intuition. Yes. Since Y is an estimator of the mean and I9 is three times the mean 4 (d) Is Earl unbiased estimator for 9? You must show your work to support your answer. E=3Eci=3n=tl (e) Consider the case 9 = 1. That is, f(3) = 1/2(1 + as) for 1 S a: g 1. Sketch the probability density curve: f (:12)