Sigma Rules (Sigma Intervals)

They state which share of a normally distributed quantity lies in symmetric intervals around the expected value: about 68.3 % of all values lie in μ ± σ, about 95.4 % in μ ± 2σ and about 99.7 % in μ ± 3σ. Example IQ scale (μ = 100, σ = 15): 68.3 % of values lie between 85 and 115, 95.4 % between 70 and 130, almost all between 55 and 145. The rules hold for every normal distribution, whatever μ and σ are, because the standardization z = (X − μ)/σ maps all of them onto the same bell curve. They are the quick tool for estimating probabilities and prediction intervals without a table or calculator.

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.

The sigma rules hold exactly only for the normal distribution. Exam success counts, however, are usually binomially distributed, i.e. discrete and, for small n or extreme p, skewed. The Laplace condition σ = √(n·p·(1−p)) > 3 checks whether the bell curve approximates the binomial distribution well enough; only then may you apply the sigma rules. Example: n = 200, p = 0.5 gives σ = √50 ≈ 7.07 > 3 ✓, so the 2σ interval [86; 114] really carries about 95.4 %. In contrast n = 20, p = 0.1 gives only σ = √1.8 ≈ 1.34; here the approximation would be badly wrong, and you compute the probabilities directly with the binomial distribution (cumulative table or calculator).

In four steps using n = 100, p = 0.3, level 95 %. First the parameters: μ = n·p = 30 and σ = √(n·p·(1−p)) = √21 ≈ 4.58. Second check the Laplace condition: 4.58 > 3 ✓. Third the radius: k·σ with k = 1.96, so 1.96·4.58 ≈ 8.98. Fourth form the interval: [30 − 8.98; 30 + 8.98] = [21.02; 38.98]; as whole success counts 22 to 38. Interpretation: with about 95 % probability the success count lies in this range; in about 5 % of cases it still falls outside, which is no contradiction. For 90 % take k = 1.64, for 99 % k = 2.58, for the quick 2σ estimate k = 2.

The sigma interval becomes the acceptance region of the null hypothesis. Idea: if H₀ holds (say p = 0.5), the success count lies within μ ± 1.96σ with 95 % probability; results outside are so unlikely under H₀ (5 % combined) that H₀ is rejected. Example: coin, n = 200 tosses: μ = 100, σ ≈ 7.07, acceptance region [86; 114]. If someone throws heads 120 times, that lies outside, and fairness is doubted at the 5 % level. Two subtleties matter: in a one-sided test the entire rejection region lies on one side (k = 1.64 instead of 1.96 for 5 %), and a result inside the acceptance region does not PROVE H₀, it merely fails to refute it.