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Chebyshev's inequality
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===Probabilistic statement=== Let ''X'' (integrable) be a [[random variable]] with finite non-zero [[variance]] ''Ο''<sup>2</sup> (and thus finite [[expected value]] ''ΞΌ'').<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 β€ 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% |}
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