Chebyshev's Inequality(2)

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Jul 24, 2020, 6:28:07 PM

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May 22, 2014, 6:29:18 PM

P r (| X - μ | \geq k σ) = \frac{1}{k^{2}}

$Pr(|X-mu| >= ksigma) = 1/ "k" ^2$

Any real number multiple of the standard deviation

$"Any real number multiple of the standard deviation"$

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[Mathematics | Probability | Statistics] Chebyshev's inequality (also known as Tchebysheff's inequality) states that in any distribution almost all of the values are close to the mean. To be more accurate, the statement means that no more than $\frac{1}{k^{2}}$ $1/k^2$ of the distribution's values will be more than k standard deviations away from the mean.

$P (| x - μ | \geq k σ) \leq \frac{1}{k^{2}}$ $P(|x - mu| >= ksigma) <= 1 / k^2$

So, this calculation tells you the probability that your data value, x, is further from the mean than the input value, k, times the standard deviation. As we expect, the probability that your data value is further from the mean than some multiple of the standard deviation decreases with that increase in that multiple. I.e, the Probability is lower that you value falls outside of six $σ$ $sigma$ from the means than is the probability of falling outside one $σ$ $sigma$ from the mean.

Variables:

$k$ $k$ = Any real number multiple of the standard deviation

Notes

There are two variants of the formula that are equally valuable. This is the second of the two which is denoted in the name of the equation.

The difference being where k is denoted.

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Chebyshev's Inequality(2)

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