Mathematics · Stochastics

Sigma Rules (Sigma Intervals)

Q: How do you set up a prediction interval for a success count?

For $n = 100$ and $p = 0.3$, $\mu{} = 30$ and $\sigma{} = \sqrt{21} \approx{} 4.58$; the Laplace condition $\sigma{} > 3$ holds. A continuous 95% normal interval uses radius $1.96\sigma{} \approx{} 8.98$. With a continuity correction, this becomes roughly 21 through 39 whole counts. The exact binomial sum $P(21 \le{} X \le{} 39)$ is about 96.25%; discrete intervals often cannot hit the nominal level exactly. The bounds 22 through 38 would cover only about 93.72% here. For a precise statement, compute the binomial tails directly. The normal approximation is a quick guide, not a guarantee of exactly 95%.

The sigma rules state with which probability a normally distributed quantity lies in the intervals $\mu\pm\sigma$ , $\mu\pm2\sigma$ and $\mu\pm3\sigma$ .

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Formula

P (μ - 2 σ \leq X \leq μ + 2 σ) \approx 95, 4 %

Probabilities dimensionless · μ and σ in the unit of the random variable

Variables & units

Variables & units
Symbol	Meaning	Unit
$μ$	Expected value (centre of the interval)	same as X
$σ$	Standard deviation (half width of the 1σ interval)	same as X
$k$	Interval factor (1, 2, 3 or 1.64 / 1.96 / 2.58)	dimensionless
$P$	Probability of the $k\cdot{}\sigma{}$ interval	dimensionless

LaTeX

P(\mu - 2\sigma \leq X \leq \mu + 2\sigma) \approx 95{,}4\,\%

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Diagram: a bell curve with a highlighted area from μ − 2σ to μ + 2σ; below it three nested double arrows for the 1σ, 2σ and 3σ neighbourhoods with their percentages. — Sigma rules: about 68.3 % of the values lie within μ ± σ, about 95.4 % within μ ± 2σ and about 99.7 % within μ ± 3σ.

Derivation & background

The rules follow from the normal distribution: 68.3 % of the values lie in $\mu\pm\sigma$ , 95.4 % in $\mu\pm2\sigma$ and 99.7 % in $\mu\pm3\sigma$ . For round levels one uses the inverse factors $\mu\pm1.64\sigma$ (90 %), $\mu\pm1.96\sigma$ (95 %) and $\mu\pm2.58\sigma$ (99 %). The rules can be applied to binomial distributions if the Laplace condition $\sigma{} = \sqrt{(n\cdot{}p\cdot{}(1-p))} > 3$ is met; then $\mu{} = n\cdot{}p$ .

Exam blueprint

Validity range

Exact for the normal distribution; admissible as an approximation for binomial distributions when the Laplace condition $\sigma{} = \sqrt{(np(1-p))} > 3$ is met. For integer success counts round the interval bounds.

Derivation steps

Areas under the bell curve between symmetric bounds around μ.

1Standardize: $z =\frac{ (X - \mu{})}{\sigma{}}$ turns every normal distribution into the standard normal distribution.
2The table gives $\Phi{}(1) - \Phi{}(-1) \approx{} 0.683$ , $\Phi{}(2) - \Phi{}(-2) \approx{} 0.954$ , $\Phi{}(3) - \Phi{}(-3) \approx{} 0.997$ .

Rearrangements

Inverse sigma rules

\mu \pm 1{,}64\sigma \; (90\,\%), \quad \mu \pm 1{,}96\sigma \; (95\,\%), \quad \mu \pm 2{,}58\sigma \; (99\,\%)

For prescribed round confidence levels.

Interval radius

r = k \cdot \sigma

Prediction interval [μ − kσ; μ + kσ]; choose k by level.

Binomial parameters

\mu = n p, \quad \sigma = \sqrt{n p (1-p)}

Compute first and check the Laplace condition $\sigma{} > 3$ .

Task variant

$n = 100$ , $p = 0.3$ : give an approximate 95% prediction interval for the success count.

$\mu{} = 30$ , $\sigma{} = \sqrt{21} \approx{} 4.58 > 3$ .
The continuous normal interval μ ± 1.96σ has bounds [21.02; 38.98].
With a continuity correction, 21 through 39 whole counts are plausible; the exact binomial sum covers about 96.25%.
Without the correction, 22 through 38 cover only 93.72%.

IQ values: $\mu{} = 100$ , $\sigma{} = 15$ . What percentage lies between 85 and 115, what percentage above 130?

85 to 115 is the 1σ interval: ≈ 68.3 %.
Above 130 (more than 2σ above μ): $\frac{(100 - 95.4)}{2} \approx{} 2.3 \%$ .

Common mistakes

Confusing the normal 2σ rule (95.45%) with the 95% factor 1.96.

For a normal distribution, $k = 2$ gives about 95.45%; exactly 95% corresponds to $k \approx{} 1.96$ . For discrete binomial data, check coverage separately after rounding.

Applying sigma rules to binomial distributions without the Laplace condition.

First check $\sigma{} = \sqrt{(np(1-p))} > 3$ , otherwise the approximation is too rough.

Mixing one-sided and symmetric questions.

Outside μ ± 2σ lie 4.6 %, above alone only 2.3 %.

Exam context

Prediction intervals and hypothesis tests in stochastics finals, quality control, interpretation of measurement series.

Formula cluster

Applying the normal distribution

Sigma rules translate μ and σ into concrete intervals and test decisions.

f(x) = 1/(σ√2π) · e^(−(x−μ)²/2σ²)

Normal Distribution (Gaussian Bell Curve)

σ = √Var(X)

Standard Deviation

P(X=k) = (n über k)·pᵏ·(1−p)ⁿ⁻ᵏ

Binomial Distribution

Worked example

Binomial with $n = 200$ , $p = 0.5$ : $\mu{} = 100$ and $\sigma{} = \sqrt{(200\cdot{}0.5\cdot{}0.5)} = \sqrt{50} \approx{} 7.07 > 3$ . The 2σ interval $100\pm14.1$ means: about 95.4 % of all success counts lie in [86; 114].

Applications

Prediction intervals in final exams, hypothesis tests (acceptance and rejection region), quality control (Six Sigma), interpretation of IQ and measurement values

How do you calculate with Sigma Rules (Sigma Intervals)?

Here is how to work through a typical Sigma Rules (Sigma Intervals) (μ±σ: 68,3 %, μ±2σ: 95,4 %, μ±3σ: 99,7 %) task step by step:

1
Task
$n = 100$ , $p = 0.3$ : give an approximate 95% prediction interval for the success count.
Solution path
1. $\mu{} = 30$ , $\sigma{} = \sqrt{21} \approx{} 4.58 > 3$ .
2. The continuous normal interval μ ± 1.96σ has bounds [21.02; 38.98].
3. With a continuity correction, 21 through 39 whole counts are plausible; the exact binomial sum covers about 96.25%.
4. Without the correction, 22 through 38 cover only 93.72%.
2
Task
IQ values: $\mu{} = 100$ , $\sigma{} = 15$ . What percentage lies between 85 and 115, what percentage above 130?
Solution path
1. 85 to 115 is the 1σ interval: ≈ 68.3 %.
2. Above 130 (more than 2σ above μ): $\frac{(100 - 95.4)}{2} \approx{} 2.3 \%$ .

Frequently asked questions about Sigma Rules (Sigma Intervals)

What exactly do the sigma rules state?+

They give the share of a normally distributed quantity in symmetric intervals: about 68.27% lies within μ ± σ, 95.45% within μ ± 2σ, and 99.73% within μ ± 3σ. For an IQ model with $\mu{} = 100$ and $\sigma{} = 15$ , about 68.27% lies from 85 to 115 and 95.45% from 70 to 130. The rules apply to every normal distribution because $z =\frac{ (X - \mu{})}{\sigma{}}$ maps it to the same standard normal curve. For binomial counts, they are only approximations: rounding to whole values changes coverage, so exact results require summing binomial probabilities. The rules remain useful for quick graphical or rough estimates.

What is the difference between the 2σ rule and the factor 1.96?+

They are two viewing directions on the same bell curve. The 2σ rule starts from the round factor $k = 2$ and asks for the probability: 95.4 % of the values lie in μ ± 2σ. The factor 1.96 starts conversely from the round level 95 % and asks for the required k: for exactly 95 %, μ ± 1.96σ suffices. Analogously 90 % belongs to 1.64σ and 99 % to 2.58σ. In tasks the wording decides: "2σ interval" means $k = 2$ and 95.4 %; "95 % prediction interval" means $k = 1.96$ . Mixing the two systematically yields slightly wrong intervals. Mnemonic: round k, non-round percentages; non-round k, round percentages.

What is the Laplace condition and why do you need it?+

The sigma rules are exact for normal distributions. Binomial counts are discrete; the Laplace condition $\sigma{} = \sqrt{(n\cdot{}p\cdot{}(1-p))} > 3$ is a rough school-level check for whether a normal approximation is plausible, not a guarantee of exact coverage. For $n = 200$ and $p = 0.5$ , $\sigma{} \approx{} 7.07$ . The integer counts 86 through 114 lie in the rounded 2σ range; their exact binomial probability is about 96.00%, while the continuous normal area is 95.45%. With $n = 20$ and $p = 0.1$ , $\sigma{} \approx{} 1.34$ , so the approximation is especially questionable. For exact tail probabilities, sum the binomial distribution or use its cumulative function.

How do you set up a prediction interval for a success count?+

For $n = 100$ and $p = 0.3$ , $\mu{} = 30$ and $\sigma{} = \sqrt{21} \approx{} 4.58$ ; the Laplace condition $\sigma{} > 3$ holds. A continuous 95% normal interval uses radius $1.96\sigma{} \approx{} 8.98$ . With a continuity correction, this becomes roughly 21 through 39 whole counts. The exact binomial sum $P(21 \le{} X \le{} 39)$ is about 96.25%; discrete intervals often cannot hit the nominal level exactly. The bounds 22 through 38 would cover only about 93.72% here. For a precise statement, compute the binomial tails directly. The normal approximation is a quick guide, not a guarantee of exactly 95%.

How do you use the sigma rules in hypothesis tests?+

The normal approximation first gives μ ± 1.96σ for a two-sided test with $\alpha{} = 5\%$ . Binomial counts are integers, however, so the actual test size depends on which tail values are included. For a fair coin tossed $n = 200$ times, $\mu{} = 100$ and $\sigma{} \approx{} 7.07$ . The rounded range 86 through 114 has exact coverage of about 96.00%, so exclusion is about 4.00%, not exactly 5%. An exact binomial test sets its limits from cumulative tail probabilities. A count of 120 heads is far outside and gives strong evidence against H₀. A result inside the acceptance range does not prove H₀.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, Introductory Statistics 2e, §6.1: The Standard Normal Distribution

More formulas: Probability and statistics

P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}

Bayes' Theorem

E(X) = \mu = \sum_{i} x_{i} \cdot p_{i}

Expected Value

P(A \mid B) = \frac{P(A \cap B)}{P(B)} \quad (P(B) > 0)

Conditional Probability

b = \frac{\sum (x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum (x_{i}-\bar{x})^{2}}, \quad a = \bar{y} - b\bar{x}

Linear Regression (Line of Best Fit)

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Sigma Rules (Sigma Intervals)

The sigma rules state with which probability a normally distributed quantity lies in the intervals

\mu\pm\sigma

\mu\pm2\sigma

and

\mu\pm3\sigma

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

μ

Expected value (centre of the interval)

same as X

σ

Standard deviation (half width of the 1σ interval)

same as X

k

Interval factor (1, 2, 3 or 1.64 / 1.96 / 2.58)

dimensionless

P

Probability of the

k\cdot{}\sigma{}

interval

dimensionless

Derivation & background

Exam blueprint

Validity range

Derivation steps

Areas under the bell curve between symmetric bounds around μ.

1Standardize: $z =\frac{ (X - \mu{})}{\sigma{}}$ turns every normal distribution into the standard normal distribution.
2The table gives $\Phi{}(1) - \Phi{}(-1) \approx{} 0.683$ , $\Phi{}(2) - \Phi{}(-2) \approx{} 0.954$ , $\Phi{}(3) - \Phi{}(-3) \approx{} 0.997$ .

How do you calculate with Sigma Rules (Sigma Intervals)?

Here is how to work through a typical Sigma Rules (Sigma Intervals) (μ±σ: 68,3 %, μ±2σ: 95,4 %, μ±3σ: 99,7 %) task step by step:

Task

$n = 100$ , $p = 0.3$ : give an approximate 95% prediction interval for the success count.

Solution path

$\mu{} = 30$ , $\sigma{} = \sqrt{21} \approx{} 4.58 > 3$ .
The continuous normal interval μ ± 1.96σ has bounds [21.02; 38.98].
With a continuity correction, 21 through 39 whole counts are plausible; the exact binomial sum covers about 96.25%.
Without the correction, 22 through 38 cover only 93.72%.

Task

IQ values: $\mu{} = 100$ , $\sigma{} = 15$ . What percentage lies between 85 and 115, what percentage above 130?

Solution path

85 to 115 is the 1σ interval: ≈ 68.3 %.
Above 130 (more than 2σ above μ): $\frac{(100 - 95.4)}{2} \approx{} 2.3 \%$ .

Frequently asked questions about Sigma Rules (Sigma Intervals)

What exactly do the sigma rules state?+

What is the difference between the 2σ rule and the factor 1.96?+

What is the Laplace condition and why do you need it?+

How do you set up a prediction interval for a success count?+

How do you use the sigma rules in hypothesis tests?+

Retain Sigma Rules (Sigma Intervals) for exams

Create a curated FSRS exam set for μ±σ: 68,3 %, μ±2σ: 95,4 %, μ±3σ: 99,7 %: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential