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Practice Problems from Chapters 4 and 5 - Faculty web pages

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use of a (transparent) methodologicalmodel in a contract: advantages data <strong>and</strong> price-fixing mechanismscan be shared between theparties of the contract by using the model, the“haggling” tends to shift away<strong>from</strong> the price <strong>and</strong> more on thekey factors that enhances priceaccountability the model gives the opportunityto use the embedded know-howeven to beginners.CustomerModelSupplierSMEF 2005 Roma / 16-17-18 Marzo 2005 6/20


(b) One fuse in part 31a is in a primary system, <strong>and</strong> the other is in a backup system thatcomes into use only if the primary system fails. The total effective length of life of thetwo fuses is then Y 1 + Y 2 . Find P (Y 1 + Y 2 ≤ 1).32. (WMS7 Exercises 5.79 <strong>and</strong> 5.93) The r<strong>and</strong>om variablesY 1 <strong>and</strong> Y 2 are uniformly distributed over the triangle atright.y 2(0, 1)(a) Find E(Y 1 Y 2 )(b) Find Cov(Y 1 , Y 2 )(c) Are Y 1 <strong>and</strong> Y 2 independent?(−1, 0)(1, 0)y 133. (WMS7 Exercise 5.80 <strong>and</strong> 5.109a) The r<strong>and</strong>om variables Y 1 <strong>and</strong> Y 2 have joint probabilitydensity given byf(y 1 , y 2 ) ={y1 + y 2 if 0 ≤ y 1 ≤ 1 <strong>and</strong> 0 ≤ y 2 ≤ 10 otherwise.Let T = 30Y 1 + 25Y 2 . Find the expected value <strong>and</strong> variance of T .34. (WMS7 Exercise 5.82) The r<strong>and</strong>om variables Y 1 <strong>and</strong> Y 2 have joint density function given byf(y 1 , y 2 ) ={1/y1 if 0 ≤ y 2 ≤ y 1 ≤ 10 otherwise.Find E(Y 1 − Y 2 ).35. (WMS7 Exercise 5.95) The discrete r<strong>and</strong>om variables Y 1 <strong>and</strong> Y 2 have the joint probabilityfunctionP (Y 1 = y 1 , Y 2 = y 2 ) = 1 if (y 1 , y 2 ) ∈ {(−1, 0), (0, 1), (1, 0)}.3(a) Find Cov(Y 1 , Y 2 ).(b) Are Y 1 <strong>and</strong> Y 2 independent?36. (WMS7 Exercise 5.103) Assume Y 1 , Y 2 , <strong>and</strong> Y 3 are r<strong>and</strong>om variables withE(Y 1 ) = 2, E(Y 2 ) = −1, E(Y 3 ) = 4,V (Y 1 ) = 4, V (Y 2 ) = 6, V (Y 3 ) = 8,Cov(Y 1 , Y 2 ) = 1, Cov(Y 1 , Y 3 ) = −1, Cov(Y 2 , Y 3 ) = 0.Find the expected value <strong>and</strong> variance of 3Y 1 + 4Y 2 − 6Y 3 .

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