Question

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

Short answer

$P[0<X<40]\ge \frac{19}{20}$

Step-by-step solution

Step 1. Given information

Let X is a random variable with mean and variance both 20.

Step 2. Denoting mean and variance

E(X) = 20

V(X) = 20

Step 3. To find

$P[0<X<40]$

Step 4. Simplification

$$\begin{gathered} P[0-20<X-20<40-20] \\ =P[-20<X-20<20] \\ =P\left[\right|X-20|<20] \end{gathered}$$

Step 5. Using Chebyshev's inequality

$1-P\left[\right|X-20|\ge 20]\ge 1-\frac{1}{20}=\frac{19}{20}$

Step 6. Final answer 

$P[0<X<40]\ge \frac{19}{20}$