hw11solns - Solutions to Homework 11 6.23 For a fixed Y y...

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Solutions to Homework 11 6.23 For a fixed [0,1] Y y , we have0 1 x y   , then the marginal pdf of Y is 1 0 ( ) ( , ) 2 2(1 ), 0 y 1 Y y f y f x y dx dx y  Then, the conditional probability density function for X given Y y is , | ( ,0.25) 2 4 ( | 0.25) , 0 0.75 (0.25) 2(1 0.25) 3 X Y X Y Y f x f x y x f The probability that chemical I comprise more than half of the mixture if a fourth of the mixture is chemical II is | 0.5 0.75 0.5 ( 0.5| 0.25) ( | 0.25) 4 3 1 3 X Y P X Y f x y dx dx 6.28 Let X denote the time of one inspector interrupting a production line in a given day. Let Y denote the time of the other inspector interrupting a production line in a given day. Then X and Y can both be modeled to have a uniform distribution in[0,8]. Because X and Y are independent, the joint pdf of X and Y is 1 ,0 8,0 8 ( , ) 64 0, otherwise x y f x y Define set   ( , ),0 8,0 8, 4 A x y x y x y , A is the shadow area in the figure below
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Thus the probability that the two interruptions will be more than four hours apart is 2 (| | 4) ( , ) 1 64 1 (Shadow) 64 1 1 2 4 64 2 1 4 A A P X Y f x y dxdy dxdy S      6.31
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hw11solns - Solutions to Homework 11 6.23 For a fixed Y y...

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