exam2-key

This preview shows pages 1–4. Sign up to view the full content.

IE 230 Seat # ________ Name ___ < KEY > ___ Please read these directions. Closed book and notes. 60 minutes. Covers through the normal distribution, Section 4.6 of Montgomery and Runger, fourth edition. Cover page and four pages of exam. Pages 8 and 12 of the Concise Notes. A normal-distribution cdf table. No calculator. No need to simplify beyond probability concepts. For example, unsimplified factorials, integrals, sums, and algebra receive full credit. Throughout, f denotes probability mass function or probability density function and F denotes cumulative distribution function. For one point of credit, write your name neatly on this cover page. Circle your family name. For one point of credit, write your name on each of pages 1 through 4. Score ___________________________ Exam #2, October 19, 2010 Schmeiser

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

View Full Document
Name _______________________ Closed book and notes. 60 minutes. 1. Suppose that the random variable X has the discrete uniform distribution over the set {0, 1,. .., 10} and that the random variable Y has the continuous uniform distribution over the set [0,10]. (a) (3 pt) T F E( X ) = E( Y ). (b) (3 pts) T F V( X ) = V( Y ). (c) (3 pts) T F F X (5) = F Y (5). (d) (3 pts) T F f X (5) = f Y (5). 2. (from Montgomery and Runger, 3–113) In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers . He used data collected over twenty years to show that the number of soldiers killed by horse kicks each year in each corps in the Prussian cavalry followed a Poisson distribution with a mean of 0.61. (a) (7 pts) For a particular corps over one year, determine the value of the probability of more than one death. ____________________________________________________________ Deaths occur with rate λ = 0.61 deaths per year per corps. Let N = "number of deaths in one year in a particular corps". Then N is Poisson with rate μ = λ t = (0.61)(1) = 0.61 deaths. Therefore, P( N > 1) = 1 [P( N = 0) + P( N = 1)] = 1 R J Q 0! e −μ μ 0 hhhhh + 1! e −μ μ 1 hhhhh H J P = 1 e 0.61 [1 + 0.61] ____________________________________________________________ (b) (7 pts) For a particular corps over five years, determine the value of the probability of no death. ____________________________________________________________ Deaths still occur with rate λ = 0.61 deaths per year per corps. Now let N denote the number of deaths during five years. Then N is Poisson with mean μ = λ t = (0.61)(5) = 3.05. Therefore, P( N = 0) = 0! e −μ μ 0 hhhhh = e 3.05 ____________________________________________________________ Exam #2, October 19, 2010 Page 1 of 4 Schmeiser
Name _______________________ 3. (from Montgomery and Runger, 4–61) The lifetime of a semiconductor laser at a constant power is normally distributed with a mean of 6000 hours and standard deviation of 500 hours. (a) (8 pts) Sketch the corresponding normal pdf. Label and scale both axes.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.