formula.expert

Chebyshev's Inequality

Also known as Chebyshev bound · Chebychev inequality · proportion within k standard deviations · 1 - 1/k^2 · distribution-free bound

P≥1−1k2P \ge 1 - \frac{1}{k^{2}}

Worked example: k = 3 → at least 8/9 = 88.9% — press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!

See also
Z-Score (Standard Score) — Where k comes from: it is the z-score of the interval's edge, (x − μ)/σ.Normal Probability Density — The bell curve this bound deliberately does not assume — for a normal population the true proportions are far tighter than Chebyshev allows.

Learning zone

Chebyshev's inequality is the promise that survives when you know nothing about the shape of your data. Whatever the distribution — skewed, bimodal, spiky, anything with a finite variance — at least 1 − 1/k² of it lies within k standard deviations of the mean. Worked example: at k = 3, 1 − 1/9 = 0.8889, so 88.9% of any data set is within three standard deviations of its mean. At k = 2 the guarantee is 1 − 1/4 = 75%, and at k = 1 it is zero, which is honest rather than useless: a data set really can have most of its mass more than one standard deviation out.

Compare that with the bell curve, where the true figures are about 68%, 95% and 99.7% within one, two and three standard deviations — the empirical rule. Chebyshev is far looser, and it has to be, because it is covering every distribution at once rather than one. That is exactly when to reach for it: quality data with a long tail, incomes, failure times, anything you have not established is normal. Irénée-Jules Bienaymé published the inequality in 1853 and Pafnuty Chebyshev proved it independently in 1867; his student Andrey Markov generalised the argument, which is why the one-line proof is usually taught through Markov's inequality applied to the squared deviation. The bound is tight, in the sense that a distribution can be built that attains it exactly for any given k — so it cannot be improved without assuming something more. The trap is reading the answer as an estimate. It is a floor, and a real data set usually does far better.

Chebyshev's Inequality
P≥1−1k2P \ge 1 - \frac{1}{k^{2}}
Where
  • PP= Minimum proportion within k σ
  • kk= Standard deviations from the mean
Missing one of these? Work it out first, then come back