winter 2006 Test3Solun

# winter 2006 Test3Solun - STAT 230 Solutions Test#3 3:30...

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STAT 230 Test #3 Feb 15, 2006 3:30 – 4:15 pm Solutions 1a) Let X = number of calls in a 3-hour shift. Then X ~ Poisson( λ t), where λ = 2 calls per hour and t = 3 hours. Therefore, . = = = 0.0025 e ! 0 t) ( e 0) P(X 6) ( 0 t) (- λ 1b) Let Y = number of days with no calls. Then Y ~ Binomial(100, p) where p = probability that there is no call during a 3-hour shift, i.e., p = e (-6) from (1a). Therefore, () ( ) () () . = = 96 4 4 100 ) 6 (- 4 ) 6 (- 9975 . 0 0025 . 0 4 100 e - 1 e 4 100 4) P(Y 1c) For n large and p small, we can approximate Binomial(n,p) with the Poisson(np). Here n = 100 and p = 0.0025 (from above). So np = 0.25 and . = 0001 . 0 4! .25) 0 ( e 4) P(Y 4 .25) 0 (- 2a) Selecting two cards out of three without replacement, there are three equally likely outcomes: (\$1, \$4), (\$1, \$7) and (\$4, \$7). Let X = total winning. Then X = 5, 8, or 11, each with probability 1/3. Let Y = net winning, then Y = X – 9. Therefore, we have

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## This note was uploaded on 03/03/2010 for the course STAT 230 taught by Professor Various during the Fall '06 term at Waterloo.

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winter 2006 Test3Solun - STAT 230 Solutions Test#3 3:30...

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