This online Chebyshev’s Theorem Calculator estimates the maximal probability Pr that a random variable X is outside of the range of k (k > 1) standard deviations σ of the mean μ.
Pr(|X – μ| ≥ kσ) ≤ 1 / k2
Chebyshev’s Theorem
Chebyshev’s Theorem or Chebyshev’s inequality, also called Bienaymé-Chebyshev inequality, is a theorem in probability theory that characterizes the dispersion of data away from its mean (average).
Chebyshev’s inequality (named after Russian mathematician Pafnuty Chebyshev) puts an upper bound on the probability that an observation is at a given distance from its mean.
Only two minimal conditions are required for the inequality to be true: (1) that the underlying distribution have a mean and (2) that the average size of the deviations away from this mean (the standard deviation) is not infinite.
Chebyshev’s inequality then states that the probability that an observation will be more than \(k\) standard deviations from the mean is at most \(1/k^{2}\).
To proof this, suppose a population of \(n\) values consists of \(n_1\) values of \(x_1\), \(n_2\) values of \(x_2\), etc. (i.e., \(n_i\) values of each different \(x_i\) in the population).
Now suppose \(σ\) is the standard deviation of this population, \(μ\) is it’s mean, and \(k>1\) is some positive real number.
Consider the variance of the population:
$$ σ^{2}=\frac{∑(x_i−μ)^{2}⋅n_i}{n},$$
where the sum ranges over all distinct \(x_i\). Then
$$ σ^{2} ≥ \frac{∑(x_i−μ)^{2}⋅n_i}{n},$$
where the sum ranges only over those \(x_i\) when \(|x_i−μ|≥kσ\), so we can also write:
$$ σ^{2} ≥ \frac{∑k^{2}σ^{2}⋅n_i}{n} = k^{2}σ^{2}⋅\frac{∑n_i}{n} = k^{2}σ^{2}⋅P_{out},$$
where \(P_{out}\) is the proportion of the population outside \(k\) standard deviations of \(μ\). So we finally get:
$$P_{out} ≤ \frac{1}{k^{2}}.$$
The inequality can be applied to any probability distribution in which the mean and variance are defined. And as the result, we get:
$$\Pr(|X-\mu| \ge k\sigma) ≤ \frac{1}{k^2}.$$
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