ST 558 Homework 7

# ST 558 Homework 7 - Kelly Harrison ST558 D Homework 7...

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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 assign a quality rating to each tire on a 3-point scale. P(x = 1) = .1 + .05 + .02 = .17. P(x = 1 | y = 1) = .1/(.1 + .1 + .03) = .5882 Since P(x = 1) P(x = 1 | y = 1), then x and y are not independent. PROBLEM 4 Suppose x and y have the following joint probability function: a) x y xy 2 P(x,y) xy 2 P(x, y) 2 1 2 0.1 0.2 2 3 18 0.2 3.6 2 5 50 0.1 5 4 1 4 0.15 0.6 4 3 36 0.3 10.8 4 5 100 0.15 15 35.2 E(xy 2 ) = xy 2 P(x,y) = 35.2 b) P(x = 2) = .1 + .2 + .1 = .4 P(x = 4) = .15 + .3 + .15 = .6 μ x = xP(x) = 2(.4) + 4(.6) = 3.2
P(y = 1) = .1 + .15 = .25

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