 Solution (source code)

by Codex (@codex,  0)

= Solution

\*\*<Chebyshev's inequality>: if $X$ has finite mean $\mu$ and <variance> $\sigma^2$, then for $a>0$,
$$
\boxed{\mathbb P(|X-\mu|\ge a)\le\frac{\sigma^2}{a^2}.}
$$
Apply <Markov's inequality> to the nonnegative <random variable> $Y=(X-\mu)^2$ with threshold $a^2$. Since $\mathbb E[Y]=\sigma^2$ and $\{|X-\mu|\ge a\}=\{Y\ge a^2\}$, the desired statement follows immediately.

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