Quiz2-2007-2- test 2- 610 question 1

Quiz2-2007-2 test 2 610 question 1

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Unformatted text preview: The Jacobian matrix is then = = Y1 − ln Y2 J = 0 1/y2 1 −1/y2 • From the definition of Y1 and Y2 , we have | det(J )| = 1/y2 • Then, fY1 ,Y2 (y1 , y2 ) = | det(J )| · fX1 ,X2 (ln y2 , y1 − ln y2 ) λ2 −λy1 = e , 0 < ln y2 < y1 < ∞ y2 10. (8%) Suppose component lifetimes are exponentially distributed with mean 10 hours. Find the probability that average lifetime of three independent components exceeds 30 hours. P X1 +X2 +X3 3 > 30 = e9 + 9e−9 + 92 e−9 2 = 0.006263. Sol) Let X1 , X2 , X3 ∼ Exp(1/10) denote the lifetime of the three independent components. Then, we want to find P X1 +X2 +X3 3 > 30 or P (X1 + X2 + X3 > 90). Since X1 , X2 , X3 ∼ Exp(1/10), X1 + X2 + X3 ∼ Γ(3, 1/10). Relating it to Poison distribution, we then have P (X1 + X2 + X3 > 90) = P (N90 ≤ 2), where N90 ∼ P oison(90 ...
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This document was uploaded on 11/01/2013.

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