Extra Problems Solutions

Extra Problems Solutions - STAT 230 Solutions To Extra...

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STAT 230 Solutions To Extra Problems 1. (a) Let X = the number of tickets required to produce 2 winners. Then X has a negative binomial distribution with parameters r = 2 and p =0 . 1. Therefore E ( X )= r p = 2 0 . 1 = 20 and Var( X )= r (1 - p ) p 2 = 2(0 . 9) (0 . 1) 2 = 180. Now W = (2)(5 . 00) - X (1 . 00) = 10 . 00 - X . Therefore E ( W )= - 10 . 00 and Var( W ) = 180. (b) Let Y 1 = number of outlets requiring 14 or fewer tickets to get 2 winners, let Y 2 = number of outlets requiring 30 or more tickets to get 2 winners, and let Y 3 =15 - Y 1 - Y 2 = the remaining number of outlets. This divides the outlets into three boxes, and the result is a multinomial (trinomial) distribution. The desired probability is P ( Y 1 =3 ,Y 2 =2 ,Y 3 = 10) = 15! 3! 2! 10! p 3 1 p 2 2 (1 - p 1 - p 2 ) 10 where p 1 = probability that 14 or fewer tickets will be required to obtain 2 winners, and p 2 = probability that 30 or more tickets will be required to obtain 2 winners. The probabilities p 1 and p 2 can be solved for explicitly from the negative binomial distribution. For example,
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Extra Problems Solutions - STAT 230 Solutions To Extra...

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