## Chebyshev’s Inequality

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