Chebyshev’s Inequality

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

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