Chebyshev's inequality: what it says and how to use it

In this article

1. Before you start
2. Why interviewers ask this
3. Chebyshev's inequality from first principles
4. Worked example
5. Trade-offs and edge cases
6. How to answer it in an interview
7. Common mistakes
8. Follow-up questions
9. Sources

Before you start

This page is one step of the probability and statistics learning path.

Why interviewers ask this

Chebyshev's inequality is the standard answer to "what can you say about a distribution when you know only its mean and variance?" Interviewers in data science, quantitative and machine learning loops ask it because the answer separates people who remember the rule of thumb "95% within two standard deviations" from people who know that rule needs a normal curve.

A weak answer recites 1/k² and stops. A strong answer says why it holds for every distribution, what it costs in tightness, and when it is the only tool available. The question often arrives hidden inside another one: "how many samples do I need?" or "how do you set an alert threshold when the data is not normal?" The worked example below answers both.

Chebyshev's inequality from first principles

Chebyshev's inequality

Chebyshev's inequality is a bound which says that for any random variable with a finite mean and a finite variance, the probability of being at least k standard deviations from the mean is at most 1/k², whatever the shape of the distribution.

Markov's inequality

Markov's inequality is a bound which says that a random variable that is never negative has probability at most its mean divided by a, for any a above zero, of being at least a.

In plain words

Imagine a town where the average wait at a clinic is 10 minutes. You do not know how the waits are spread. You can still say that at most one patient in five waits 50 minutes or more, because a crowd that long would drag the average above 10. That is a bound: you give up the exact share and keep a ceiling. The analogy breaks in one place. A bound is only a ceiling, and the real share is usually far below it.

In one precise sentence: for any k above 1, no more than 1/k² of a distribution can sit k or more standard deviations from its mean, and the mean and variance are all you need to know.

Step 1: Start from Markov's inequality

Markov's inequality bounds a tail using only the mean. For a quantity Y that is never negative and any a above zero, the probability that Y is at least a is at most the mean of Y divided by a. The reason is simple: if more than that share of the values were as large as a, they alone would push the mean above its true value.

In the clinic example, the mean is 10 minutes and a is 50, so the bound is 10 ÷ 50, one in five (illustrative numbers). The bound needs only one fact, the mean, and one condition, that the quantity is never negative. Input: a mean and a threshold. Output: a ceiling on a one-sided tail. The next step makes the quantity non-negative on purpose.

Step 2: Apply Markov to the squared distance from the mean

Squaring a distance makes it never negative, so Markov applies to any quantity X. Take Y equal to (X − μ)², the squared distance from the mean μ. The mean of Y is the variance σ² by definition. Choose a equal to (kσ)².

Markov then says that the probability that (X − μ)² is at least k²σ² is at most σ² divided by k²σ². The event "the squared distance is at least k²σ²" is the same event as "the distance is at least kσ", because square roots keep the order of non-negative numbers. Cancelling σ² leaves a number that depends only on k: the probability of being at least kσ away from μ is at most 1/k² (Wikipedia, read October 2026). Here σ is the standard deviation, the square root of the variance.

Input: any X with a finite variance. Output: a ceiling on both tails together. What you now know: one line of algebra turns a variance into a distribution-free bound.

Step 3: The bound is exact for one distribution and loose for most

Chebyshev's bound is exact for one distribution and loose for most others. At least 75% of any distribution lies within two standard deviations of its mean.¹ At least 88.89% lies within three and at least 96% within five.¹ A normal distribution is far more concentrated, with 4.55% beyond two standard deviations, and an exponential distribution has 4.98% there, so the bound is generous for both.¹

It is still exact for one distribution. For k of 2, put probability 1/8 on −1, 3/4 on 0 and 1/8 on +1. Its mean is 0, its standard deviation is 0.5, and the share at least two standard deviations away, the two end points, is exactly 25%.¹ The Wikipedia article states that for every k at or above 1 such a three-point distribution gives equality, so no better bound is possible without more assumptions. A simulation of 200,000 draws from this distribution put 25.14% beyond two standard deviations.¹

1 · Markov2 · Square the distance3 · Read the boundReplay

Step-by-step description

1. A tail is at most the mean divided by a.
2. The variance cancels, only k is left.
3. Two standard deviations: at most 25%.

• Markov's inequality bounds a tail of a non-negative variable by its mean divided by a.
• The mean of the squared distance from the mean is the variance, and cancelling it leaves at most 1/k squared.
• At k of 2 the bound is 25%.

Worked example

Three questions use the same inequality: how loose is it, what does it say about a service's response times, and how many samples does an average need. All numbers come from one script, with fixed seeds.¹

Now a response-time question with illustrative numbers: the mean is 120 milliseconds and the standard deviation 30 milliseconds. A response of 210 milliseconds or more is 90 away, which is three standard deviations. Chebyshev allows at most 1/9, or 11.11%, of responses to be at least that far from the mean in either direction.¹ For the slow side alone, Cantelli's inequality is the one-sided version of Chebyshev's, and it gives at most 1/(1 + k²), which is 10% at k of 3.¹

The last question is a sample size. A fair coin has variance 0.25, so the sample mean of n flips has variance 0.25/n. Chebyshev says the chance that the mean is 0.05 or more away from 0.5 is at most 0.25 ÷ (n × 0.05²). To keep that at or below 0.05 needs n of at least 2,000.¹ The exact binomial chance stays at or below 5% for every count from 401 flips up, five times fewer.¹ It is not smooth in n: 371 flips already give 4.84% and 400 flips give 5.10%.¹ At 2,000 flips the true chance is 0.0008%.¹ The bound is safe and wasteful.

Text description of the diagram

• At three standard deviations Chebyshev allows up to 11.11% of values beyond, while a normal distribution has 0.27%.
• At two standard deviations a normal distribution has 4.55% beyond and an exponential distribution has 4.98%.

Trade-offs and edge cases

The inequality has two forms and a short history. The Wikipedia article gives the distance form: for any a above zero, the probability that the distance is at least a is at most σ²/a². Only k above 1 is useful, because for k of 1 or less the bound is 1 or more, and every probability is already at most 1. Bienaymé stated and proved the result in 1853, Chebyshev proved it more generally in 1867, and Markov gave another proof in his 1884 thesis, according to the same article.

Chebyshev's inequality trades tightness for generality, and four cases decide whether it is the right tool.

First, shape knowledge. If you know the data is normal, the rule of 68%, 95% and 99.7% on the Gaussian page is far sharper. With only a mean and a variance, Chebyshev is the best bound that holds for everything.

Second, one tail. Halving the two-sided bound is a common shortcut and is wrong in general: at k of 3 it gives 5.56%,¹ while the true worst case for one tail is Cantelli's 10%. Halving is safe only when the distribution is symmetric.

Third, infinite variance. The bound needs a finite variance, so it says nothing for heavy-tailed data such as a Cauchy distribution, whose variance does not exist.

Fourth, estimated parameters. In practice μ and σ come from a sample, and the bound then holds only approximately; the Wikipedia article lists a sample-based variant by Saw, Yang and Mo (1984). For alert thresholds on skewed data, see anomaly detection in time series, and for the sample-based side of the same ideas, the t-test page.

Text description of the diagram

• Markov's inequality needs a non-negative variable and bounds one tail.
• Chebyshev's inequality needs a finite variance and bounds both tails by 1/k².
• Cantelli's inequality bounds one tail by 1/(1 + k²), which is 10% at k of 3.

How to answer it in an interview

The spoken structure: state the bound, say it holds for every distribution, give the proof in one line, then say how loose it is.

Common mistakes

• "Chebyshev says 68% of data is within one standard deviation." (wrong) At k of 1 the bound is 0%; 68% is the normal-curve rule. (right) "Chebyshev says at least 75% lies within two standard deviations, for any distribution."
• "It only works for normal data." (wrong) It needs only a finite mean and variance. (right) "It holds for every distribution with a finite variance, which is why it is loose."
• "The probability equals 1/k²." (wrong) 1/k² is a ceiling; the real share was 4.55% against 25% for a normal at k of 2. (right) "The probability is at most 1/k²."
• "For one tail, divide by two." (wrong) That gives 5.56% at k of 3, below Cantelli's 10%. (right) "Use 1/(1 + k²) for one tail, or halve only if the distribution is symmetric."

Follow-up questions

What is the difference between Markov and Chebyshev inequality?

Markov's inequality bounds a one-sided tail of a non-negative variable using its mean. Chebyshev's inequality is Markov applied to the squared distance from the mean, so it bounds both tails using the variance.

How does Chebyshev's inequality prove the law of large numbers?

The mean of n independent samples with variance σ² has variance σ²/n, so the chance that it is ε or more from the true mean is at most σ²/(nε²), which goes to zero as n grows. That is the weak law of large numbers, which the MIT course lists with Chebyshev's inequality as its tool (MIT 6.041SC, lecture 19).

Is Chebyshev's inequality tight?

For the worst case, yes: a three-point distribution on −1, 0 and +1 reaches the bound exactly. For a specific distribution such as a normal one it is far from tight.

What if the variance is infinite?

Then the inequality does not apply. Use a bound that needs a weaker moment, or a distribution-specific tail result, and say which assumption you are making.

How many samples do I need for a given error?

Set σ²/(nε²) equal to the failure chance and solve for n. For a fair coin, error 0.05 and failure 0.05 that gives 2,000, a safe upper bound, against 401 from the exact binomial tail, the count from which every larger sample also passes.¹

Sources

1. Chebyshev's inequality — Wikipedia (date = read date; used for the statement, the history, the tight case and Cantelli's bound, each of which the page attributes), 24 Oct 2026
2. Lecture 19: Weak Law of Large Numbers (6.041SC, Probabilistic Systems Analysis and Applied Probability) — MIT OpenCourseWare (Tsitsiklis, Fall 2013 term; the page gives no day, so the date is the start of the term), 1 Sep 2013
3. Lecture 15 notes: limit theorems, Chebyshev inequality, convergence in probability (6.041) — MIT OpenCourseWare (Dahleh, Spring 2006 term; the page gives no day, so the date is the start of the term), 1 Feb 2006