Normal Distribution Empirical Rule: 68-95-99.7

Q: What is the empirical rule for a normal distribution?

It states that about 68%, 95% and 99.7% of values lie within 1, 2 and 3 standard deviations of the mean of a normal distribution. The exact figures are 68.27%, 95.45% and 99.73%.

Q: What is the relationship between the normal distribution and the empirical rule?

The empirical rule is a consequence of the normal distribution’s shape. It does not hold in general for other shapes; for those, Chebyshev’s inequality gives a weaker guaranteed bound.

Q: Is the empirical rule the same as the 68-95-99.7 rule?

Yes. The empirical rule, the 68-95-99.7 rule and the three-sigma rule name the same statement.

Q: Why is 95% not exactly 2 standard deviations?

Two standard deviations cover 95.45%. The multiplier that covers exactly 95% is 1.96.

Q: Can I use the empirical rule on skewed data?

Not reliably. Check the shape first, or use Chebyshev’s lower bounds of 75% (2 SD) and 88.9% (3 SD). For a closed-form look at normal probabilities, see the probability distribution function of the normal distribution .

The empirical rule says that in a normal distribution about 68% of values fall within one standard deviation of the mean, about 95% within two, and about 99.7% within three. It is also called the 68-95-99.7 rule or the three-sigma rule. This guide goes beyond the memorised percentages: it gives the exact coverage values, a worked example on a small dataset with the arithmetic shown, and the point where the rule stops being valid and standard deviation alone is all you can rely on.

Empirical rule of the normal distribution: definition

The empirical rule is a statement about how a normal distribution spreads probability around its mean. If a variable is normally distributed with mean μ and standard deviation σ, then roughly 68% of observations lie between μ − σ and μ + σ, roughly 95% between μ − 2σ and μ + 2σ, and roughly 99.7% between μ − 3σ and μ + 3σ. OpenStax’s Introductory Statistics 2e states the rule in exactly these three bands. The rule is a shortcut: it lets you judge how unusual a value is without a table, provided the data are approximately bell-shaped. For the full definition of the distribution itself, see the guide on the normal and standard normal distribution.

Formula: intervals and exact coverage

The empirical rule has no single formula; it is three intervals of the form μ ± kσ, with k = 1, 2 or 3. The normal density that sits behind it is given in the NIST/SEMATECH e-Handbook of Statistical Methods as f(x) = exp(−(x − μ)² / (2σ²)) / (σ√(2π)), where μ is the location parameter and σ is the scale parameter. The case μ = 0, σ = 1 is the standard normal distribution.

μ (mu): the mean, the centre of the distribution.
σ (sigma): the standard deviation, the scale. For a sample you estimate it with s.
k: how many standard deviations either side of the mean the interval extends.
x: a value of the variable. Its distance from the mean in standard-deviation units is its z-score, z = (x − μ) / σ.

The percentages are the area under the curve between z = −k and z = +k. Computed from the normal cumulative distribution function, the exact coverage differs slightly from the rounded figures:

Interval	Rule of thumb	Exact coverage
μ ± 1σ	68%	68.27%
μ ± 2σ	95%	95.45%
μ ± 3σ	99.7%	99.73%

Two exact figures worth remembering alongside the rule: 95.00% of a normal distribution lies within 1.96 standard deviations of the mean, and 99.00% within 2.576. So “95%” in the rule really means 2 SD only approximately; the cut-off that gives exactly 95% is 1.96. Confidence-interval and significance work uses 1.96, not 2.

Worked example with the arithmetic shown

The following is an illustrative dataset, not real measurements: 25 reaction times in milliseconds from a simple timing task. Sorted, they are 231, 238, 242, 245, 247, 249, 250, 251, 252, 253, 254, 255, 256, 257, 258, 259, 260, 262, 264, 266, 268, 271, 274, 279, 291.

Mean. The values sum to 6,432, so the mean is 6,432 / 25 = 257.28 ms.
Standard deviation. The squared deviations from 257.28 sum to about 4,123. With n − 1 = 24 in the denominator, the sample variance is 4,123 / 24 ≈ 171.8 and s = √171.8 ≈ 13.11 ms.
Build the intervals.
- 1 SD: 257.28 ± 13.11 gives 244.17 to 270.39 ms.
- 2 SD: 257.28 ± 26.21 gives 231.07 to 283.49 ms.
- 3 SD: 257.28 ± 39.32 gives 217.96 to 296.60 ms.
Count the values inside each. 18 of 25 fall in the 1 SD interval (72%), 23 of 25 in the 2 SD interval (92%; 231 and 291 fall outside), and all 25 in the 3 SD interval (100%).

The observed 72%, 92% and 100% are close to 68%, 95% and 99.7% for a sample this small. A sample of 25 cannot reproduce the percentages exactly: one value moves the 2 SD figure by 4 percentage points. The rule describes the distribution; a small sample only approximates it. The value 291 is (291 − 257.28) / 13.11 = 2.57 standard deviations above the mean, which the rule says to expect in about 1 case in 200 on the high side, so it is worth a look but not alarming in a sample of 25.

Interpretation: reading the result

Where a value falls	Share of a normal distribution outside that band	How to read it
Within 1 SD of the mean	about 31.7% outside	Ordinary; roughly two in three values are here.
Between 1 and 2 SD	about 4.6% outside 2 SD	Somewhat high or low, still common.
Between 2 and 3 SD	about 0.27% outside 3 SD	Unusual; expect about 1 in 22 values beyond 2 SD in total.
Beyond 3 SD	0.27% in total, or 0.135% per tail	Rare: about 1 in 370 values. Check for error or a different population.

The 1-in-370 figure is simply 1 / 0.0027. Because the normal curve is symmetric, the tail beyond 3 SD splits evenly: about 1 in 740 values above μ + 3σ.

Assumptions and common mistakes

It requires an approximately normal distribution

The percentages are properties of the normal curve, not of data in general. Skewed or heavy-tailed data can have far more or fewer values within 1, 2 or 3 SD. The guide on normality of distribution covers how to check before you rely on the rule.

Empirical rule versus Chebyshev’s inequality

If the data are not normal, Chebyshev’s inequality still guarantees that at least 1 − 1/k² of values lie within k standard deviations of the mean for any distribution with finite variance. That is at least 75% within 2 SD and at least 88.9% within 3 SD. The guarantee is weaker than the empirical rule’s 95% and 99.7% because it must hold for every shape. Use the empirical rule only when normality is defensible; otherwise Chebyshev’s lower bound is the safe statement.

Sample SD is an estimate

The rule is stated for the population σ. With a sample you substitute s, and with small n both the mean and s are uncertain. The percentages remain approximate descriptions, not tests.

Describing data is not describing the mean

The rule covers individual observations. Where a sample mean is concerned, the relevant spread is the standard error, and the distribution is the sampling distribution of the mean, justified by the central limit theorem. A 95% interval for a mean is far narrower than a 95% interval for individual values.

Rounding 95% to 2 SD

Use 1.96 when you need 95% exactly (see the formula section).

How to report it (APA example)

Report the descriptive statistics, then say how you assessed whether the rule applied. In APA 7th edition style, M and SD are italicised and reported to a consistent number of decimals. For the example above:

Reaction times were approximately normally distributed (M = 257.28 ms, SD = 13.11, n = 25). Consistent with the empirical rule, 72% of observations fell within one standard deviation of the mean, 92% within two and 100% within three.

Do not write that the data “follow the empirical rule” as if it were a test; it is a description. If you need a formal check, report a normality test or a Q-Q plot alongside it.

How to calculate it in Excel, R or Python

Each snippet gives the exact coverage for k = 1, 2, 3 from the normal CDF.

Excel

=NORM.S.DIST(1,TRUE)-NORM.S.DIST(-1,TRUE)   returns 0.682689
=NORM.S.DIST(2,TRUE)-NORM.S.DIST(-2,TRUE)   returns 0.954500
=NORM.S.DIST(3,TRUE)-NORM.S.DIST(-3,TRUE)   returns 0.997300

To count the share of your own data in a band, with the mean in B1 and SD in B2: =COUNTIFS(A2:A26,">="&B1-B2,A2:A26,"<="&B1+B2)/COUNT(A2:A26).

R

k <- 1:3
pnorm(k) - pnorm(-k)
# 0.6826895 0.9544997 0.9973002

x <- c(231,238,242,245,247,249,250,251,252,253,254,255,256,
       257,258,259,260,262,264,266,268,271,274,279,291)
mean(abs(x - mean(x)) <= sd(x))   # 0.72

Python

from statistics import NormalDist, mean, stdev
N = NormalDist()
for k in (1, 2, 3):
    print(k, round(N.cdf(k) - N.cdf(-k), 6))   # 0.682689, 0.9545, 0.9973

x = [231,238,242,245,247,249,250,251,252,253,254,255,256,
     257,258,259,260,262,264,266,268,271,274,279,291]
m, s = mean(x), stdev(x)
print(sum(abs(v - m) <= s for v in x) / len(x))   # 0.72

FAQ

What is the empirical rule for a normal distribution?