Editing Chebyshev's inequality (section)

===Probabilistic statement===
Let ''X'' (integrable) be a [[random variable]] with finite non-zero [[variance]] ''σ''<sup>2</sup> (and thus finite [[expected value]] ''μ'').<ref>Feller, W., 1968. An introduction to probability theory and its applications, vol. 1. p227 (Wiley, New York).</ref> Then for any [[real number]] {{nowrap|''k'' > 0}},
: <math>
    \Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}.
  </math>

Only the case <math>k > 1</math> is useful. When <math>k \leq 1</math> the right-hand side <math> \frac{1}{k^2} \geq 1 </math> and the inequality is trivial as all probabilities are&nbsp;≤&nbsp;1.

As an example, using <math>k = \sqrt{2}</math> shows that the probability values lie outside the interval <math>(\mu - \sqrt{2}\sigma, \mu + \sqrt{2}\sigma)</math> does not exceed <math>\frac{1}{2}</math>. Equivalently, it implies that the probability of values lying within the interval (i.e. its [[Coverage_probability|"coverage"]]) is ''at least'' <math>\frac{1}{2}</math>.

Because it can be applied to completely arbitrary distributions provided they have a known finite mean and variance, the inequality generally gives a poor bound compared to what might be deduced if more aspects are known about the distribution involved.

{|class="wikitable" style="background-color:#FFFFFF; text-align:center"
|-
! k
! Min. % within ''k'' standard<br />deviations of mean
! Max. % beyond ''k'' standard<br />deviations from mean
|-
|  1
|| 0%
|| 100%
|-
|  {{sqrt|2}}
|| 50%
|| 50%
|-
|  1.5
|| 55.55%
|| 44.44%
|-
|  2
|| 75%
|| 25%
|-
|  2{{sqrt|2}}
|| 87.5%
|| 12.5%
|-
|  3
|| 88.8888%
|| 11.1111%
|-
|  4
|| 93.75%
|| 6.25%
|-
|  5
|| 96%
|| 4%
|-
|  6
|| 97.2222%
|| 2.7778%
|-
|  7
|| 97.9592%
|| 2.0408%
|-
|  8
|| 98.4375%
|| 1.5625%
|-
|  9
|| 98.7654%
|| 1.2346%
|-
|  10
|| 99%
|| 1%
|}