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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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 Winter '08
 Makarov,N
 Math, Differential Equations, Statistics, Equations, Probability

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