What is the probability that in at least 2 of the next 10 years less than 20 inches of rain are recorded in the region?

Question

Information: The annual rainfall (in inches) in a certain region is normally distributed with $\mu=40$ and $\sigma=5$.

My attempt is: probability of less than 20 inches of rain $$P(x<20)=P(Z<-4)=1-\phi(4).$$

I'm not sure how to apply this to the part where at least 2 of the next ten years.

Answer 1

Its a binomial distribution with 10 years. (assume each year is independent of previous year.)

So, $X_i$~Norm(40,25) (this is the mean, and variance) represents the annual rainfall of a year.

P(X<20) which is aprx $\phi$($\frac{20-40}{5}$) which $\phi(-4)$

Y, composed of the $X_{i}$s, is ~Binomial(10,$\phi(-4)$). This represents the amount of years that have this rain or less.

So, we want P(Y>1) = 1 - P(Y=1) - P(Y=0) which you get by plugging in the binomial distribution here.

$$P(X=x)= \begin{pmatrix} n\\x \end{pmatrix} p^x(1-p)^{n-x}$$

Answer 2

Let $X$ be the annual rainfall, then $X\sim N(40,5)$. Then you were right about $$P(X<20)=P\left(Z<\frac{20-40}{5}\right)=P(Z<-4)=1-\Phi(4)$$

Let $Y$ be the number of years, in the next $10$ that will have less than 20 inches of rainfall. If we suppose that the annual rainfall is independant over year, then $Y$ follow a binomial distribution. $Y\sim \text{Binom}(10,1-\Phi(4))$

We are looking for $P(Y\geq 2)$. It will be simplier to look for the opposite. $$P(Y\geq2)=1-P(Y<2)=1-P(Y=1)-P(Y=0)$$ $$P(Y\geq2)=1-{10\choose1}(1-\Phi(4))^1(\Phi(4))^9-{10\choose0}(1-\Phi(4))^0(\Phi(4))^{10}$$ We could simplify this last equation to $$P(Y\geq2)=1-10(\Phi(4))^9+9(\Phi(4))^{10}$$ Hope it helps