# Solved (Free): Suppose the amount of rain that falls on a particular day in New York is a random variable with mean 20 mm

#### ByDr. Raju Chaudhari

Mar 31, 2021

Suppose the amount of rain that falls on a particular day in New York is a random variable with mean 20 mm.

(a) What is an upper bound for the probability that the amount of rain exceeds 60 mm?
(b) Suppose you also know the variance is 10 mm$^2$. Now what can you say about the probability above?
(c) Given the information in part (b), what can you say about the probability that the amount of rainfall lies between 10 mm and 30 mm?

#### Solution

Let $X$ denote the amount of rain that falls on a particular day in New York. Let $\mu = E(X) =20$mm.

(a) An upper bound for the probability that the amount of rain exceeds 60 mm is

 $$\begin{eqnarray*} P(X > 60) &\leq & \frac{E(X)}{a} \\ &\leq & \frac{20}{60}\\ &\leq & \frac{1}{3} =0.3333. \end{eqnarray*}$$

(b) Suppose the variance of $X$ is 10 mm$^2$. That is $V(X) =\sigma^2 = 10$ mm$^2$.

The one-sided Chebyshev's inequality is

$P(X-\mu\geq a) \leq \frac{\sigma^2}{\sigma^2 + a^2}$.

 $$\begin{eqnarray*} P(X > 60) &=& P(X-20 > 60-20) \\ &=& P(X-20 > 40)\\ &\leq &\frac{10}{10+40^2}\\ &\leq & 0.006211. \end{eqnarray*}$$

(c) The probability that the amount of rainfall lies between 10 mm and 30 mm is

 $$\begin{eqnarray*} P(10 < X < 30) &=& P(10-\mu < X-\mu < 30-\mu) \\ &=& P(10-20 < X-\mu < 30-20) \\ &=& P(-10 < X-\mu < 10)\\ &=& P(|X-\mu| < 10)\\ &\geq & 1-\frac{\sigma^2}{k^2}\\ &\geq & 1- \frac{10}{10}\\ & \geq & 1. \end{eqnarray*}$$

But probability can not be greater than 1, so $P(10 < X < 30) =1$.