Asst4-suggested

# Asst4-suggested - .44 or X ≥ 5.72 = P(X = 6 =.04...

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Statistics 427: Homework #4 Solutions to Suggested Problems 3.36 P(x) .8 .1 .08 .02 x 0 1,000 5,000 10,000 H(x) 0 500 4,500 9,500 E[h(X)] = 600. E(Premium –h(X))=100. Premium should be \$100 plus expected value of damage minus deductible (h(X)),that is \$700. 3.44 a. k 2 3 4 5 10 2 1 k .25 .11 .06 .04 .01 b. 64 . 2 ) ( 6 0 = = = x x p x μ , , 37 . 2 ) ( 2 6 0 2 2 = = = μ σ x x p x 54 . 1 = σ Thus μ - 2 σ = -.44, and μ + 2 σ = 5.72, so P(|x- μ | 2 σ ) = P(X is at least 2 s.d.’s from μ ) = P(X
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Unformatted text preview: -.44 or X ≥ 5.72) = P(X = 6) = .04. Chebyshev’s bound of .25 is much too conservative. For k = 3,4,5, and 10, P(|x-μ | ≥ k σ ) = 0, here again pointing to the very conservative nature of the bound 2 1 k . c. μ = 0 and 3 1 = , so P(|x-μ | ≥ 3 σ ) = P(| X | ≥ 1)= P(X = -1 or +1) = 9 1 18 1 18 1 = + , identical to the upper bound. d. Let p(-1) = 25 24 50 1 50 1 ) ( , ) 1 ( , = = + p p...
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