O Question Details My Notes Ask YourT The probability that a pet owner gives her
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O Question Details My Notes Ask YourT The probability that a pet owner gives her/his pet a gift for the holidays is .08. Suppose we randomly sample pet owners 1 by one. Let X equal the number of pet owners sampled until the first owner who gives her/his pet a gift for the holidays. Let W equal the number of owners sampled until the 2nd gift giving owner is found. a. What is the expected value of X? b. What is the variance of X? c, what is the probability that X = 12 d. what is the probability that X > 121 e. What is the probability that X > 12 given that X > 11? f, what is the probability that w Add any comments below I 207 My Notes Ask Your Teacher 5 Question Details If it snows on game day there is a 52% chance the Giants will win. If it does not snow on game day there is a 28% chance the Giants will win. There is a 29% chance that it will snow on game day. a. What is the probability the Giants will win? b. What is the probability that it snows on game day and the Giants win? PaExplanation / Answer
Q.4 The given dstribution is Geometric distribution. where paramenter p = 0.08
where X is the number of pet owners sampled until the first owner gives her pet for the gift.
(a) E(X) = (1 -p)/p = (1 - 0.08)/ 0.08 = 11.5
(b) Var(X) = (1 -p)/p2 = (1 - 0.08)/ 0.082 = 143.75
(c) Pr(X =12) = (1 -p)X p = (1 - 0.08)12 0.08 = 0.0294
(d) Pr(X >12) = 1 - Pr(X <= 12) = 1 - [ 1 - (1-p)X +1 ] = (1 - 0.08)13 = 0.3383
(e) Pr(X >12 l X> 11) = (1- 0.08)13 / (1- 0.08)12 = 0.92
(f) Pr(W = 20) that means we will get out second owner in 20th attempt so that also means that in earlier 19 attempt we get one pet owner who gifter her pet
so, Pr(W = 20) = 19C1 (0.08)1(0.92)18 * 0.08 = 0.0271
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