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Unformatted text preview: Statistics 427: Homework #5 Solutions to Assigned Problems 3.48 [10 pts (2,2,2,2)] (a) We are dealing with X ∼ Bin(25 , . 05). We can use the table to determine P ( X ≤ 2) by going to the table for n = 25, the row for x = 2 and the column for p = 0 . 05 and read off P ( X ≤ 2) = 0 . 873 . (b) Write P ( X ≥ 5) = 1 P ( X ≤ 4) which we can read off the table as P ( X ≥ 5) = 1 . 993 = . 007. (c) We can compute this as P (1 ≤ X ≤ 4) = P ( X ≤ 4) P ( X ≤ 0) = 0 . 993 . 277 = 0 . 716 . (d) The probability that none of the 25 boards is defective is P ( X = 0) = P ( X ≤ 0) = 0 . 277. (e) From our results on the binomial distribution, we know that E [ X ] = np = (25)(0 . 05) = 1 . 25 and Var [ X ] = np (1 p ) = 1 . 1875, and so the standard deviation is SD( X ) = √ 1 . 1875 = 1 . 0897. 3.65 [6 pts (3,3)] (a) This experiment represents n = 100 independent Bernoulli trials, where the probability of “success” for each trial is p = P ( B ) = 0 . 2. If X is the number out of the next 100 customers who...
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 Spring '08
 Staff
 Statistics, Standard Deviation, Variance, Probability theory, Binomial distribution, independent Bernoulli trials

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