Chebyshev's Theorem Calculator

Chebyshev's theorem, also written Chebychev or Tchebysheff and known as Chebyshev's inequality or the Bienaymé-Chebyshev inequality, is a guarantee about how spread out the values of a data set or a probability distribution can be. For any k greater than 1, at least 1 − 1/k² of the values lie within k standard deviations of the mean. It needs only the mean and the standard deviation, not the shape of the distribution, so it holds for skewed, bimodal and unknown distributions alike.

Said the other way round, at most 1/k² of the values are k or more standard deviations from the mean. For k = 2 at least 75% of the values lie within two standard deviations of the mean, and for k = 3 at least 88.89% (8/9) lie within three. Irénée-Jules Bienaymé proved it in 1853 and Pafnuty Chebyshev more generally in 1867.

For k of 1 or less the bound is useless: 1 − 1/k² is 0 or negative, and all of the values could lie k or more standard deviations from the mean. The calculator says so instead of returning a negative share.

Choose Load example to fill in a worked example for the mode you picked. The steps under the results show every substitution, so you can copy them into a solution.

The empirical rule (68-95-99.7) gives the shares of a normal distribution. Chebyshev's theorem gives the shares every distribution must have at least, so it is much smaller, and it is the only one of the two you can use when the distribution is skewed or unknown.

Learn how the normal shares arise in the empirical rule article, and use the normal distribution calculator for any other probability of a normal variable.

The scores of a class have mean 86 and standard deviation 3, and their distribution is unknown. What share of the scores is guaranteed to lie between 77 and 95? Choose Share within an interval and Load example.

• The distances from the mean are 86 − 77 = 9 and 95 − 86 = 9, so both limits are 9 / 3 = 3 standard deviations away: k = 3.
• The share outside is at most 1/k² = 1/9 and the share within at least 1 − 1/9 = 8/9.
• At least 88.89% of the scores lie between 77 and 95, and at most 11.11% are below 77 or above 95.

Values have mean 100 and standard deviation 15. Which interval around the mean holds at least 90% of them? Choose Standard deviations needed for a share and load the example.

• k = 1 / √(1 − 0.9) = 1 / √0.1 = √10 = 3.1623.
• The interval is 100 ± √10 × 15 = 100 ± 47.4342, from 52.5658 to 147.4342.
• A normal distribution needs only 1.6449 standard deviations for 90%: an interval of about 100 ± 24.67, from 75.33 to 124.67. The wider Chebyshev interval is the price of assuming nothing about the shape.

A sample of 200 measurements has mean 100 and standard deviation 10. How many are guaranteed to lie between 75 and 125? Choose Share within k standard deviations, enter k = 2.5, the mean, the standard deviation and n = 200.

• k = 2.5, so k² = 6.25, and 1 − 1/6.25 = 1 − 4/25 = 21/25 = 84%.
• At most floor(200 / 6.25) = 32 measurements lie 2.5 or more standard deviations from the mean, so at least 200 − 32 = 168 lie between 75 and 125.

Counts are whole numbers, so the number outside is rounded down. The calculator does this division exactly: in double precision, 121 values and k = 1.1 give 121 / 1.2100000000000002 = 99.99999999999999, which rounds down to 99 values outside when 100 are possible.

Chebyshev's bound covers both tails together. To bound one tail, Cantelli's inequality says that at most 1/(1 + k²) of the values are k or more standard deviations above the mean, and at most as many are below it. For k = 2 that is at most 20% above and at most 20% below, and it cannot be improved for a general distribution.

Half of the two-sided bound, 12.5% for k = 2, is only right when the distribution is symmetric. For a skewed distribution one tail can hold more than half of the 25%, which is why the one-sided bound is larger.

• The distribution is unknown or not normal. Skewed incomes, waiting times and other data with heavy tails do not follow the empirical rule, but the theorem still holds.
• You need a guarantee, not an estimate. Quality control limits, risk bounds and proofs, such as the proof of the weak law of large numbers, use the inequality because it never fails.
• Only the mean and the standard deviation are known. No other information about the data is needed.

The guarantee is conservative. Real data usually have far more values within k standard deviations than the minimum, and the bound is attained only by a three-value distribution with the values −kσ, 0 and kσ (with probabilities 1/(2k²), 1 − 1/k² and 1/(2k²)), which real data rarely resemble.

For a data set, use its own mean and standard deviation. With the population standard deviation (divisor n) at most n/k² of the values are k or more standard deviations from the mean. The sample standard deviation (divisor n − 1) is a little larger, so an interval built with it holds even more of the values and the guarantee still stands. To bound a new observation when only sample statistics are known, finite-sample versions of the inequality by Saw, Yang and Mo exist; this calculator does not cover them.

• k must be greater than 1. For k = 1 the guarantee is 0%, and for k below 1 the formula would give a negative share.
• The result is a minimum. "At least 75%" does not mean 75%; the true share is usually much larger.
• The standard deviation must be positive and finite. A distribution without a finite variance, such as the Cauchy distribution, has no Chebyshev bound.
• An interval that is not centred on the mean uses its nearer limit. The calculator applies the theorem to the largest interval around the mean that fits inside yours, so the guarantee is correct but not the best possible for a lopsided interval. If the mean is not between the limits there is no guarantee.
• Use the standard deviation of the same values as the mean. Mixing a mean from one sample and a standard deviation from another gives a meaningless guarantee.