Chebyshev’s inequality is a safety net for situations where you know the mean and standard deviation of a variable but nothing about the shape of its distribution. Whatever the shape – skewed, bimodal or heavy-tailed – at least 1 − 1/k² of the values lie within k standard deviations of the mean. So at least 75% lie within 2σ, at least 88.9% within 3σ and at least 93.75% within 4σ.
You can enter the number of standard deviations k directly together with the mean and standard deviation, or enter a distance from the mean and the variance (the form P(|X − μ| ≥ d) ≤ σ²/d²). The calculator returns the upper bound on the probability of being that far out, the lower bound on the probability of being inside, the interval itself, and – for comparison – what a normal distribution would give.
The bound is deliberately conservative. For a normal distribution 95.45% lies within 2σ, well above Chebyshev’s 75%. For k ≤ 1 the theorem says nothing useful, since the bound on the tail probability is 100% or more.
How to use the Chebyshev’s Theorem Calculator
- Input
- Mean (μ)
- Number of standard deviations (k)
- Standard deviation (σ)
- Distance from the mean (d)
- Variance (σ²)
Results update as you type, and everything is calculated in your browser.
How it works
Convert the distance into standard deviations, k = d ÷ σ, then the tail bound is 1/k² (capped at 100%) and the coverage bound is 1 − 1/k² (at least 0). The interval is mean ± k standard deviations.
Worked example
Exam scores have mean 50 and standard deviation 10, shape unknown. With k = 2 the interval is 50 ± 20 = 30 to 70, and at least 1 − 1/2² = 75% of students scored inside it; at most 25% scored outside. If the scores were normal the share would be 95.45%.
Distance form: with variance 100 (σ = 10) and distance 15, k = 1.5, so at most 100/15² = 44.44% of values are 15 or more from the mean and at least 55.56% are within 35 to 65.
Chebyshev vs normal coverage
| k | Chebyshev: at least | Normal distribution |
|---|---|---|
| 1.5 | 55.56% | 86.64% |
| 2 | 75.00% | 95.45% |
| 2.5 | 84.00% | 98.76% |
| 3 | 88.89% | 99.73% |
| 4 | 93.75% | 99.994% |
Inputs explained
| Field | What to enter |
|---|---|
| Input | Choose one: Number of standard deviations k, Distance from the mean and variance |
| Mean (μ) | A number |
| Number of standard deviations (k) | A number |
| Standard deviation (σ) | A number |
| Distance from the mean (d) | A number |
| Variance (σ²) | A number |
What each result means
- At least this share lies within the interval = max(0, 1 – 1÷[k]^2)×100
- k (distance in standard deviations) = [k]
- At most this share lies outside (|X − μ| ≥ kσ) = min(1, 1÷[k (distance in standard deviations)]^2)×100
- Share within the interval if the data were normal = (1 – 2×(1 – ncdf([k (distance in standard deviations)])))×100
Formula
P(|X − μ| ≥ kσ) ≤ 1 ÷ k² · P(|X − μ| < kσ) ≥ 1 − 1 ÷ k²
In terms of a distance d: P(|X − μ| ≥ d) ≤ σ² ÷ d²
Worked example
Inputs — Input: Number of standard deviations k; Mean (μ): 50; Number of standard deviations (k): 2; Standard deviation (σ): 10.
Results — At least this share lies within the interval = 75 %; k (distance in standard deviations) = 2; At most this share lies outside (|X − μ| ≥ kσ) = 25 %; Interval lower end = 30; Interval upper end = 70; Share within the interval if the data were normal = 95.45 %.
Key terms
- Chebyshev’s inequality
- P(|X − μ| ≥ kσ) ≤ 1/k² for any distribution with finite variance.
- k
- The distance from the mean measured in standard deviations.
- Empirical rule
- The 68–95–99.7 percentages that apply to normal distributions only.
Frequently asked questions
What does Chebyshev’s theorem say?
For any distribution with a finite variance, at least 1 − 1/k² of the values lie within k standard deviations of the mean.
What percentage lies within 2 standard deviations?
At least 75% (1 − 1/4). Within 3 standard deviations at least 88.89%, within 1.5 at least 55.56%.
How is it different from the empirical rule?
The 68–95–99.7 rule only holds for normal (bell-shaped) data. Chebyshev’s bound holds for every distribution but is much looser.
Why does the theorem need k > 1?
For k ≤ 1 the bound 1/k² is 100% or more, which tells you nothing.
Can I use a distance instead of k?
Yes – P(|X − μ| ≥ d) ≤ σ²/d². With variance 100 and distance 15, at most 44.44% of values are 15 or more away from the mean.
Is the bound ever reached exactly?
Yes. A distribution with mass only at μ and μ ± kσ can hit it, so the theorem cannot be improved without more information.
Does it work for samples?
It applies to any data set when you use the data’s own mean and population standard deviation; for sample estimates it is approximate.