08Ma02bMidtermSolns

08Ma02bMidtermSolns - MATH 2B 2009 MIDTERM SOLUTIONS...

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Unformatted text preview: MATH 2B 2009 MIDTERM SOLUTIONS Problem 1. Let H be the event that the plant has been watered and N that the plant has not be watered. We are given that P ( H ) = 0 . 9 and P ( N ) = 0 . 1. Also, let S , W and D be the events that the plant is sick, well or has died. We are given that P ( D | N ) = 0 . 8, P ( D | H ) = 0 . 5, P ( S | N ) = 0 . 2, P ( S | H ) = 0 . 35, P ( W | N ) = 0, P ( W | H ) = 0 . 15. a) The probability that the plant is alive is P ( S W ) = 1- P ( D ). Then by the rule of average conditional probabilities P ( D ) = P ( D | H ) P ( H ) + P ( D | N ) P ( N ) = 0 . 5 . 9 + 0 . 8 . 1 = 0 . 45 + 0 . 08 = 0 . 53 . So, the probability that the plant is alive is P (alive) = 0 . 47 . b) By Bayes rule, the probability that the plant has not been watered if it died is P ( N | D ) = P ( D | N ) P ( N ) P ( D ) = . 8 . 1 . 53 = 0 . 1509 . Problem 2. (a) Calculate the mean and variance of X . Solution. E ( X ) = ( t )(2 e- 2 t ) dt = 2 t (- 1 2 ) e- 2 t ] - (- 1 2 ) e- 2 t dt = (- 1 2 ) e- 2 t ] = 1 2 . Similarly, E ( X 2 ) = ( t 2 )(2 e- 2 t ) dt = 1 2 Therefore, V ar ( X ) = E ( X 2 )- E ( X ) 2 = 1 4 . (b) Calculate a density function for X + 2 Y . Solution. Note f 2 Y ( t ) = 1 2 f Y ( t 2 ) = e- t if 0 t , and 0 otherwise. Using the convolution formula, we get for t > 0, f X +2 Y ( t ) = - f X ( x ) f 2 Y ( t- x ) dx = t (2 e- 2 x )( e- ( t- x ) ) dx = t 2 e x- t dx = 2 e- t t 2 e- x dx =- 2 e- t e- x ] t =- 2 e- 2 t + 2 e- t For t 0, either f X ( x ) or f 2 Y ( t- x ) vanishes, so f X +2 Y ( t ) = 0 (c) Calculate a density function for W = X 1+ Y 1 2 MATH 2B 2009 MIDTERM SOLUTIONS Solution. Note f X 1+ Y ( t ) = d dt F X 1+ Y ( t ) = d dt P ( X 1+ Y t ) = d dt P ( X t + tY ) By the independence of X and Y , P ( X t + tY ) = x t + ty f X ( x ) f Y ( y ) dxdy .....
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08Ma02bMidtermSolns - MATH 2B 2009 MIDTERM SOLUTIONS...

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