What is Chebyshev's inequality?

Question

What is Chebyshev's inequality?

1 Answer

Impact of this question

6860 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-21T23:23:40

Chebyshev’s inequality says that at least $1 - \frac{1}{K^{2}}$ $1-1/K^2$ of data from a sample must fall within K standard deviations from the mean, where K is any positive real number greater than one.

Explanation:

Let play with a few value of K:

$K = 2$ $K = 2$ we have $1 - \frac{1}{K^{2}} = 1 - \frac{1}{4} = \frac{3}{4} = 75 %$ $1-1/K^2 = 1 - 1/4 = 3/4 = 75%$ . So Chebyshev’s would tell us that 75% of the data values of any distribution must be within two standard deviations of the mean.
$K = 3$ $K = 3$ we have $1 – 1/K^2 = 1 - 1/9 = 8/9 = 89%$ . This time we have 89% of the data values within three standard deviations of the mean.
$K = 4$ we have $1 – 1/K^2 = 1 - 1/16 = 15/16 = 93.75%$ . Now we have 93.75% of the data within four standard deviations of the mean.

This is consistent to saying that in Normal distribution 68% of the data is one standard deviation from the mean, 95% is two standard deviations from the mean, and approximately 99% is within three standard deviations from the mean. The difference is Chebyshev's theorem extends this principle to any distribution.

Science

Math

Humanities

... and beyond

What is Chebyshev's inequality?

1 Answer

Explanation:

Related questions

Impact of this question