CEE304__exam1_2004 - CEE 304 Uncertainty Analysis in...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CEE 304 - Uncertainty Analysis in Engineering First Examination October 6, 2004 You may use text, your notes and calculators. There are 50 points in total, one per minute. 1. (6 pts) Consider two events denoted Q and ‘P, where Pr[Q] =03; Pr[‘l’l =0.2; P[ QU‘I’] =0.45 (a) What is the probability that BOTH Q and ‘l’occur when the experiment is conducted? (b) What IS the probability that 9 does not occur, but at the same time 'i’ does? (c) For a normal random variable X with mean 10 and standard deviation 2, what IS Pr[ X 2 12 ] ? 2. (10 pts) Sue and Jill love to play tennis. Jill has a lucky hat, and when she wears it, she wins 75% of the time. When Jill fails to wear her lucky hat, the probability that she wins falls to just 55%. For some reason, when Sue and Jill play tennis, Jill wears her lucky hat only 80% of the time. (a) How often does Sue win? (b) If Sue won her tennis game with Jill today, what is the probability Jill wore her lucky hat? (c) Starting tomorrow, consider only games when Jill wears her lucky hat giving Jill a 75% chance of winning. What is the mean and variance of the number of games they would play until Sue wins a game? 3. A 4 meter rod broke into 2 or more pieces, and the joint probability density function of the length of two long pieces that were retained is fXY(x,y) = 0.125 within the right triangle with 0 s x, 0 s y, x+y s 4 x+y“ = zero otherwise a) (6 pts) What is the marginal density function for x, and the value of E{X}? x b) (2 pts) What is the conditional probability density function for Y given the value of X? 4. (20 pts) A electric utility in Florida is dealing with the aftermath of a hurricane. And what a mess! In a flat residential area, trees falling on power lines result in about 6 line-breaks per mile. Assuming that these line—breaks occur as a Poisson process: a) What is the mean and variance of the length of line a crew will inspect until they find 20 line-breaks? b) What is the mean and variance the number of line-breaks in a 15 mile segment of line? c) What is the probability that a 0.5 mile segment of line has AT MOST ONE line-break? d) Is a Poisson process likely to be a reasonable model of these electric power line breaks? e) The foreman has found that 60% of such breaks can be fixed in 2 hours or less. If an emergency crew will repair 4 breaks, what is the mean and variance of the NUMBER of these breaks that take more-than-Z hours to fix? What is the probability that exactly 3 take more-than-Z hours to fix? 5. (6 pts) For a decorative display, a company produces glass balls that have radiiwhich are lognormally distributed with mean 5 and standard deviation 0.4. However, interest is not in R, but the weight of the balls given by W = 6 R3. Use the approximate moment formulas to compute approximately the mean and variance of W. CEE 304 - Uncertainty Analysis in Engineering Solutions First Examination October 6, 2004 1. Axioms, sets, normal distribution 1a) Plfln‘I’] = 0.05 = Pm] + P[‘P] - PlQU ‘1'] Look at Venn Diagram 2 pts 1b) P[Q’n‘l’] = 0.15 = P[‘P] — P[Qn‘l’] From P[‘I’]= PIQ’n‘I’] + P[Qn‘P] 2 pts 1c) Pr[ X 2 12] = Pr[ (X-u)/O' 2 (12—10)/2 ] = Pr[ Z 2 +1 ] = 0.1587 2 pts 2. Bayes Theorem /--Iill (0.75) __ / --- Hat (0.8) --0---Sue (0.25) DRAW THE PICTURE! \ --- NoHat (0.2) —-- °—--]ill (0.55) \ —--- Sue (0.45) a) P(Sue) = P(Sue l Hat) P(Hat) + P(Sue l NoHat) P(NoHat) = (0.25) 0.80 + (0.45) 0.20 = 0.29 3 pts b) P[Hat | Sue ] = P(Sue I Hat) P(Hat)/ P(Sue) = (0.25) 0.80 / 0.29 = 0.69 3 pts c) Geometric distribution (p=0.25): u = 1/ p = 4 ; ol= (1—p)/ p2 = 12 (Okay!) 4 pts 3. Joint Distributions, marginal distributions, conditional pdfs, moments 4» a) First get fx(x) =1 fxy(x,y) dy I x 0.125 dy = 0.125 (4 - x) for 0 S x S 4 3 pts E[X] =J4 0.125x(4—x) dx=0.125 [4xZ/2—x3/3] =4—8/3=4/3 » 3pts b) f,,x(y I x) = fxy(x,y)/ fx(x) = 0.125/ [0.125(4 — x) ] = 1/ (4—x) for o < y < 4-x => Uniform 2 pts 4. Poisson process -— Poisson, Gamma, Binomial distributions Setup: 7t = 6 per mile a) Time to 20th Gamma: u --= 20/ it = 3.33 miles; 0.2 = 20/ k” = 0.556 miles2 [G= 0.745 miles] 4 pts b) # in 15 miles => Poisson dist. => moments: u = 0'2 = v = N‘t = 90 4 pts c) # in 0.5 miles => Poisson dist. => Pr[0 or 1] = (1 + Mt) exp(-7t*t) = 0.049 + 0.149 = 0.199 ~ 20% 4 pts d) Poisson process reasonable if 3 condition are satisfied: (i) The probability of an arrival in a short interval At equals Mt. (ii) The arrival rate A is constant. (iii) The number of arrivals in non-overlapping intervals is independent —- and these are all reasonable for appropriately defined ”breaks,” in homogeneous neighborhoods, with independence among breaks at different locations. OR: No, and give an example of how assumptions are likely to be violated. 4 pts e) Binomial!!! n = 4, p = 0.4, u = np = 1.6 ; 02: np(l-p) = 0.96; P(X=3) = 4 0 0.43 0 0.6 = 0.1536 4 pts 5. Approximate moment formulas with ER] = 5, and Var[R] = (0.4)2 yields EIW ] z W(E[R]) + 0.5 (d2W/ dRz) Var[R] = 6053 + 0.506030205°(0.4)2 = 750 + 0.5 028.8 = 765 3 pts Var[W] z (dW/ CIR): Var[R] = (603.25)2 (0.4)2 ’ 32,400 = (180)2 3 pts ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern