This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: (5) (c) What is the probability that you must call four times to obtain the first answer in less than 30 seconds? (5) (d) What is the mean number of calls until you are answered in less than 30 seconds? Solution . (a) Let X be the number of calls in the 10 calls that are answered in less than 30 seconds. Then, X has a binomial distribution with n = 10 and p = 0 . 60. The desired probability is P ( X ≥ 6) = 1 P ( X ≤ 5) = 1 5 X x =0 10 x (0 . 6) x (0 . 4) 10 x = 1 . 3669 = 0 . 6331 . (b) E ( X ) = np = 10 · (0 . 6) = 6 . (c) Let Y be the number of calls needed to obtain an answer in less than 30 seconds. Then Y has a geometric distribution with parameter p = 0 . 6. So the desired probability is P ( Y = 4) = (1 . 6) 3 (0 . 6) = 0 . 0384. (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 . 3679 = 0 . 6321 . (c) The expected number of errors in 100 pages is equal to (0 . 1)(100) = 10....
View
Full Document
 Summer '08
 STAFF
 Statistics, Probability, Probability theory, Florida Atlantic University, Hongwei Long

Click to edit the document details