68–95–99.7 rule

We have that

$${\displaystyle {\begin{aligned}\Pr(\mu -n\sigma \leq X\leq \mu +n\sigma )=\int _{\mu -n\sigma }^{\mu +n\sigma }{\frac {1}{{\sqrt {2\pi }}\sigma }}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}dx,\end{aligned}}}$$

doing the change of variable ${\displaystyle u={\frac {x-\mu }{\sigma }}}$, we have

$${\displaystyle {\begin{aligned}{\frac {1}{\sqrt {2\pi }}}\int _{-n}^{n}e^{-{\frac {u^{2}}{2}}}du\end{aligned}},}$$

and this integral is independent of ${\displaystyle \mu }$ and ${\displaystyle \sigma }$. We only need to calculate each integral for the cases ${\displaystyle n=1,2,3}$.

$${\displaystyle {\begin{aligned}&\Pr(\mu -1\sigma \leq X\leq \mu +1\sigma )={\frac {1}{\sqrt {2\pi }}}\int _{-1}^{1}e^{-{\frac {u^{2}}{2}}}du\approx 0.6827\\&\Pr(\mu -2\sigma \leq X\leq \mu +2\sigma )={\frac {1}{\sqrt {2\pi }}}\int _{-2}^{2}e^{-{\frac {u^{2}}{2}}}du\approx 0.9545\\&\Pr(\mu -3\sigma \leq X\leq \mu +3\sigma )={\frac {1}{\sqrt {2\pi }}}\int _{-3}^{3}e^{-{\frac {u^{2}}{2}}}du\approx 0.9973.\end{aligned}}}$$

Diagram showing the cumulative distribution function for the normal distribution with mean (μ) 0 and variance (σ²) 1

These numerical values "68%, 95%, 99.7%" come from the cumulative distribution function of the normal distribution.

The prediction interval for any standard score z corresponds numerically to (1 − (1 − Φ_μ,σ²(z)) · 2).

For example, Φ(2) ≈ 0.9772, or Pr(X ≤ μ + 2σ) ≈ 0.9772, corresponding to a prediction interval of (1 − (1 − 0.97725)·2) = 0.9545 = 95.45%. This is not a symmetrical interval – this is merely the probability that an observation is less than μ + 2σ. To compute the probability that an observation is within two standard deviations of the mean (small differences due to rounding):

$${\displaystyle \Pr(\mu -2\sigma \leq X\leq \mu +2\sigma )=\Phi (2)-\Phi (-2)\approx 0.9772-(1-0.9772)\approx 0.9545}$$

This is related to confidence interval as used in statistics: ${\displaystyle {\bar {X}}\pm 2{\frac {\sigma }{\sqrt {n}}}}$ is approximately a 95% confidence interval when ${\displaystyle {\bar {X}}}$ is the average of a sample of size ${\displaystyle n}$.

The "68–95–99.7 rule" is often used to quickly get a rough probability estimate of something, given its standard deviation, if the population is assumed to be normal. It is also used as a simple test for outliers if the population is assumed normal, and as a normality test if the population is potentially not normal.

To pass from a sample to a number of standard deviations, one first computes the deviation, either the error or residual depending on whether one knows the population mean or only estimates it. The next step is standardizing (dividing by the population standard deviation), if the population parameters are known, or studentizing (dividing by an estimate of the standard deviation), if the parameters are unknown and only estimated.

To use as a test for outliers or a normality test, one computes the size of deviations in terms of standard deviations, and compares this to expected frequency. Given a sample set, one can compute the studentized residuals and compare these to the expected frequency: points that fall more than 3 standard deviations from the norm are likely outliers (unless the sample size is significantly large, by which point one expects a sample this extreme), and if there are many points more than 3 standard deviations from the norm, one likely has reason to question the assumed normality of the distribution. This holds ever more strongly for moves of 4 or more standard deviations.

One can compute more precisely, approximating the number of extreme moves of a given magnitude or greater by a Poisson distribution, but simply, if one has multiple 4 standard deviation moves in a sample of size 1,000, one has strong reason to consider these outliers or question the assumed normality of the distribution.

For example, a 6σ event corresponds to a chance of about two parts per billion. For illustration, if events are taken to occur daily, this would correspond to an event expected every 1.4 million years. This gives a simple normality test: if one witnesses a 6σ in daily data and significantly fewer than 1 million years have passed, then a normal distribution most likely does not provide a good model for the magnitude or frequency of large deviations in this respect.

In The Black Swan, Nassim Nicholas Taleb gives the example of risk models according to which the Black Monday crash would correspond to a 36-σ event: the occurrence of such an event should instantly suggest that the model is flawed, i.e. that the process under consideration is not satisfactorily modeled by a normal distribution. Refined models should then be considered, e.g. by the introduction of stochastic volatility. In such discussions it is important to be aware of the problem of the gambler's fallacy, which states that a single observation of a rare event does not contradict that the event is in fact rare. It is the observation of a plurality of purportedly rare events that increasingly undermines the hypothesis that they are rare, i.e. the validity of the assumed model. A proper modelling of this process of gradual loss of confidence in a hypothesis would involve the designation of prior probability not just to the hypothesis itself but to all possible alternative hypotheses. For this reason, statistical hypothesis testing works not so much by confirming a hypothesis considered to be likely, but by refuting hypotheses considered unlikely.

Because of the exponentially decreasing tails of the normal distribution, odds of higher deviations decrease very quickly. From the rules for normally distributed data for a daily event:

Range	Expected fraction of population inside range	Expected fraction of population outside range	Approx. expected frequency outside range		Approx. frequency for daily event
μ ± 0.5σ	0.382924922548026	6.171E-01 = 61.71 %	3 in	5	Four or five times a week
μ ± σ	0.682689492137086[4]	3.173E-01 = 31.73 %	1 in	3	Twice or thrice a week
μ ± 1.5σ	0.866385597462284	1.336E-01 = 13.36 %	1 in	7	Weekly
μ ± 2σ	0.954499736103642[5]	4.550E-02 = 4.550 %	1 in	22	Every three weeks
μ ± 2.5σ	0.987580669348448	1.242E-02 = 1.242 %	1 in	81	Quarterly
μ ± 3σ	0.997300203936740[6]	2.700E-03 = 0.270 % = 2.700 ‰	1 in	370	Yearly
μ ± 3.5σ	0.999534741841929	4.653E-04 = 0.04653 % = 465.3 ppm	1 in	2149	Every 6 years
μ ± 4σ	0.999936657516334	6.334E-05 = 63.34 ppm	1 in	15787	Every 43 years (twice in a lifetime)
μ ± 4.5σ	0.999993204653751	6.795E-06 = 6.795 ppm	1 in	147160	Every 403 years (once in the modern era)
μ ± 5σ	0.999999426696856	5.733E-07 = 0.5733 ppm = 573.3 ppb	1 in	1744278	Every 4776 years (once in recorded history)
μ ± 5.5σ	0.999999962020875	3.798E-08 = 37.98 ppb	1 in	26330254	Every 72090 years (thrice in history of modern humankind)
μ ± 6σ	0.999999998026825	1.973E-09 = 1.973 ppb	1 in	506797346	Every 1.38 million years (twice in history of humankind)
μ ± 6.5σ	0.999999999919680	8.032E-11 = 0.08032 ppb = 80.32 ppt	1 in	12450197393	Every 34 million years (twice since the extinction of dinosaurs)
μ ± 7σ	0.999999999997440	2.560E-12 = 2.560 ppt	1 in	390682215445	Every 1.07 billion years (four occurrences in history of Earth)
μ ± 8σ	0.999999999999999	6.221E-16	1 in	2000000000000000	Once every 4 trillion years (never in the history of the Universe, once during the life of a red dwarf)
μ ± xσ	${\displaystyle \operatorname {erf} \left({\frac {x}{\sqrt {2}}}\right)}$	${\displaystyle 1-\operatorname {erf} \left({\frac {x}{\sqrt {2}}}\right)}$	1 in	${\displaystyle {\tfrac {1}{1-\operatorname {erf} \left({\frac {x}{\sqrt {2}}}\right)}}}$	Every ${\displaystyle {\tfrac {1}{1-\operatorname {erf} \left({\frac {x}{\sqrt {2}}}\right)}}}$ days

• p-value
• Six Sigma § Sigma levels
• Standard score
• t-statistic

• "The Normal Distribution" by Balasubramanian Narasimhan
• "Calculate percentage proportion within x sigmas at WolframAlpha