Question

4. (30 points) A statistics exam has five easy questions and three hard questions. You have six friends who are in the class, five are moderately prepared and one is very prepared. Probability Easy Questions Correct 80% Chance 95% Probability Hard Questions Correct 66.666% (two-thirds) 5 Moderately Prepared Friends 1 Very Prepared Friend 80% The probability of answering any one question correctly is completely independent of the probability of answering any other question correctly for every student. (a) What is the probability your very prepared friend gets all eight questions right? (b) What is the probability that a randomly chosen moderately prepared friend gets all eight of them correct? (C) The professor randomly chooses one exam to grade first. The exam happens to belong to one of your six friends and it happens to get all eight questions correct. What is the probability it is the exam of your one very prepared friend. (d) Let the random variable VP be the number of questions answered correctly by your very prepared friend. Find E(VP] (e) Let M1, M2, ..., My be the scores received by your five moderately prepared friends. Let D be the difference between the average score of your five moderately prepared friends and the score of your one very prepared friend: D_Mi + M2 + M3 + M4 + M5 --VP Find the variance of D (hint: Var(VP) = .7175 and Var(M) = 1.4667)

Answer 1

a) P(all correct) = (0.955)(0.8%) = 0.3961; 6) P(all correct by mod prep) = (0.8°) (0.6666°) = 0.09706; P(all correct|VP)P(VP) c) P(V Pall correct) = – P(all correct) P(all correct|VP) = 0.3961; P(VP) = (1/6); P(all correct) = P(all correct|VP)P(VP) + P(all correct MP)P(MP); P(MP) = 5/6: P(all correct MP) = 0.09706; P(all correct) = 0.1469; P(V Pall correct) = 0.44939; d) Let, X; = 1; if ith easy que is correct by well prepared stud. X; = 0; otherwise, Vi=1,2,...,5; Y; = 1; ifith hard que is correct by well prepared, Y; = 0; otherwise, Then, 53 VP = x + Y; i=1 =1 E(X) = 1 * 0.95 = 0.95; E(Y) = 1 *0.8 = 0.8; So, E(VP) = 5*0.95 + 3*0.8 = 7.15

Mit.. + M5 e) Var(D) = Var(VP) +Var( = Var(VP) + (1/25)Var(M1 + ... + M5); For i = 1, ...5 M = 1+...+Y5+Z+...+Zg; Yį=lif easy question ith is answered correctly by a moderately prepared student.; else Yį = 0 Vi= 1, ..., 5; Now, define Z; = 1; if hard question jth is answered correctly by moderate student, else Zi = 0; j = 6,7,8. Then, Vi= 1, ...5, E(Y) = 1* P(Y; = 1) = 0.8; Vj = 6,7,8 E(Z) = 1 * 0.6666 = 0.66666 So, E(M) = 5*0.8 +3*0.6666 = 5.999998;

5 + 2 i=1 E(M)= E (E(Y1Y2) + ... + E(Y4Y5)) j=6 + 2(E(Y126) + E(Y127) + ... + E(Y5Z3)) + 2(E(2627) + ...E(2728)) Since, Y's and Zs are independent, EY;Y;) = E(Y) E(Y;) = 0.64; E(Z;Z;) = E(Z)E(Z;) = 0.4444 ; E(Y,Z) = E(Y) E(Z;) = 0.8 *0.66666 = 0.5333 E(Y) = 1* P(Y; = 1) = E(Y) = 0.8; similarly, E(Z) = 0.66666 So, E(M ) = 5 * 0.8 +3 * 0.66666 + 2 * 10 * (0.64) + 2 * 15 * 0.5333 + 2 * 30.4444 = 37.465; Var (Mi) = E(M) - (E(M)2 = 1.4654 Now Var(VP) can be similarly calculated by using appropriate probabilities. So, Var(VP) = 0.7175; Now, Var(D) = 0.7175+ (1/25) * 5 * 1.4654 = 1.01058