The annual rainfall (in inches) in a certain region is normally distributed with $\mu =40\text{and}\sigma =4$. What is the probability that starting with this year, it will take more than $10$years before a year occurs having a rainfall of more than $50$ inches? What assumptions are you making?

Here, it is given that the annual rainfall in a region is normally distributed and$\mu =40\text{and}\sigma =4$.

Let the random variable $X$denote the annual rain fall in inches in a certain region.

$$\begin{gathered} P(X>50)=P\left(\frac{X-\mu }{\sigma }>\frac{50-\mu }{\sigma }\right) \\ =P\left(z>\frac{50-40}{4}\right) \\ =P\left(z>2.5\right) \\ =1-P\left(z\le 2.5\right) \\ =1-0.9938 \\ \therefore P(X>50)=0.0062 \end{gathered}$$

Let the random variable $Y$denote the number of years it will take before a year occurs having rainfall above $50$inches.

The random variable $Y$follows the geometric distribution with parameter $p=0.0062$.

Then the probability mass function of random variable $Y$is expressed as:

$P(Y=y)={\left(1-0.0062\right)}^y\left(0.0062\right),y=0,1,2,......$

The probability that, starting with this year, it will take over $10$years before a year occurs having a rainfall of over $50$inches $=P\left(Y>10\right)$.

$$\begin{gathered} P\left(Y>10\right)=\sum _{y=10}^{\infty }\left(0.0062\right){\left(1-0.0062\right)}^y \\ =\left(0.0062\right){\left(1-0.0062\right)}^{10}\sum _{y=10}^{\infty }{\left(1-0.0062\right)}^{y-10} \\ =\left(0.0062\right){\left(1-0.0062\right)}^{10}\left\{1+\left(1-0.0062\right)+{\left(1-0.0062\right)}^2+.......\right\} \\ =\left(0.0062\right){\left(1-0.0062\right)}^{10}{\left\{1-\left(1-0.0062\right)\right\}}^{-1} \\ ={\left(1-0.0062\right)}^{10} \\ ={\left(0.9938\right)}^{10} \\ =0.9397 \end{gathered}$$

Therefore, the probability that starting with this year, it will take more than 10 years before a year occurs having a rainfall of more than 50 inches is$0.9397$.