STA4032S12Solumidterm

# 10 b find the mean variance and standard deviation of

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(10) (b) Find the mean, variance, and standard deviation of X . Solution . (a) It is easy to find that F (1) = f (1) = 1 / 16, F (2) = F (1) + f (2) = 1 / 16 + 3 / 16 = 1 / 4, F (3) = F (2) + f (3) = 1 / 4 + 5 / 16 = 9 / 16, and F (4) = F (3) + f (4) = 9 / 16 + 7 / 16 = 1. So, the cumulative distribution function is given by F ( x ) = 0 , x < 1 1 16 , 1 x < 2 1 4 , 2 x < 3 9 16 , 3 x < 4 1 , x 4 . (b) μ = E ( X ) = 4 X x =1 xf ( x ) = 1 · (1 / 16) + 2 · (3 / 16) + 3 · (5 / 16) + 4 · (7 / 16) = 3 . 125 , E ( X 2 ) = 1 2 · (1 / 16) + 2 2 · (3 / 16) + 3 2 · (5 / 16) + 4 2 · (7 / 16) = 10 . 625 , σ 2 = E ( X 2 ) - μ 2 = 10 . 625 - 3 . 125 2 = 0 . 859375 . σ = σ 2 = 0 . 859375 = 0 . 927 . 4. The probability that your call to a service line is answered in less than 30 seconds is 0.60. Assume that your calls are independent. (5) (a) If you call 10 times, what is the probability that at least 6 calls are answered in less than 30 seconds? (5) (b) If you call 10 times, what is the mean number of calls that are answered in less than 30 seconds? (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 - 0 . 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.

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Then Y has a geometric distribution with parameter p = 0 . 6. So the desired probability is P ( Y = 4) = (1 - 0 . 6) 3 (0 . 6) = 0 . 0384.

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