The sample mean is:
\(\bar X_n=\dfrac{1}{n}\sum_{i=1}^nX_i\)
The variance of this is:
\(Var[\bar X_n]=Var[\dfrac{1}{n}\sum_{i=1}^nX_i]\)
\(Var[\bar X_n]=\dfrac{1}{n^2}nVar[X]\)
\(Var[\bar X_n]=\dfrac{\sigma^2}{n}\)
We know from Chebyshev’s inequality:
\(P(|X-\mu | \ge k\sigma )\le \dfrac{1}{k^2}\)
Use \(\bar X_n\) as \(X\):
\(P(|\bar X_n-\mu | \ge \dfrac{k\sigma }{\sqrt n})\le \dfrac{1}{k^2}\)
Update \(k\) so \(k:=\dfrac{k\sqrt n}{\sigma}\)
\(P(|\bar X_n-\mu | \ge k)\le \dfrac{\sigma^2}{nk^2}\)
As \(n\) increases, the chance that the sample mean lies outside a given distance from the population mean approaches \(0\).