Chebyshev inequality comparison Given $n$ distinct values $(x_i)_{i=1}^{n}$ with mean $\mu$ and standard deviation $s$, for all $i$, we have $|x_i−\mu| ≤ s \sqrt{n − 1}$.
How does this inequality compare with Chebyshev inequality as $n$ increases?
Chebyshev's inequality states that
$$
\Pr[|x_i - \mu| &gt; c\sigma] &lt; \frac{1}{c^2}.
$$
Choosing $c = \sqrt{n}$, we deduce that with positive probability,
$$
|x_i - \mu| \leq \sqrt{n} \cdot \sigma \text{ for all } i.
$$