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Assume a binomial probability distribution with n = 40 and p = .55. Compute the following: a. The mean and standard deviation of the random variable. b. The probability that x is 25 or greater. c. The probability that x is 15 or less. d. The probability that x is between 15 and 25 inclusive. a) The mean is given by np, or, in this case (40)(.55) = 22. The standard deviation = sqrt(np(1-p)) = sqrt((40)(.55)(.45))=3.146 approx. b) The Z score at 24 is(24 - 22)/3.146 = 0.64 approx Therefore, according to the standard normal table, the probability that x is 25 or greater is 1 - 0.7389 = 0.2611 c) The Z score at 15 is (15-22)/3.146 = -2.22 approx Therefore, according to the standard normal table, the probability that x is 15 or less is 1 - 0.9868 = 0.0132. d)This would be equal to the probability that it is 25 or less minus the probability that it is 14 or less. 25 or less (Z score = 0.95) ==> 0.8289 14 or less (Z score =-2.54) ==>1-0.9945=0.0055. Therefore, the difference is 0.8289-0.0055=0.8234