09_06ans - STAT 400 1. Outcomes x TT 0 HT TH 1 HH A...

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STAT 400 Examples for 09/06/2011 Fall 2011 1. A balanced (fair) coin is tossed twice. Outcomes x f ( x ) Let X denote the number of H's. Construct the probability distribution of X. S = { HH , HT , TH , TT } X = 2 1 1 0 TT 0 1 / 4 HT TH 1 1 / 2 HH 2 1 / 4 1.00 Just for fun: Outcomes x f ( x ) Suppose P(H) = 0.60, TT 0 0.40 × 0.40 = 0.16 P(T) = 0.40. HT TH 1 0.60 × 0.40 + 0.40 × 0.60 = 0.48 HH 2 0.60 × 0.60 = 0.36 1.00 2. Suppose a random variable X has the following probability distribution: x f ( x ) 10 0.20 11 0.40 12 0.30 13 0.10 a) Find the expected value of X, E(X). x f ( x ) x f ( x ) 10 0.2 2.0 11 0.4 4.4 12 0.3 3.6 13 0.1 1.3 1.0 11.3 μ X = E(X) = x x f x all ) ( = 11.3 .
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b) Find the variance of X, Var(X). x f ( x ) ( x - μ X ) ( x - μ X ) 2 f ( x ) 10 0.2 1.3 1.69 0.2 = 0.338 11 0.4 0.3 0.09 0.4 = 0.036 12 0.3 0.7 0.49 0.3 = 0.147 13 0.1 1.7 2.89 0.1 = 0.289 0.810 2 X σ = Var(X) = ( ) - x x f x all 2 X ) ( μ = 0.81 . OR x f ( x ) x 2 f ( x ) 10 0.2 20.0 11 0.4 48.4 12 0.3 43.2 13 0.1 16.9 128.5 2 X σ = Var(X) = [ ] 2 all 2 E(X) ) ( x x f x - = 128.5 – (11.3) 2 = 0.81 . c) Find the standard deviation of X, SD(X). X σ = SD(X) = 2 X σ = 0.9 . 3. Suppose E(X) = 7, SD(X) = 3. a) Y = 2 X + 3. Find E(Y) and SD(Y). E(Y) = 2 E(X) + 3 = 17 . SD(Y) = | 2 | SD(X) = 6 . b) W = 5 – 2 X. Find E(W) and SD(W). E(W) = 5 – 2 E(X) = 9 . SD(Y) = | 2 | SD(X) = 6 .
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4. Suppose a discrete random variable X has the following probability distribution: P( X = 0 ) = e 2 - , P( X = k ) = ! 2 1 k k , k = 1, 2, 3, … a) Find E ( X ). E ( X ) = x x f x all ) ( = 0 ( ) 2 1 2 e - + = 1 ! 2 1 k k k k = ( ) = - 1 ! 1 2 1 k k k = ( ) = - - 1 1 ! 1 2 1 2 1 k k k = = 0 ! 2 1 2 1 n n
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09_06ans - STAT 400 1. Outcomes x TT 0 HT TH 1 HH A...

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