ST 558 Homework 7

ST 558 Homework 7 - Kelly Harrison ST558 D 3/8/10 Homework...

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Kelly Harrison ST558 D 3/8/10 Homework 7 PROBLEM 1 x P(x) xP(x) x 2 P(x) -3 0.16666 7 -0.5 1.5 6 0.5 3 18 9 0.33333 3 3 27 5.5 46.5 E(x) = xP(x) = 5.5 E(x 2 ) = x 2 P(x) = 46.5 (2x+1) 2 = 4x 2 + 4x + 1 E[(2x+1) 2 ] = E[4x 2 + 4x + 1] = 4E(x 2 ) + 4E(x) + 1 = 4(46.5) + 4(5.5) + 1 = 209. PROBLEM 2 E[(x – 1) 2 ] = 10 E[(x – 2) 2 ] = 6 E[(x – 1) 2 ] = 10 E[x 2 – 2x + 1] = 10 E[x 2 ] – 2E(x) + 1 = 10 E[x 2 ] – 2E(x) = 9 E[(x – 2) 2 ] = 6 E[x 2 – 4x + 4] = 6 E[x 2 ] – 4E(x) + 4 = 6 E[x 2 ] – 4E(x) = 2 Subtract the following: E[x 2 ] – 2E(x) = 9
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E[x 2 ] – 4E(x) = 2 2E(x) = 7 E(x) = 7/2 μ = 7/2 = 3.5 Plug E(x) = 3.5 into either equation. I choose the top one: E[x 2 ] – 2(3.5) = 9 E[x 2 ] – 7 = 9 E[x 2 ] = 16 σ 2 = E[x 2 ] – μ 2 = 16 – 3.5 2 = 3.75 σ = √σ 2 = 3.75 = 1.93649 PROBLEM 3 Two tire quality experts examine stacks of tires and as ign a quality rating to each tire on a 3-point scale. P(x = 1) = .1 + .05 + .02 = .17.
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This note was uploaded on 05/06/2010 for the course ST 558 taught by Professor Staff during the Spring '08 term at NMT.

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ST 558 Homework 7 - Kelly Harrison ST558 D 3/8/10 Homework...

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