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Asst6-solved

# Asst6-solved - Statistics 427 Homework#6 Solutions to...

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Statistics 427: Homework #6 Solutions to Assigned Problems 3.82 [6 pts (2,2,2)] (a) We are assuming that X Pois( λ = 0 . 2). The probability that the disk has exactly one missing pulse is P ( X = 1) = (0 . 2) 1 e - 0 . 2 1! = 0 . 2 e - 0 . 2 = 0 . 1637 . (b) It is easier to work with the complement: P ( X 2) = 1 - P ( X < 2) = 1 - P ( X = 0) - P ( X = 1) = 1 - e - 0 . 2 - (0 . 2) e - 0 . 2 = 1 - 0 . 982477 = 0 . 0175 . (c) Now we have X i Pois(0 . 2) for i = 1 , 2 and we know that X 1 and X 2 are independent. “Neither” means that both do not contain a missing pulse. Because they are independent, we can compute P ( X 1 = 0 and X 2 = 0) = P ( X 1 = 0) × P ( X 2 = 0) = e - 0 . 2 × [ e - 0 . 2 ] = e - 0 . 4 = 0 . 67032 . 3.83 [6 pts (3,3)] (a) Here n = 1000 > 50 and np = 1000 / 200 = 5, which is not strictly less than five, but we will use the Poisson approximation to the binomial distribution anyway: X Pois(5). We are asked to compute P (5 X 8) = 8 x =5 5 x e - 5 x ! = F X (8) - F X (4) = 0 . 932 - 0 . 440 = 0 . 492 . This was found using the table in the back of the book. 1

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(b) Now we compute P ( X 8) = 1 - P ( X 7) = 1 - F X (7) = 1 - 0 . 867 = 0 . 133 . 3.86 [6 pts (2,2,2)] (a) We know that X Pois(5 t ), where t is measured in hours. The probability that exactly
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