# HW5 - Section 4.2 Variance 4.34 Solution: = (2) 0.3 + 3 0.2...

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Section 4.2 Variance 4.34 Solution: μ = ( - 2) · 0 . 3 + 3 · 0 . 2 + 5 · 0 . 5 = 2 . 5 E ( X 2 ) = ( - 2) 2 · 0 . 3 + 3 2 · 0 . 2 + 5 2 · 0 . 5 = 15 . 5 σ 2 = E ( X 2 ) - μ 2 = 15 . 5 - (2 . 5) 2 = 9 . 25 Thus, σ = 9 . 25 = 3 . 041 . 4.35 Solution: μ = 2 · 0 . 01 + 3 · 0 . 25 + 4 · 0 . 4 + 5 · 0 . 3 + 6 · 0 . 04 = 4 . 11 E ( X 2 ) = 2 2 · 0 . 01 + 3 2 · 0 . 25 + 4 2 · 0 . 4 + 5 2 · 0 . 3 + 6 2 · 0 . 04 = 17 . 63 Thus, σ 2 = E ( X 2 ) - μ 2 = 0 . 7379 4.36 Solution: μ = 0 · 0 . 4 + 1 · 0 . 3 + 2 · 0 . 2 + 3 · 0 . 1 = 1 E ( X 2 ) = 0 2 · 0 . 4 + 1 2 · 0 . 3 + 2 2 · 0 . 2 + 3 2 · 0 . 1 = 2 σ 2 = E ( X 2 ) - μ 2 = 2 - 1 = 1 . 4.38 Solution: μ = R 1 0 x 2 5 ( x + 2) dx = 8 / 15 E ( X 2 ) = R 1 0 x 2 2 5 ( x + 2) dx = 11 / 30 Thus, σ 2 = E ( X 2 ) - μ 2 = 11 / 30 - (8 / 15) 2 = 37 / 450 . 4.43 Solution: EY = R 0 (3 x - 2) 1 4 e - x 4 dx = - R 0 (3 x - 2) de - x 4 = - (3 x - 2) e - x 4 | 0 + R 0 e - x 4 d (3 x - 2) = - 2 - 12 e - x 4 | 0 = - 2 + 12 = 10 . EY

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## This note was uploaded on 02/04/2009 for the course STATISTIC 3011 taught by Professor Sally during the Fall '08 term at University of Minnesota Crookston.

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HW5 - Section 4.2 Variance 4.34 Solution: = (2) 0.3 + 3 0.2...

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