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 = 0.409 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	.0684
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

The probability mass function of geometric distribution can be defined as,

$$ P(Y=y)=q^{y} p $$

a)

Find the value of \(P(Y=3)\).

$$ \begin{aligned} P(Y=3) &=(1-0.409)^{3}(0.409) \\ &=0.206425(0.409) \\ &=0.084 \end{aligned} $$

Find the value of \(P(Y \leq 3)\).

$$ \begin{aligned} P(Y \leq 2) &=P(Y=0)+P(Y=1)+P(Y=2)+P(Y=3) \\ &=\left[\begin{array}{l} (1-0.409)^{0}(0.409)+(1-0.409)^{1}(0.409)+(1-0.409)^{2}(0.409) \\ +(1-0.409)^{3}(0.409) \end{array}\right] \\ &=0.878 \end{aligned} $$

b)

Find the brobability that the length of a drought exceeds its mean value by at least one standard deviation. First, find the mean value of \(Y\).

$$ \mu_{Y}=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445 $$

The standard deviation of \(Y\) is.

$$ \sigma_{Y}=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{0.409^{2}}}=\overline{1.880} $$

The probability that the length of a drought exceeds its mean value by at least one standard deviation is,

$$ \begin{aligned} P\left(Y>\mu_{Y}+\sigma_{Y}\right) &=P(Y>1.445+1.880) \\ &=P(Y>3.325) \\ &=1-P(Y \leq 3) \\ &=1-0.878 \quad(\mathrm{Usin} \end{aligned} $$

(Using normal tables)

$$ =0.122 $$