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# ps09_sol - University of Illinois Spring 2010 ECE 313...

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University of Illinois Spring 2010 ECE 313: Problem Set 9: Problems and Solutions Probability Density, Poisson Processes Due: Wednesday March 31 at 4 p.m. Reading: Ross, Chapter 5; Powerpoint Lecture Slides, Sets 22-25 Noncredit Exercises: Chapter 5: Problems 1-3, 5, 6, 15-19, 23-25, 32-34; Theoretical Exercises 1, 8; Self-Test Problems 1-4 1. [Probability Density] Random variable X has the following PDF: f X ( u ) = 1 5 0 < u 1 1 2 1 < u 2 1 10 2 < u 5 0 otherwise Answer each of the following questions by shading and labeling the area of a region on a graph. Do not perform any integrals. (a) What is P (2 < X 3)? Solution: 1 / 10. The plot should show shading between u = 2 and u = 3. (b) What is P ( - 1 X 1 . 5)? Solution: 9 / 20 (c) What is P (2 X 6)? Solution: 3 / 10 (d) What is P (1 . 173 X 1 . 174)? Solution: 0 . 0005. The graph should show a shaded rectangle of width 0.001 and height 0.5. (e) What is F X ( v )? Solution: F X ( v ) = 0 v 0 v 5 0 v 1 1 5 + v - 1 2 1 v 2 7 10 + v - 2 10 2 v 5 1 5 v 2. [Black Swans] A recent census conducted in the country of Metasylvania found that the body weight of adult Metasyl- vanians, measured in pounds, is a random variable X with the following probability density function: f X ( u ) = Ae - 0 . 02( u - 80) 80 u 0 else The same survey found that the personal wealth of adult Metasylvanians, measured in Metasylvanian dollars, is a random variable Y with the following probability density function: f Y ( u ) = B ( u 100 ) - 1 . 5 100 u 0 else (a) Find the constants A and B . Solution: The PDFs must integrate to 1, therefore A = 0 . 02, B = 0 . 5. (b) Notice that all Metasylvanians weigh at least 80 pounds.

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i. What is the probability that an individual Metasylvanian weighs at least 160 pounds? So- lution: Z 160 0 . 02 e - 0 . 02( u - 80) du = e - 80 / 50 0 . 2 ii. What is the probability that an individual Metasylvanian weighs at least 800 pounds? So- lution: Z 800 0 . 02 e - 0 . 02( u - 80) du = e - 720 / 50 5 . 6 × 10 - 7 iii. What is the expected weight of a Metasylvanian? Solution: Z = X - 80 is an exponential random variable with expectation of 50, therefore E [ X ] = 80 + 50 = 130 pounds. (c) Notice that every Metasylvanian has a personal wealth of at least \$100. i. What is the probability that a Metasylvanian has a personal wealth of greater than \$200? Solution: Z 200 0 . 5 u 100 - 1 . 5 du = 200 100 - 0 . 5 0 . 7 ii. What is the probability that a Metasylvanian has a personal wealth of greater than \$1000?
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ps09_sol - University of Illinois Spring 2010 ECE 313...

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