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PracticeMidterm2 Solutions

PracticeMidterm2 Solutions - Practice Midterm 2 1 1 Suppose...

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Practice Midterm 2 1
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1. Suppose that the probability mass function of a discrete random variable X is given by the following table x P ( X = x ) -3 0.2 -1 0.3 1.5 0.4 2 0.1 Find and graph the corresponding distribution function F ( x ). Solution: F ( x ) = 0 x < - 3 0 . 2 - 3 x < - 1 0 . 5 - 1 x < 1 . 5 0 . 9 1 . 5 x < 2 1 x 2 2
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2. Let X be a random variable with distribution function F ( x ) = 0 x < - 2 0 . 2 - 2 x < 0 0 . 3 0 x < 1 0 . 7 1 x < 2 1 x 2 Determine the probability mass function of X . Solution: x P ( X = x ) -2 0.2 0 0.1 1 0.4 2 0.3 3
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3. Suppose that the probability mass function of a discrete random variable X is given by the following table. x P ( X = x ) -2 0.1 -1 0.4 0 0.3 1 0.2 (a) Find EX . (b) Find EX 2 . (c) Find E [ X ( X - 1)]. (d) Find var ( X ). Solution: (a) EX = X x x · p ( x ) = ( - 2)(0 . 1) + ( - 1)(0 . 4) + (0)(0 . 3) + (1)(0 . 2) = - 0 . 4 (b) EX 2 = X x x 2 · p ( x ) = ( - 2) 2 (0 . 1) + ( - 1) 2 (0 . 4) + (0) 2 (0 . 3) + (1) 2 (0 . 2) = 1 . 0 (c) E [ X ( X - 1)] = X x x ( x - 1) · p ( x ) = ( - 2)( - 3)(0 . 1) + ( - 1)( - 2)(0 . 4) + (0)( - 1)(0 . 3) + (1)(0)(0 . 2) = 1 . 4 Alternatively: E [ X ( X - 1)] = E [ X 2 - X ] = EX 2 - EX = 1 . 0 - ( - 0 . 4) = 1 . 4 (d) var ( X ) = EX 2 - ( EX ) 2 = 1 . 0 - ( - 0 . 4) 2 = 0 . 84 4
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4. Let X and Y be two random variables with joint distribution X = 0 X = 1 Y=0 0.3 0.1 Y=1 0.2 0.4 (a) Find P ( X = 1 , Y = 0). (b) Find P ( X = 1). (c) Find P ( Y = 0). (d) Find P ( Y = 0 | X = 1). Solution: (a) P ( X = 1 , Y = 0) = 0 . 1 (b) P ( X = 1) = 0 . 1 + 0 . 4 = 0 . 5 (c) P ( Y = 0) = 0 . 3 + 0 .
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