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.