# Instructor's Solution Manual for Probability and Statistics - download pdf or read online By Sharon L. Myers, Keying Ye

Instructor's resolution guide for the eighth variation of Probability and facts for Engineers and Scientists by means of Sharon L. Myers, Raymond H. Myers, Ronald E. Walpole, and Keying E. Ye.

Note: a number of the routines within the more moderen ninth variation also are present in the eighth variation of the textbook, in basic terms numbered in a different way. This answer handbook can frequently nonetheless be used with the ninth variation by way of matching the routines among the eighth and ninth versions.

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Instructor's Solution Manual for Probability and Statistics - download pdf or read online

Instructor's answer handbook for the eighth version of chance and statistics for Engineers and Scientists by means of Sharon L. Myers, Raymond H. Myers, Ronald E. Walpole, and Keying E. Ye.

Note: the various routines within the more moderen ninth variation also are present in the eighth variation of the textbook, in basic terms numbered in a different way. This resolution guide can usually nonetheless be used with the ninth version via matching the routines among the eighth and ninth variants.

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Extra info for Instructor's Solution Manual for Probability and Statistics for Engineers and Scientists (8th Edition)

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1 x(3x2 + 1) 0 2 3 1 y 14 + y2 2 0 (b) E(X + Y ) = E(X) + E(Y ) = 2E(X) = E(XY ) = 23 = 32 14 12 1 0 + 1 0 1 2 xy(x2 + y 2 ) dx dy = 1 4 dx = 3/4 + 1/2 = 5/4. dy 3 . 8 = 2 1 7 25 73 (c) V ar(X) = E(X 2 ) − [E(X)]2 = 12 0 x2 (3x2 + 1) dx − 58 = 15 − 64 = 960 , and 2 73 3 5 1 V ar(Y ) = 960 . Also, Cov(X, Y ) = E(XY ) − E(X)E(Y ) = 8 − 8 = − 64 . 77 (a) E(Y ) = (b) E(Y 2 ) = ∞ ye−y/4 dy = 4. 0 ∞ 2 −y/4 y e dy = 32 0 29 . 240 and V ar(Y ) = 32 − 42 = 16. 78 (a) The density function is shown next. 32.

67 (a) x f (x) x2 + 3 4 = 1 2 = 1 12 3 2 ∞ 2 1/2 0 3 2 dx = 1 3/4 ∞ −y e 2 ye−y(1+x) dx dy = − 1/2 2 (x 0 + y 2 ) dxdy = 1/2 0 3 2 ∞ −3y e dy 2 ∞ e−yx |2 dy = x2 y + 1/2 y3 3 0 dx 1 . 16 x2 + 1 3 dx = 53 . 0120 (b) A histogram is shown next. 5 x2 2 1 2 + xy dy = 38 . 1)5−x , for x = 0, 1, 2, 3, 4, 5. 70 (a) g(x) = h(y) = 2 1 3 1 3x−y 9 3x−y 9 dy = 3xy−y 2 /2 9 dx = 4 3 2 1 = x 3 − 16 , for 1 < x < 3, and − 29 y, for 1 < y < 2. (b) No, since g(x)h(y) = f (x, y). (c) P (X > 2) = 3 2 x 3 − 1 6 dx = x2 6 − x 6 3 2 = 23 .

1353. 6321. 1 2 dx2 = 2(1 − x1 ), for 0 < x1 < x1 x2 2 dx1 = 2x2 , for 0 < x2 < 1. 5 0 2 dx1 dx2 (d) fX1 |X2 (x1 |x2 ) = 10 (1 10 = 1 , x2 1. 2. for 0 < x1 < x2 . 78 (a) fX1 (x1 ) = 0 1 6x2 dx2 = 3x21 , for 0 < x1 < 1. Apparently, fX1 (x1 ) ≥ 0 and 1 1 f (x1 ) dx1 = 0 3x21 dx1 = 1. So, fX1 (x1 ) is a density function. 0 X1 (b) fX2 |X1 (x2 |x1 ) = f (x1 ,x2 ) fX1 (x1 ) = 6x2 3x21 = 2 xx22 , for 0 < x2 < x1 . 5 x2 0 dx2 = 25 . 79 (a) g(x) = 9 (16)4y ∞ x=0 1 4x = 9 1 (16)4y 1−1/4 = 34 · 41x , for x = 0, 1, 2, .

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### Instructor's Solution Manual for Probability and Statistics for Engineers and Scientists (8th Edition) by Sharon L. Myers, Keying Ye

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