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exam2sol - Rensselaer Polytechnic Institute Department of...

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Rensselaer Polytechnic Institute Department of Electrical, Computer, and Systems Engineering ECSE 2500: Engineering Probability, Fall 2011 Exam 2 Solutions The boxed numbers beside each problem give the mean (left) and median (right) score for each problem. If your score on a problem is lower than the median, you need to learn the concept better. The overall mean was 72.3 and the median was 73.5. There were 5 scores in the 90’s. 23.6 24.5 1. (30 points.) A continuous random variable X has the PDF given by: f X ( x ) = ( 1 9 ( x + 1) 2 x [ - 1 , 2] 0 otherwise 6.2 7 (a) (8 points.) Compute the CDF F X ( x ) of X . Be sure to specify F X ( x ) for all values of x . We get the CDF at x by integrating the PDF up to x. Clearly when x < - 1 , F X ( x ) = 0 and when x > 2 , F X ( x ) = 1 . When x [ - 1 , 2] , we have: F X ( x ) = Z x - 1 1 9 ( u + 1) 2 du = 1 27 ( u + 1) 3 u = x u = - 1 = 1 27 ( x + 1) 3 Thus F X ( x ) = 0 x < - 1 1 27 ( x + 1) 3 x [ - 1 , 2] 1 x > 2 A big problem here was using the wrong limits of integration. For example, many people com- puted the integral F X ( x ) = Z 2 - 1 1 9 ( u + 1) 2 du = 1 Of course, we already know this integral equals 1 since we were told f X ( x ) was a PDF. The idea of the CDF is that we want to know how much of this unit area has been accrued up to point x . A minor (but frequent) issue was not including the values of the CDF outside [ - 1 , 2] or just saying “0 elsewhere” instead of noting that F X ( x ) = 1 for x 2 . 1
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6.9 8 (b) (8 points.) Compute E ( X ) . E ( X ) = Z -∞ xf X ( x ) dx = Z 2 - 1 1 9 x ( x + 1) 2 dx = 1 9 Z 2 - 1 x 3 + 2 x 2 + x dx = 1 9 1 4 x 4 + 2 3 x 3 + 1 2 x 2 x =2 x = - 1 ! = 1 9 4 + 16 3 + 2 - 1 4 + 2 3 - 1 2 (1) = 1 9 45 4 = 5 4 = 1 . 25 (2) As a sanity check, since X only ranges between -1 and 2, we know that the expected value has to be in this range! 6.2 7 (c) (7 points.) Compute P ( X > 0) . An easy way to do this is to note that P ( X > 0) = 1 - P ( X 0) = 1 - F X (0) Since we computed F X ( x ) in part (a), we can just plug in to get P ( X > 0) = 1 - 1 27 = 26 27 = 0 . 96 Another approach is to integrate the PDF from 0 to 2, similar to part (b). 4.4 4 (d) (7 points.) Determine the conditional PDF f X ( x | X > 0) . Be sure to specify the conditional PDF for all values of x . We get the conditional PDF simply by dividing the PDF over the interval X > 0 by the proba- bility P ( X > 0) (which we computed as 26 27 in part (c)). The conditional PDF is 0 outside the interval X > 0 . Thus, f X ( x | X > 0) = ( 3 26 ( x + 1) 2 x [0 , 2] 0 otherwise For some reason, several people computed the CDF instead of the PDF, or arrived at the answer by determining the conditional CDF and taking the derivative to get the conditional PDF. This will get to the answer at the expense of a lot of extra work.
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