ECSE 2500 S11 Exam 3b Solutions

ECSE 2500 S11 Exam 3b Solutions - Exam 3v2 Answers,...

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Page 1 Exam 3v2 Answers, 5/17/2011, ECSE-2500, Engineering Probability, RPI (Eng Prob Spring 2011) 5/18/2011 1:49:34 PM http://www.ecse.rpi.edu/Homepages/wrf/pmwiki/EngProbSpring2011/Exam3v2Sol To print higher -resolution math symbols, click the Hi-Res Fonts for Printing button on the jsMath control panel. E XAM 3 V 2 A NSWERS , 5 / 17 / 2011 , ECSE - 2500 , E NGINEERING P ROBABILITY , RPI ECSE-2500 Engineering Probability, RPI, Spring 2011 Home Page (for grader use only) Finish order _________ Grade _______________ /60 NAME: ___WRF Answers_______________________________________________EMAIL:___________________________ RIN:____________ _____ There are eight pages with eight questions. Answer any six questions. (If you answer more, then we will grade the first ones.) We record the order in which you hand in your exam the better to keep track of the exams, and sometimes to plot scatterplots of order vs grade. (There is never any significant correlation.) Your order does not affect your grade. This exam is closed book but a calculator and two 3-sided letter-paper-size note sheets are allowed. You may not use other computers or communication devices, or share material with other students. For full points, your answers must be simplified as much as reasonable. Blue marks an answer. Green marks a comment to the answer. Numbering error: There were two question #1 1. [_____/10] Evaluate and give me the answer as an expression perhaps involving and . Show your work. is a Gaussian pdf with and integrates to 1. So, 2. [_____/10] Random variable X is continuous U[0,2]. Random variable Y is continuous U[0,3]. Random variable Z is continuous U[0,4]. Random variable W min(X,Y,Z). Compute P[0<W<3]. You could do it the detailed way. Or, you could notice that min(X,Y,Z) <=2 and so P[0<W<3]=1.
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ECSE 2500 S11 Exam 3b Solutions - Exam 3v2 Answers,...

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