For a certain river suppose the drought length Y is the numb

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 = 0.392 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?

Solution

(a) P(exactly 3 intervals) = .392^3 * (1-.392) = 0.0366
P(at most 3 intervals) = P(0 intervals) + P(1 interval) + P(2 intervals) + P(3 intervals)

[.392^0 * (1 - .392)] + [.392^1 * (1 - .392)] + [.392^2 * (1 - .392)] + [.392^3 * (1 - .392)]

= 0.9763


(b) mean = (1-p)/p = 0.608/.392 = 1.551
standard deviation = sqrt[(1-p)/p^2] = sqrt(.608 / .392^2) = 1.989

P(Y > (1.55 + 1.989)) = P(Y > 3.539) = P(Y at least 4) = 1 - P(Y at most 3)

= 1 - 0.9763

=0.0237