## Chebyshev’s Inequality

Let $X$ be a random variable with mean $\mu$ and finite variance $\sigma^2$. Then for any real constant $k>0$,
```
$$
\begin{equation*}
P[|X-\mu| \geq k\sigma] \leq \frac{1}{k^2}\quad \text{ and }\quad P[|X-\mu|< k\sigma] \geq 1-\frac{1}{k^2}
\end{equation*}
$$
```