---
tag: 0EYQ03XCR3
kind: example
name: 68–95–99.7 rule
aliases: [empirical-rule, three-sigma-rule]
msc: ["60E05"]
instance_of: [05AQE0QR0D]
uses: [07RXBVTVZB, 0VTN5RKNC5, 05RZ4YZG4E]
sources:
  - url: https://en.wikipedia.org/wiki/68–95–99.7_rule
    locator: Table of numerical values
same_as:
  wikidata: Q847822
  wikipedia: 68–95–99.7 rule
---

## Example

Let $X \sim \mathcal{N}(\mu, \sigma^2)$ with $\sigma > 0$, and let $\Phi$ be the cumulative distribution function of $\mathcal{N}(0, 1)$.
Since $Z = (X - \mu)/\sigma \sim \mathcal{N}(0, 1)$, $P(|X - \mu| \le k\sigma) = 2\Phi(k) - 1$ for each $k > 0$. Numerical values of $\Phi$ give:
$$P(|X - \mu| \le \sigma) \approx 0.6827, \quad P(|X - \mu| \le 2\sigma) \approx 0.9545, \quad P(|X - \mu| \le 3\sigma) \approx 0.9973.$$
So about $68\%$, $95\%$ and $99.7\%$ of the mass of a normal distribution lie within one, two and three standard deviations of the mean.

## Remarks

The rule holds for normal distributions only. For an arbitrary distribution with finite variance, Chebyshev's inequality gives only $P(|X - \mu| \le k\sigma) \ge 1 - 1/k^2$, for example $3/4$ for $k = 2$.
The interval $\mu \pm 1.96\sigma$ contains $95\%$ of the mass. It is the base of the usual $95\%$ confidence intervals.

## Explanations

### L1 human

For data that follow a bell curve, almost all values are within three standard deviations of the mean.
If adult heights have mean $170$ cm and standard deviation $10$ cm, then about $95\%$ of the heights are between $150$ cm and $190$ cm.