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mont4e_sm_ch05_sec04

mont4e_sm_ch05_sec04 - 2 2 2 2 Also U = E[U E(U]2 = a 2 E X...

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5-32 Section 5-4 5-46 a) 1 0 - 2 0.00 x 0 - 1 0.01 0.02 0 0.03 0.04 1 2 z(0) 3 - 10 4 y b)
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5-33 1 0 - 2 0.00 x 0.01 0 - 1 0.02 0.03 0.04 0 0.05 0.06 0.07 1 2 z(.8) 3 - 10 4 y c) 1 0 - 2 0.00 x 0.01 0 - 1 0.02 0.03 0.04 0 0.05 0.06 0.07 1 2 z(-.8) 3 - 10 4 y 5-47 Because 0 = ρ and X and Y are normally distributed, X and Y are independent. Therefore,
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5-34 (a) P(2.95 < X < 3.05) = ) ( 04 . 0 3 05 . 3 04 . 0 3 95 . 2 < < Z P = 0.7887 (b) P(7.60 < Y < 7.80) = ) ( 08 . 0 70 . 7 80 . 7 08 . 0 70 . 7 60 . 7 < < Z P = 0.7887 (c) P(2.95 < X < 3.05, 7.60 < Y < 7.80) = P(2.95 < X < 3.05) P(7.60 < Y < 7.80) = 6220 . 0 7887 . 0 ) ( ) ( 2 08 . 0 70 . 7 80 . 7 08 . 0 70 . 7 60 . 7 04 . 0 3 05 . 3 04 . 0 3 95 . 2 = = < < < < Z P Z P 5-48 (a) ρ = cov(X,Y)/ σ x σ y =0.6 cov(X,Y)= 0.6*2*5=6 (b) the marginal probability distribution of X is normal with mean μ x , σ x . (c) P(X<116) =P(X-120<-4)=P((X_120)/5<-0.8)=P(Z<-0.8) = 0.21 (d) The conditional probability distribution of X given Y=102 is bivariate normal distribution with mean and variance μ X|y=102 = 120 – 100*0.6*(5/2) +(5/2)*0.6(102) = 123 σ 2 X|y=102 = 25(1-0.36) =16 (e) P(X<116|Y=102)=P(Z<(116-123)/4)=0.040 5-49 Because ρ= 0 and X and Y are normally distributed, X and Y are independent. Therefore, μ X = 0.1 mm, σ X =0.00031 mm, μ Y = 0.23 mm, σ Y =0.00017 mm Probability X is within specification limits is 0.8664 ) 5 . 1 ( ) 5 . 1 ( ) 5 . 1 5 . 1 ( 00031 . 0 1 . 0 100465 . 0 00031 . 0 1 . 0 099535 . 0 ) 100465 . 0 099535 . 0 ( = < < = < < = < < = < < Z P Z P Z P Z P X P Probability that Y is within specification limits is 0.9545 ) 2 ( ) 2 ( ) 2 2 ( 00017 . 0 23 . 0 23034 . 0 00017 . 0 23 . 0 22966 . 0 ) 23034 . 0 22966 . 0 ( = < < = < < = < < = < < Z P Z P Z P Z P X P Probability that a randomly selected lamp is within specification limits is (0.8664)(0.9594) = 0.8270 5-50 a) By completing the square in the numerator of the exponent of the bivariate normal PDF, the joint PDF can be written as 2 ) 1 ( 2 ) 1 ( )) ( ( ( 1 2 | 2 2 2 2 2 2 2 1 1 2 1 ) ( ) , ( + + = = = x x x x x X Y Y Y x x x x y y x X XY x X Y e e x f y x f f σ μ ρ π πσ Also, f x (x) = 2 2 1 2 x x
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