ECSE 2500 S10 Exam 2 Solutions

ECSE 2500 S10 Exam 2 Solutions - ECSE 2500: Engineering...

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ECSE 2500: Engineering Probability Exam 2 Solutions April 13, 2010 1. Solution: (a) X is binomial random variable with n = 5 and p = 0 . 3 P [ X = k ] = 5 k ! (0 . 3) k (0 . 7) 5 - k , k = 0 , 1 ,..., 5 P [ X 2] = 1 - P [ X = 0] - P [ X = 1] = 0 . 4718 (b) E [ X ] = np = 5(0 . 3) = 1 . 5 (c) V ar [ X ] = np (1 - p ) = 5(0 . 3)(1 - 0 . 3) = 1 . 05 2. Solution: (a) X is Poisson random variable with α = 1( call per minute ) × 10( minutes ) = 10. The pmf of X is P [ X = k ] = 10 k k ! e - 10 , k = 0 , 1 ,... (b) E [ X ] = α = 10 (c) V ar [ X ] = α = 10 1
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(d) F X ( x ) = P [ X x ] = 0 x < 0 b x c X k =0 10 k k ! e - 10 x 0 where b x c is the largest integer less than or equal to x . (e) P [ X = 0] = 10 0 0! e - 10 = 1 1 e - 10 = e - 10 = 4 . 54 × 10 - 5 3. Solution: (a) For uniform random variable X , f X ( x ) = ( 1 2 - 1 < x < 1 0 otherwise (b) F X ( x ) = 0 x ≤ - 1 x +1 2 - 1 < x < 1 1 x 1 (c) E [ X ] = 1 + ( - 1) 2 = 0 (d) V ar [ X ] = (1 - ( - 1))
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ECSE 2500 S10 Exam 2 Solutions - ECSE 2500: Engineering...

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