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 = 0.375 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?
Let \(Y\) denote the length of the drought. Then \(Y \sim \operatorname{Geo}(p=0.375)\)
\(P(Y=y)=(1-0.375)^{x} 0.375=(0.625)^{x} 0.375, x=0,1,2, \ldots\)
a)
P(at most 3 intervals \()=P(Y \leq 3)=P(Y=0)+P(Y=1)+P(Y=2)+P(Y=3)\)
\(=(0.625)^{0} 0.375+(0.625)^{1} 0.375+(0.625)^{2} 0.375+(0.625)^{3} 0.375=0.847\)
b)
\(=P(Y-\mu>\sigma)=P\left(\frac{Y-\mu}{\sigma}>1\right)=P(Z>1)=1-P(Z<1)\)
Required probability
\(=1-0.8413=0.1587\)
Y - geometric distribution with p = 0.375
a)
P(Y = 3) = 0.146484
P(Y<= 3) =0.755859
b)
mean = 2.666666
sd = 2.108185
mean + sd = 2.666666 + 2.108185 = 4.774851
P(X > 4.774851) =
P(X>=5) = 0.152587