Mont4e_sm_ch05_sec04

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Unformatted text preview: ;Z< ⎟ 0.00031 0.00031 ⎠ ⎝ = P(−1.5 < Z < 1.5) = P( Z < 1.5) − P( Z < −1.5) = 0.8664 Probability that Y is within specification limits is 0.23034 − 0.23 ⎞ ⎛ 0.22966 − 0.23 P(0.22966 < X < 0.23034) = P⎜ <Z< ⎟ 0.00017 ⎠ ⎝ 0.00017 = P(−2 < Z < 2) = P( Z < 2) − P( Z < −2) = 0.9545 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 1 fY|X =x 2 f XY ( x , y ) 2πσ x σ y 1 − ρ = = f X ( x) e 2 ⎡1⎡ ⎛ x−μ x ⎤ σ ⎢ ( y − ( μ Y + ρ Y ( x − μ x )) ⎥ + (1− ρ 2 ) ⎜ ⎜σ 2⎢ ⎢σ Y ⎣ σX x ⎦ ⎝ ⎣ − 2 2 (1− ρ ) 1 2π σ x Also, fx(x) = 1 2π σ x e ⎡ x−μx ⎤ ⎢ ⎥ σx ⎦ −⎣ 2 e ⎡ x−μ x ⎤ ⎢ ⎥ σx ⎦ −⎣ 2 2 By definition, 5-34 2 ⎞ ⎟ ⎟ ⎠ 2 ⎤ ⎥ ⎥ ⎦ 1 fY | X = x 2 f ( x, y ) 2πσ x σ y 1 − ρ = XY = f X ( x) e 2 ⎡1⎡ ⎛ x −μ x ⎤ σY ⎢ ( x −μ x )) ⎥ + (1−ρ 2 )⎜ ⎢ ( y − (μ Y + ρ ⎜σ 2 ⎢ σY ⎣ σX ⎦ ⎝x −⎣ 2 2(1− ρ ) 2π σ x 1 2π σ y 1 − ρ2 e 2⎤ ⎥ ⎥ ⎦ 2 1 = ⎞ ⎟ ⎟ ⎠ ⎤ σ 1⎡ ( y − (μY + ρ Y ( x −μ x )) ⎥ 2⎢ σX σY ⎣ ⎦ − 2(1−ρ 2 ) e ⎡ x −μ x ⎤ ⎥ ⎢ σx ⎦ −⎣ 2 2(1−ρ ) 2 Now fY|X=x is in the form of a normal distribution. b) E(Y|X=x) = μ y + ρ σy σx ( x − μ x ) . This answer can be seen from part a). Since the PDF is in the form of a normal distribution, then the mean can be obtained from the exponent. c) V(Y|X=x) = σ2 (1 − ρ2 ) . This answer can be seen from part a). Since the PDF is in the form of a y normal distribution, then the variance can be obtained from the exponent. 5-51 ⎡ ( x − μ X ) 2 ( y − μY ) 2 ⎤ ⎡ ⎤ −1⎢ + 2 2 σ Y 2 ⎥ ⎥ dxdy = ⎦ 1 f XY ( x, y )dxdy = ∫ ∫ ⎢ 2πσ X σ Y e ⎣ σ X ∫∫ ⎥ − ∞− ∞ − ∞− ∞ ⎢ ⎣ ⎦ ∞∞ ⎡ ∫⎢ −∞⎢ ⎣ ∞∞ ∞ 1 e 2πσ X ⎡ ( x−μ X )2 ⎤ −1⎢ 2 2⎥ X ⎣ ⎦ σ ⎤ ∞⎡ ⎥ dx ∫ ⎢ ⎥ −∞⎢ ⎦ ⎣ 1 e 2πσ...
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This document was uploaded on 02/09/2014.

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