Math 511 Exam #2 Show All Work 1. A package of 200 seeds contains 40 that are defective and will not grow (the rest are fine). Suppose that you choose a sample of 10 seeds from the box without replacement. Let X denote the number of seeds in the sample that will not grow. (a). The value of the mean of X is = 2 µ = E( X) = 10 40 200 = 2 (b). What is an expression for the exact value of P(X = 4)? P( X = 4) = 40 160 4 6 200 10 2. (a). Suppose that the moment generating function for X is M (t) = 2et +1 3 E(X) = 8 Var(X) = 2.667 (Reduce to integer or decimal values.) X is a binomial random variable with n = 12, and p = 2 3. Thus, µ = 12 2 3 = 8, and σ 2 = 2 2 3 1 3 = 8 3. (b). Suppose that the moment generating function for X is M (t) = e t 4 3e t. E(X) = 4 Var(X) = 12 (Reduce to single fractions) 12. X is a geometric random variable with p = 1 4. Thus, µ = 1 1 4 = 4, and σ 2 = q p 2 = 3 4 1 16 = 12 (c). Suppose X is a random variable with P(X = 0) = 0.3, P(X = 2) = 0.1, and P(X = 4) = 0.6. The moment generating function for X is M(t) = M (t) = 0.3+ 0.1e 2t + 0.6 4t
3. A box contains 12 prizes. Six are worth $8, two are worth $6 and four are worth $3. A contestant chooses one of the prizes at random. Let X denote the value of the selected prize. Determine the mean and variance of X. µ = 6 σ 2 = 5 (Reduce each value to an integer.) = 1 2, P( X = 6) = 1 6, and P( X = 3) = 1 3. P X = 8 So, E( X) = 1 2 8 + 1 6 6 + 1 3 3 = 6. E( X 2 ) = 1 2 64 + 1 6 36 + 1 3 9 = 41. So, σ 2 = E( X 2 ) E( X) 2 = 41 36 = 5. 4. A certain booklet has an average of 1 misprint every 400 characters. Page ten of the booklet contains 2000 characters. Let X denote the number of misprints on page ten. You may assume that the distribution of the misprints is a Poisson distribution. (a). P(X = 4) = (Answer should be exact and in terms of e.) We have λ = 5, and so P( X = 4) = e 5 5 4 24 (b). P(X 3 X 6) = (Approximate using tables.) P(X 3 X 6) = P 3 X 6 P X 6 P( X 2) = P X 6 P X 6 = 0.762 0.125 0.762
5. (a). A box contains two red cards and 4 blue cards. Cards are randomly drawn one at a time with replacement until a red card is followed by a blue card. Let X denote the number of cards drawn. Examples: BBBRRB 6- draws, RB 2- draws, RRRB 4- draws, BRRRRRB 7- draws. E(X) = 4.5 (Reduce to a decimal) Let X R = # draws needed to get a Red card, and X B = # draws needed to get a Blue card. Then X = X R + X B, and so E( X) = E( X R ) + E( X B ) = 1 1 + 1 = 3 +1.5 = 4.5 2 3 3 (b). Suppose that X is a Poisson random variable and P( X = 1) = P( X = 2). Determine the value of P(X = 3) = (Exact in terms of e) From P( X = 1) = P( X = 2), we get that λe λ = λ 2 e λ 2 λ = 2, and so P( X = 3) = 8e 2 6 = 4 3e 2, and so we get 6. An archer hits the target on 60% of his shots. Let X equal the number of times the archer hits the target in 20 shots. (a). Determine the mean and variance of X. µ = 12 σ 2 = 4.8 (Reduce to decimals) µ = np = 20 0.6 = 12, σ 2 = npq = 12 0.4 = 4.8 (b). P( X = 9) = (exact value don t simplify) = 20 9 P X = 9 ( 0.6)9 0.4 11
7. (a). A box contains six cards numbered 1 6. Cards are drawn at random with replacement until the lowest numbered card (initially a one) gets drawn, then that card is thrown away and the drawings begin anew (so after getting the one, we discard it and begin drawing again with replacement until we get the two and then we discard the two, etc.). This continues until all six cards have been drawn. Let X denote the number of drawings needed to get all the cards. E(X) = 21 (Answer should be an integer.) Let X k denote the number of draws needed to get card number k. Then, X = X 1 + X 2 + X 3 + X 4 + X 5 + X 6, and so E X So, E( X) = 6 + 5 + 4 + 3 + 2 +1 = 21 6 = E( X k ). (b). Suppose that the Gamecocks have a 70% chance to win any baseball game against Clemson (this is actually about right based on the last 28 games). Suppose that the Gamecocks and Clemson play a super series where the first team to win 5 games wins the series. What is the probability the gamecocks win by a score of 5 2? For the Gamecocks to win 5 2, they must win exactly 4 of the first 6 games and then win the 7 th game. So the answer is 6 4 (0.7)4(0.3) 2 0.7 = 6 4 (0.7)5 (0.3) 2 8. Suppose that the moment generating function for Y is M(t) = ae t + be 2t + 0.2e 5t and that the mean of Y is 2.3, then a = 0.3 b = 0.5 Hint: Find two simple equations in a and b. Since M 0 = 1, we get, a + b + 0.2 = 1, which reduces to a + b = 0.8 Now note that, M (t) = ae t + 2be 2t + e 5t Since So, since M ( 0) = E( X), we get, a + 2b +1 = 2.3, which reduces to a + 2b = 1.3 Solving the two equations simultaneously gives a = 0.3, and b = 0.5 k=1
9. Let X be a binomial random variable with n = 20, p = 0.2 Name (a). P( 2 X 5) = 0.735 (Express your answer as a decimal using tables.) P( 2 X 5) = P( X 5) P( X 1) = 0.804 0.069 (b). P( X > 4) = 0.370 (Express your answer as a decimal using tables.) P( X > 4) = 1 P( X 4) = 1 0.630 = 0.370 (c). Approximate X by a Poisson random variable and express P(0 X 2) in terms of the constant e. (Simplify to the form Ae B where A and B are numbers.) P(0 X 2) = e 4 + 4e 4 + 8e 4 = 13e 4