Chapter 8, Problem 8.1P

Suppose that X is a random variable with mean and variance both equal to 20. What can be said about $P\left\{0<X<40\right\}$?

To determine

To find: $P\left(0<X<40\right)$

Answer to Problem 8.1P

  $P\left(0<X<40\right)=\frac{19}{20}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Population mean}\left(\mu \right)=20\\ \text{Population variance}\left({\sigma }^2\right)=40\end{array}$

Calculation:

  $P\left(0<X<40\right)$can be calculated as:

  $\begin{array}{c}P\left(0<X<40\right)=P\left(0-\mu <X-\mu <40-\mu \right)m\\ =P\left(0-20<X-\mu <40-20\right)\\ =P\left(-20<X-\mu <20\right)\\ =P\left(\left|X-\mu \right|\le 20\right)\end{array}$

Now,

  $\begin{array}{c}P\left(\left|X-\mu \right|\le 20\right)=P\left(P\left(\left|X-\mu \right|\le \sqrt{20}\times \sqrt{20}\right)\right)\\ \ge 1-\frac{1}{{\left(\sqrt{20}\right)}^2}\\ \ge \frac{19}{20}\end{array}$

Thus, the required probability is $\frac{19}{20}$ .