First error 1 3 the way down the first page is n = 1

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Unformatted text preview: first error 1 / 3 the way down the first page, is N = 1 or N = 1 / 3? I accepted both interpretations. Interpretation: N is a real number. The occurrence of errors is governed by a Poisson arrival process. We only notice 90% of errors and, assuming that this is independent for each error, the occurrence of errors we find is a Poisson arrival process with rate 0 . 9 · 1 / 4. So N is exponentially distributed with rate 0 . 9 / 4 and E ( N ) = 1 . 9 / 4 = 4 . 9 ≈ 4 . 44 . Interpretation: N is an integer. The number of misprints we find per page is given by a Poisson distribution with rate 0 . 9 / 4, so the probability that we find at least one misprint on a single page is 1- e- . 9 / 4 . Therefore, N has geometric distribution with p = 1- e- . 9 / 4 and E ( N ) = 1 1- e- . 9 / 4 ≈ 4 . 96 . (c) [6 points] What is the probability that at least 2 of the first 50 pages contain at least 3 misprints each? Solution: Using the Poisson distribution, the probability that a single page contains at least three misprints is p = 1- P (none)- P (one)- P (two) = 1- e- 1 / 4 (1 / 4) 0!- e- 1 / 4 (1 / 4) 1 1!- e- 1 / 4 (1 / 4) 2 2! ≈ . 00216 . Using the binomial distribution, the probability that at least two of the first 50 pages contain at least one misprint is 1- P (none)- P (one) = 1- (1- p ) 50- 50 1 (1- p ) 49 p ≈ . 534% . 3. The type of lightbulb used in your refrigerator has a lifetime that is exponentially dis- tributed with a mean of 2 years. You replace the bulb every time it fails. (a) [6 points] What is the probability that the first lightbulb lasts less than a year? Page 2 Math 431: Exam 2 Solution: The c.d.f. for the exponential distribution with rate 1 / 2 is F ( t ) = 1- e- t/ 2 . So the probability that first lightbulb dies within the first year is 1- e- 1 / 2 ....
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first error 1 3 the way down the first page is N = 1 or N =...

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