STAT 400 Lecture AL1 1 Spring 2015 Dalpiaz Answers for 5 3 5 4 5 5 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 82 2 BQ 92 2 DK 64 9 81 4 900 VP 30 300 270 P VP 300 P Z P Z 1 00 1 0 8413 0 1587 30 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 0 2 1 P Z 1 40 P Strength Torque 0 P Z 1 5 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 1500 1430 P Total 1500 P Z P Z 2 50 1 0 9938 0 0062 28 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 n P W n P X k P Y n k k 0 e 1 2 n n k 0 n k e 1 1 k k 0 n 1k n2 k k n k n2 k e 2 n k 1 2 n e 1 2 n Therefore W is a Poisson random variable with mean 1 2 OR M W t M X t M Y t e 1 e t 1 e 2 e t 1 e 1 2 e 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 P X k Y n k P X k W n P W n P W n 1k e 1 n2 k e 2 k n k 1 2 n e 1 2 n k 1 2 n k n k 1 2 1 2 n k X W n has a Binomial distribution p 1 1 2 t 1 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 p 1 p k 1 a Find P X Y n k 1 2 3 n 2 3 4 n 1 P X Y n P X k P Y n k k 1 n 1 p 1 p k 1 p 1 p n k 1 n 1 k 1 p 2 1 p n 2 k 1 n 1 p 2 1 p n 2 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 p e t M X Y t M X t M Y t 1 1 p e t b Find P X k X Y n P X k X Y n k 1 2 3 n 1 2 n 2 3 4 P X k Y n k P X k X Y n P X Y n P X Y n p 1 p k 1 p 1 p n k 1 n 1 p 2 1 p n 2 1 n 1 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 Hint First find P X Y p X k p Y k P X Y k 1 p2 k 1 p 2 1 k 1 p2 k 1 p2 1 1 p 2 p 1 p k 1 p 1 p k 1 1 p 2 n n 0 p 2 p P X Y P X Y P X Y 1 Since P X Y P X Y P X Y p 1 1 1 P X Y 1 2 2 p 2 1 p 2 p OR P X Y p 1 p x 1 p 1 p y 1 y 1 x y 1 p 1 p y 1 2 y 1 1 p x 1 x y 1 p 2 1 p y 1 y 1 p 1 p 1 p y 1 1 p 1 p 2 n 0 n p 1 p …
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