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?

Solution :

a). P(X = x)

     =    for x = 0, 1, 2, ....

   P(x ≤ 3) = 0.837

b). Expectation =

                        = 1.7397

  Variance =

                 = 4.7663726

  Standard deviation = 2.1832

 Therefore, mean + standard deviation

                  = 1.7397 + 2.1832

                  = 3.9229

So the required P = 2 x 0.1626

                             = 0.325

figure c

step-by-step explanation:

hope this .

step-by-step explanation:

f(x)=2x^2 -1 

translation is a function transformation that transform the function in a way of translating. by translating the function its graph is moved  up an down. moving the function f(x) 2 units to the left will be the new function g(x) with the equation: g(x)=f(x+2)=2(x+2)^2 -1