exam2sol - Rensselaer Polytechnic Institute Department of...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 ² 4 + 16 3 + 2 - 1 4 + 2 3 - 1 2 ³ (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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/29/2011 for the course ECSE 2500 taught by Professor Radke during the Spring '11 term at Rensselaer Polytechnic Institute.

Page1 / 9

exam2sol - Rensselaer Polytechnic Institute Department of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online