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 = 0.397 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?

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?

Answer 1

An article proposes a geometric distribution with p = 0.397. The probability that a drought lasts exactly 3 intervals is, P(Y 3)b (3;1,0.397) (1-0.397) (0.397) = 10.087 Find P(Y s3) nb (0:1,0.397)+nb(1;1,0.397)+nb(2;1,0.397)+ nb (3:1,0.397) -(1-0.397) (0.397)+(1-0.397) (0.397)+(1-0.397) (0.397) (1-0.397) (0.397) 0.8678 The mean value of Yis, 1-0.397 0.397 1.519

The standard deviation of Yis, 磊 1-0.397 0.397- = 1.956 The probability that the length of a drought exceeds its mean value by at least one standard deviation is, PlY >My+g:)-PlY > 1.5 1941.956) P(Y> 3.475) 0.1322