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hw4 fall 07 solns

# hw4 fall 07 solns - 1 ISyE 3770 Fall 2007 Homework#4(Covers...

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1 ISyE 3770 – Fall 2007 Homework #4 (Covers Modules 14–15) — Solutions 1. Suppose that X is a discrete random variable with p.m.f. Pr ( X = - 1) = 0 . 5, Pr ( X = 0) = 0 . 2, and Pr ( X = 2) = 0 . 3. Find E [ X ], Var ( X ), and E [ X 3 ]. Solution: By definition of expectation and variance for discrete random variables: E [ X ] = Pr ( X = - 1) · ( - 1) + Pr ( X = 0) · 0 + Pr ( X = 2) · 2 = - 0 . 5 + 0 + 0 . 6 = 0 . 1 Var [ X ] = E [ X 2 ] - E [ X ] 2 = 0 . 5 · ( - 1) 2 + 0 . 2 · (0) 2 + 0 . 3 · (2) 2 - (0 . 1) 2 = 1 . 69 E [ X 3 ] = Pr ( X = - 1) · ( - 1) 3 + Pr ( X = 0) · (0) 3 + Pr ( X = 2) · (2) 3 = - 0 . 5 + 0 + 2 . 4 = 1 . 9 . 2. Suppose that X is a continuous random variable with p.d.f. f ( x ) = cx for 0 < x < 1. Find c , E [ X ], Var ( X ), and E [ X 3 ]. Solution: Use the fact that the p.d.f. of a random variable, integrated over its domain, equals one: Z 1 0 cx dx = c 2 = 1 . Thus, c = 2 and so, using the definition of expectation and variance for a continuous variable, we obtain: E [ X ] = Z 1 0 xf ( x ) dx = Z 1 0 2 x 2 dx = 2 3 Var [ X ] = E [ X 2 ] - E [ X ] 2 = Z 1 0 x 2 f ( x ) dx - 4 9 = 1 2 - 4 9 = 1 18 E [ X 3 ] = Z 1 0 x 3 f ( x ) dx = 2 5 . 3. Suppose that X Unif( - 1 , 6). Compare the upper bound on the probability Pr ( | X - μ | ≥ 1 . 5 σ ) obtained from Chebychev’s inequality with the exact probability.

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hw4 fall 07 solns - 1 ISyE 3770 Fall 2007 Homework#4(Covers...

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