Empirical Rule Calculator

This empirical rule calculator shows what share of a normal distribution lies within one, two and three standard deviations of the mean. Enter the mean and standard deviation: for 100 and 15 the ranges run from 85 to 115, 70 to 130 and 55 to 145.

Example: Mean μ 100 · Standard deviation σ 15 · Single value x (optional) 130 → Result: 68 %: 85–115 · 95 %: 70–130 · 99.7 %: 55–145. Source: NIST/SEMATECH e-Handbook – Normal distribution. Updated: .

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Result

Result
68 %: 85–115 · 95 %: 70–130 · 99.7 %: 55–145
All values
RangeShare of valuesFrom – to
μ ± 1σ≈ 68 %85 – 115
μ ± 2σ≈ 95 %70 – 130
μ ± 3σ≈ 99.7 %55 – 145
Calculation steps
  • μ ± 1σ = 100 ± 1 · 15: from 85 to 115 (≈ 68 % of values).
  • μ ± 2σ = 100 ± 2 · 15: from 70 to 130 (≈ 95 % of values).
  • μ ± 3σ = 100 ± 3 · 15: from 55 to 145 (≈ 99.7 % of values).
  • z = (x − μ) / σ = (130 − 100) / 15 = 2.
  • The value lies within μ ± 2σ, the range that holds ≈ 95 % of values.
The percentages are rounded; the exact figures are 68.27 %, 95.45 % and 99.73 %. The rule only holds for roughly normal data.

How it is calculated

The empirical rule (68-95-99.7 rule) describes how far from the mean the values of a normal distribution fall. About 68 % lie within one standard deviation of the mean, about 95 % within two and about 99.7 % within three.

Formula and steps

The limits are μ − k·σ and μ + k·σ for k = 1, 2, 3. For a single value x you also form the z-score z = (x − μ) / σ; its absolute value shows which of the three ranges the value falls into. More precise than the rounded percentages are 68.27 %, 95.45 % and 99.73 %.

Worked example

An IQ test is scaled to a mean of 100 and a standard deviation of 15. Then about 68 % of people score between 85 and 115, about 95 % between 70 and 130 and about 99.7 % between 55 and 145. A score of 130 has z = (130 − 100) / 15 = 2, so it just sits inside the 95 % range.

Frequently asked questions

What does the 68-95-99.7 rule say?

In a normal distribution about 68 % of values lie within one standard deviation of the mean, 95 % within two and 99.7 % within three.

How do I apply the rule to one value?

Compute z = (x − μ) / σ. If the absolute value of z is at most 2, the value is inside the 95 % range; with mean 100, σ = 15 and x = 130 you get z = 2.

Does the rule work for data that is not normal?

Only approximately, and only for bell-shaped data. For any distribution Chebyshev’s inequality gives much more cautious minimum shares, namely 75 % within 2σ.

Sources and legal basis

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