Chebyshev Inequality

Apply Markov's Inequality with $a \equiv k^2$ to obtain

$$P[(x-\mu)^2\geq k^2] \leq {\left\langle{(x-\mu)^2}\right\rangle{}\over k^2} = {\sigma^2\over k^2}.$$ (1)

Therefore, if a Random Variable $x$ has a finite Mean $\mu$ and finite Variance $\sigma^2$, then $\forall\ k \geq 0$,

$$P(\vert x-\mu\vert\geq k)\leq {\sigma^2\over k^2}$$ (2)

$$P(\vert x-\mu\vert\geq k\sigma) \leq {1\over k^2}.$$ (3)

References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 11, 1972.