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STA4032S12Solumidterm

D e y 1 p 1 0 6 1 667 5 the number of errors in a

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(d) E ( Y ) = 1 /p = 1 / (0 . 6) = 1 . 667. 5. The number of errors in a textbook follows a Poisson distribution with a mean of 0.1 error per page. (8) (a) What is the probability that there are three or less errors in 10 pages? (7) (b) What is the probability of at least one error in 10 pages? (5) (c) What is the expected number of errors in 100 pages? Solution . (a) Let X be the number of errors in 10 pages. Then X has a Poisson distribution with λ = (0 . 1)(10) = 1. So, the desired probability is P ( X 3) = P ( X = 0) + P ( X = 1) + P ( X = 2) + P ( X = 3) = e - 1 + e - 1 (1) 1 1! + e - 1 (1) 2 2! + e - 1 (1) 3 3! = 0 . 9810 . (b) P ( X 1) = 1 - P ( X = 0) = 1 - e - 1 = 1 - 0 . 3679 = 0 . 6321 . (c) The expected number of errors in 100 pages is equal to (0 . 1)(100) = 10.
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