For a certain river, suppose the drought length Y is the number of consecutive t
ID: 2978319 • Letter: F
Question
For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains below a critical value y0 (a deficit), preceded by and followed by periods in which the supply exceeds this critical value (a surplus). An article proposes a geometric distribution with p = 0.396 for this random variable. (Round your answers to three decimal places.) (a) What is the probability that a drought lasts exactly 3 intervals? At most 3 intervals? (b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?Explanation / Answer
(a)for calculating exactly 3 intervals : =0 .369^3 * (1-.369) = .0317 for at most three intervals = P(0 intervals) + P(1 interval) + P(2 intervals) + P(3 intervals)=[.369^0 * (1 - .369)] + [.369^1 * (1 - .369)] + [.369^2 * (1 - .369)] + [.369^3 * (1 - .369)] = .98146 (b) mean = (1-p)/p = .631/.369 = 1.71 S.D= sqrt[(1-p)/p^2] = sqrt(.631 / .369^2) = sqrt(4.63421) = 2.1527 P(Y > (1.71 + 2.1527)) = P(Y > 3.8627) = P(Y at least 4) = 1 - P(Y at most 3)= 1 - 98146= .01854