The number of calls they receive is independent so

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distributed with mean 10, and therefore parameter 10. The number of calls they receive is independent, so the probability they both receive exactly 5 calls is P ( N X = 5 and N Y = 5 ) = P ( N X = 5 ) P ( N Y = 5 ) = e - 6 6 5 5 ! e - 10 10 5 5 ! 3 of 7
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3. ( 20 points ) The average number of points scored in a game by a certain basketball team is 65. Assume that the number of points scored is Poisson distributed. ( a ) What is the standard deviation in the number of points scored? The parameter of the Poisson distribution is its mean, in this case 65. This is also the variance, so the standard deviation is 65. ( b ) What is the likelihood of this particular team scoring 75 points in a given game? e - 65 65 75 75 ! ( c ) What is the likelihood of this particular team scoring 10 points in a game? e - 65 65 10 10 ! ( d ) A smudge makes it hard to read score of the latest game: the team got either 70 points or 78 points. What is the likelihood that the team scored 78, given that the team scored either 70 or 78? P ( X = 78 | X = 78 or X = 70 ) = P ( X = 78 ) P ( X = 78 or X = 70 ) = P ( X = 78 ) P ( X = 78 ) + P ( X = 70 ) = e - 65 65 78 78 ! e - 65 65 78 78 ! + e - 65 65 70 70 ! 4 of 7
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4. ( 30 points ) X is a random variable distributed according to the exponential distribution with parameter 2. Y is a random variable with pdf g ( x ) = 16 xe - 4 x if x > 0 0 if x 0 X and Y are independent. ( a ) For t 0, what is P ( X > t )? (Your answer should be a function of t .) This the exponential distribution, so the answer is 1 - F ( t ) where F is the cdf of X , so e - 2t ( b ) For t 0, what is P ( Y > t )? (Your answer should be a function of t .) R t g ( x ) dx = R t 16xe - 4x dx . We apply integration by parts with u = x v = - 4e - 4x du = dx dv = 16e - 4x dx so lim a →∞ Z a t 16xe - 4x dx = lim a →∞ - 4xe - 4x a
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