STAT 400 Lecture AL1 Answers for 5.3, 5.4, 5.5 Spring 2015 Dalpiaz 1. Models of the pricing of stock options often make the assumption of a normal distribution. An investor believes that the price of an Burger Queen stock option is a normally distributed random variable with mean $18 and standard deviation $3. He also believes that the price of an Dairy King stock option is a normally distributed random variable with mean $14 and standard deviation $2. Assume the stock options of these two companies are independent. The investor buys 8 shares of Burger Queen stock option and 9 shares of Dairy King stock option. What is the probability that the value of this portfolio will exceed $300? BQ has Normal distribution, BQ = $18, BQ = $3. DK has Normal distribution, DK = $14, DK = $2. Value of the portfolio VP = 8 BQ + 9 DK. Then VP has Normal distribution. VP = 8 BQ + 9 DK = 8 18 + 9 14 = $270. 2 VP = 8 2 2 BQ + 9 2 2 DK = 64 9 + 81 4 = 900. VP = $30. P( VP > 300 ) = 30270300ZP = P( Z > 1.00 ) = 1 – 0.8413 = 0.1587.2. A machine fastens plastic screw-on caps onto containers of motor oil. If the machine applies more torque than the cap can withstand, the cap will break. Both the torque applied and the strength of the caps vary. The capping machine torque has the normal distribution with mean 7.9 inch-pounds and standard deviation 0.9 inch-pounds. The cap strength (the torque that would break the cap) has the normal distribution with mean 10 inch-pounds and standard deviation 1.2 inch-pounds. The cap strength and the torque applied by the machine are independent. What is the probability that a cap will break while being fastened by the capping machine? That is, find the probability P ( Strength < Torque ). P ( Strength < Torque ) = P ( Strength – Torque < 0 ). E ( Strength – Torque ) = 10 – 7.9 = 2.1. Var ( Strength – Torque ) = ( 1 ) 2 1.2 2 + ( – 1 ) 2 0.9 2 = 2.25. SD ( Strength – Torque ) = 1.5. ( Strength – Torque ) is normally distributed. P ( Strength – Torque < 0 ) = 5.11.20ZP = P ( Z < – 1.40 ) = ( – 1.40 ) = 0.0808.3. In Neverland, the weights of adult men are normally distributed with mean of 170 pounds and standard deviation of 10 pounds, and the weights of adult women are normally distributed with mean of 125 pounds and standard deviation of 8 pounds. Six women and four men got on an elevator. Assume that all their weights are independent. What is the probability that their total weight exceeds 1500 pounds? Total = W 1 + W 2 + W 3 + W 4 + W 5 + W 6 + M 1 + M 2 + M 3 + M 4. E(Total) = E(W 1) + E(W 2) + E(W 3) + E(W 4) + E(W 5) + E(W 6) + E(M 1) + E(M 2) + E(M 3) + E(M 4) = 125 + 125 + 125 + 125 + 125 + 125 + 170 + 170 + 170 + 170 = 1430. Var(Total) = Var(W 1) + Var(W 2) + Var(W 3) + Var(W 4) + Var(W 5) + Var(W 6) + Var(M 1) + Var(M 2) + Var(M 3) + Var(M 4) = 64 + 64 + 64 + 64 + 64 + 64 + 100 + 100 + 100 + 100 = 784. SD(Total) = 784 = 28. Total has Normal distribution. P( Total > 1500 ) = 2814301500ZP = P( Z > 2.50 ) = 1 – 0.9938 = 0.0062. Note: It is tempting to set Total = 6 W + 4 M, but that would imply that the six women who got on the elevator all have the same weight, and so do the four men, which is most likely not the case here._ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 4. Let X and Y be two independent Poisson random variables with mean 1 and 2 , respectively. Let W = X + Y. a) What is the probability distribution of W? P( W = n ) = nkknk0 YP XP = nkknkknkee0 21!)(λ!λ 2 1λ λ = nkknkknknne021 21λλ!)( !!! )λλ( = !λλ )λλ(2121 nen. Therefore, W is a Poisson random variable with mean 1 + 2 . OR M W ( t ) = M X ( t ) M Y ( t ) = ) 1 (λ) 1 (λ 2 1 tteeee = ) 1 ( )λλ 21 ( tee. Therefore, W is a Poisson random variable with mean 1 + 2 . b)* What is the conditional distribution of X given, W = n? P( X = k | W = n ) = WPWXPnnk = WPYXPnknk = !λλ!)(λ!λ 2 1)λλ( λ λ 2121 21 nknkeeenknk = knkknkn 212211 λλλ λλλ!)(!! . X | W = n has a Binomial distribution, p = 211λλλ.5. Let X and Y be independent random variables, each geometrically distributed with the probability of “success” p, 0 < p < 1. That is, p X ( k ) = p Y ( k ) = 1 1kpp, k = 1, 2, 3, … , a) Find P ( X + Y = n ), n = 2, 3, 4, … . P ( X + Y = n ) = 11 YP XP nkknk = 1111 1 1 nkknkpppp = 1122 1 nknpp = 22 11 nppn, n = 2, 3, 4, … . If X and Y both have Geometric ( p ) distribution and are independent, then X + Y has Negative Binomial distribution with r = 2. OR M X + Y ( t ) = M X ( t ) M Y ( t ) = 2 11 tteepp. b) Find P ( X = k | X + Y = n ), k = 1, 2, 3, … , n – 1, n = 2, 3, 4, … . P ( X = k | X + Y = n ) = YXPYXXPnnk = YXPYXPnknk = 2211 11 1 1 nknkppnpppp = 11n, k = 1, 2, 3, … , n – 1. X | X + Y = n has a Uniform distribution on integers 1, 2, 3, … , n – 1.c) Find P ( X > Y ). …
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