Week 5 Wed Sept 29 (1)

Week 5 Wed Sept 29 (1) - WEEK 5, Wednesday, Sept 29 Today:...

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Unformatted text preview: WEEK 5, Wednesday, Sept 29 Today: Cover material from Sections 3-12, 4-1, 4-2, 4-3, and 4-4 . Exponential Distribution with Parameter > 0 (Sec 3-12) . Here is the density function: x e - , x 0 = ) ( x f 0, x < 0 Graph the density function. Important Results for Exponential ( ): = E ( X ) = 1/ 2 = V ( X ) = 1/ 2 , = SD ( X ) = 1/ . There are some useful formulas for probabilities of events involving an exponentially distributed random variable X . Let X be distributed Exponential( ). Consider these useful formulas (p. 131): x x t x t e e dt e x X P x F ---- =- = = = 1 | ) ( ) ( , x > x e x X P - = ) ( , x > 0 1 b a e e b X a P --- = ) ( , 0 a b Example 1. Light Bulbs . A manufacturer of 100W light bulbs has data that suggest that the lifetimes X of its light bulbs follow an exponential distribution with mean 2000 hours. Use this exponential model to answer the following questions about a randomly selected light bulb from the manufacturer. (a) Write the density function of X . f ( x ) = _____________, x > 0. Sketch f . .0005 0 2000 4000 x (b) What is the probability that it will fail in the first 1500 hours, i. e., what is P ( X 1500)? Ans. Example 2. More Light Bulbs . A plant has a large assembly area with many light bulbs that are turned on 24 hours per day. An electronic device records how many hours each new light bulb burns until it fails and finds across a large sample that the mean time to failure for new light bulbs is 2,732 hours. Based on the failure time data, the plant engineer models the failure time X of a new light bulb as Exponential( ) with mean 2,732 hours. Use this model to answer the following questions. (a) What is the value of ? Ans. 1/2732 (b) Plot the density function for X . (c) E ( X ) = 2,732 hours; determine SD ( X ). Ans. 2732 hours (d) P( X > 3,000 hours) = Ans. 0.33 Shade in the area on your plot of Part (b). Exercise 3-51 . Hint . Model time to failure X as an exponential random variable with parameter = .1742 days-1 and mean = 1/ = 5.74 days. (The textbook uses the term Mean Time Between Failures (MTBF) for what we can take as meaning E ( X ) = .) Exercise 3-52 . The mean = 38.36 months is assumed for the model. Ans. (a) .269 2 (b) about .27 (that is, 27%) (c) Perhaps not. (d) 1.97 months (say, 2 months) (e) 58.49 months Hint on (d): Solve for x . Normal (Gaussian) Probability Distributions (Chap 4) The Normal probability distributions have two parameters, usually, denoted by the symbols and . We denote the distribution by N ( , ). The textbook uses the notation N ( , 2 ). Figure 4-2 in the textbook provides graphs of the density functions of N (50, 2), N (50, 5) and N (60, 2). This are the familiar bell-shaped density functions....
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Week 5 Wed Sept 29 (1) - WEEK 5, Wednesday, Sept 29 Today:...

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