Chebyshev's Theorem Calculator

What the theorem promises, in plain terms

Chebyshev's theorem answers a question that comes up whenever you know two numbers about a group and nothing else: how much of the group can be far from the average? The answer is a cap. If the average delivery runs 42 minutes with a standard deviation of 6, then no matter what the schedule looks like, at most 25 percent of deliveries take under 30 or over 54 minutes, and at least 75 percent land in between.

The statement needs no assumption about the shape of the data. Skewed, bimodal, spiky, or smooth, discrete or continuous, it makes no difference: the theorem works for any distribution that has a mean and a standard deviation. That is the whole appeal. When you cannot defend a bell curve, for delivery times, repair bills, incomes, hospital stays, or anything else that cannot go below zero and lumps up on one side, Chebyshev still gives you a number you can put in a promise.

The price of that generality is that the number is weak. The 75 percent floor at k = 2 sits far below the 95.45 percent a bell curve delivers, and most real data beats the floor by a wide margin. Treat the theorem as a guarantee, the way a warranty is a guarantee: it tells you the worst case, and the worst case is the only thing you can promise without knowing more.

One restriction matters in practice. The bound is empty at k = 1 and meaningless below it, since 1 minus 1 over 1 squared is zero. There is no free lunch at one standard deviation; the theorem starts saying something at 1.25, and gets useful from 1.5 upward.

The formula, and the two shares it produces

The theorem comes in one line. For any value X from a distribution with mean μ and standard deviation σ:

P(|X − μ| ≥ kσ) ≤ 1/k²

Read it from the right: the share of values at least k standard deviations away from the mean, on either side combined, is at most 1/k². Flip it around and you get the two numbers people want most. The share inside the band is at least 1 − 1/k², and the share outside is at most 1/k².

A detail worth keeping straight: the bound counts values at or beyond k standard deviations, so the guaranteed share strictly between the edges is at least 1 − 1/k². For continuous data the difference is zero; for data piled exactly on the edge it can matter by a rounding step.

The result falls out of Markov's inequality applied to the squared distance from the mean, which is why it needs nothing but the variance. The Wikipedia article on the inequality walks through the derivation, and the NIST handbook section on measures of scale covers the standard deviation the k multiplies.

A worked example: 400 deliveries against a promise

Take a courier service with a mean delivery time of 42 minutes and a standard deviation of 6 minutes, and a manager who wants to know what can be promised about a day of 400 deliveries without assuming anything about the shape of the rush hours.

At k = 2 the band runs from 42 − 12 = 30 minutes to 42 + 12 = 54 minutes. The theorem guarantees at least 1 − 1/4, which is 75 percent of deliveries inside it, which is 300 of the 400, and it caps the deliveries outside at 1/4 = 25 percent, which is 100. At k = 3 the band widens to 24 to 60 minutes, the floor climbs to 1 minus 1/9, about 88.89 percent, so at least 355.56, which means at least 356 deliveries, and at most 44.44, which means at most 44, sit outside.

The counts come with a built-in test. If a real day puts 120 deliveries outside the two standard deviation band, the theorem is not broken; the inputs are. Either the standard deviation is bigger than 6 minutes, the mean is not 42, or the day mixed in something the averages do not describe, and the numbers should be recomputed before anyone quotes the promise again.

Notice what the theorem does not say. It does not say 75 percent of deliveries arrive near 30 to 54 minutes on a typical day, and it does not say anything about which side the stragglers fall on. It is a floor and a cap, and that is the whole contract.

Running it backwards: the k a coverage target needs

Most people arrive at this theorem with a coverage in mind and need the k. Solve 1 − 1/k² = p for k and you get k = 1/√(1 − p), with p written as a fraction. The square root is what makes the numbers grow fast at the top of the range.

The 95 percent row is the one that changes minds. Guaranteeing that at least 95 percent of anything falls in a band, without assuming a shape, takes 4.472 standard deviations on each side, and at mean 42 with SD 6 that is a promise covering 15.2 to 68.8 minutes, a window wide enough that a customer might not find it worth much. Ninety-nine percent takes ten standard deviations, 18 to 102 minutes, which is a number most operators would refuse to publish.

That is not a flaw in the theorem; it is the cost of knowing almost nothing. If the delivery times were bell shaped, 95 percent would fit inside 1.96 standard deviations. The distance between 1.96 and 4.472 is what the shape assumption is worth, and it is why measuring the actual distribution, even roughly, usually pays for itself before a service level is signed.

The one-sided version: when only the late side matters

Deliveries cannot be early in a way anyone complains about, and repair costs cannot be too low. When only one tail matters, the two-sided bound wastes half its allowance on the side you do not care about, and a sharper result called the Cantelli inequality takes over:

P(X − μ ≥ kσ) ≤ 1/(1 + k²)

The share above mean plus k standard deviations is at most 1 over (1 + k²), and the same cap applies below mean minus kσ. Solving 1/(1 + k²) = α for a tail allowance α gives k = √(1/α − 1).

At the 5 percent level the one-sided bound saves little, 4.359 against 4.472, but the reading changes: 4.359 standard deviations above the mean keeps the upper tail at 5 percent while saying nothing about the lower one. For a promise like no more than 5 percent of orders later than X minutes, that is the right tool and the two-sided k quietly overcharges you.

A courier with the 42 minute mean and 6 minute standard deviation can therefore promise that at most 5 percent of deliveries run past 42 + 4.359 × 6 = 68.2 minutes, without any shape assumption. That number, 68 minutes, is the honest version of a 95 percent late guarantee.

Next to the 68, 95, 99.7 rule

The empirical rule and Chebyshev's theorem answer the same question with different amounts of courage. The empirical rule commits to a shape, the bell curve, and pays for it with tight numbers. Chebyshev assumes nothing and pays with loose ones.

The gap is enormous at every k, which is why the choice is worth a moment of thought. Use the empirical rule when the data is known to be roughly bell shaped, measurement errors, heights, averages of many small causes, or when the numbers are for description rather than commitment. Use Chebyshev when the shape is unknown or untrustworthy, or when the sentence will appear in a contract, a service level, or a compliance filing where being wrong in one direction costs more than being vague.

Income is the classic case. Household income is right skewed with a long tail, so the empirical rule understates how many values sit far above the mean. Chebyshev's 75 percent at two standard deviations still holds, and that is the number to quote. The MathWorld entry on the inequality states the general result and the one-sided form together.

A distribution that spends the whole allowance

A bound is only as good as its worst case, and Chebyshev's worst case is easy to build. Put a value exactly k standard deviations above the mean with probability 1/(2k²), the same distance below with probability 1/(2k²), and the mean itself with the remaining probability. The mean is still μ, the variance works out to exactly σ², and the share at or beyond k standard deviations is 1/k², the full allowance the theorem permits.

At k = 2 the construction gives 12.5 percent of the values at mean minus 2σ, 12.5 percent at mean plus 2σ, and 75 percent sitting on the mean. Every number in the theorem's promise is met with nothing to spare, which proves the constant cannot be improved without adding assumptions.

Real data rarely looks like that, which is why observed shares usually beat the floor comfortably. But the construction explains the theorem's caution in a useful way: a distribution can hide a quarter of its values two standard deviations out while looking unremarkable in its mean and spread. Long-tailed business data, insurance claims, and system response times all have a bit of this character, and the theorem is built so that no amount of weirdness can push past the cap.

Reporting it without overclaiming

Most errors around this theorem are errors of wording rather than arithmetic.

• Dropping the at least. The output is a floor, not a measurement. At least 75 percent is the claim; 75 percent of our orders is not.
• Using it with k at or below 1. The bound is zero or negative there and carries no information. Start at 1.25 and expect nothing useful before 1.5.
• Feeding it a standard deviation from a tiny sample. The theorem is exact for a known mean and spread. With 8 observations the sample standard deviation wobbles, and a floor computed from a wobbly input is a suggestion, not a guarantee. Compute the spread on hundreds of values before quoting the bound.
• Confusing it with a confidence interval. A confidence interval pins down the mean from a sample; Chebyshev describes where individual values fall. The two use some of the same words and answer different questions.
• Quoting the two-sided bound for a one-sided worry. Late deliveries, high costs and long outages are one-sided problems, and the Cantelli form is tighter. Reaching for 1 − 1/k² there leaves precision on the table.

The clean sentence has all three pieces: the share, the at least, and the band. At least 88.89 percent of the 400 deliveries fell between 24 and 60 minutes, where 24 and 60 are three standard deviations either side of the 42 minute mean. Anyone auditing that sentence can reproduce it from the mean, the standard deviation, and k, which is the point of quoting a theorem instead of a hunch.