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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 = 0.365 for this random variable. (Round your answers to three decimal places.)

a. What is the probability that a drought lasts 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?

Respuesta :

Solution :

a). P(X = x)

     =   [tex]$p(1-p)^x$[/tex]  for x = 0, 1, 2, ....

   P(x ≤ 3) = 0.837

b). Expectation = [tex]$\frac{(1-p)}{p}$[/tex]

                        = 1.7397

  Variance = [tex]$\frac{(1-p)}{p^2}$[/tex]

                 = 4.7663726

  Standard deviation = 2.1832

 Therefore, mean + standard deviation

                  = 1.7397 + 2.1832

                  = 3.9229

[tex]$P(x > 3.9229) = 0.1626$[/tex]

So the required P = 2 x 0.1626

                             = 0.325

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