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

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 < u 1 1 2 1 < u 2 1 10 2 < u 5 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: . 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 ) = v v 5 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- . 02( u- 80) 80 u 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 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. i. What is the probability that an individual Metasylvanian weighs at least 160 pounds? So- lution: Z 160 . 02 e- . 02( u- 80) du = e- 80 / 50 . 2 ii. What is the probability that an individual Metasylvanian weighs at least 800 pounds? So- lution: Z 800 . 02 e- . 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....
View Full Document

This note was uploaded on 04/20/2010 for the course ECE ECE 313 taught by Professor S during the Spring '10 term at University of Illinois at Urbana–Champaign.

Page1 / 6

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

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