Statistical Distributions in Six Sigma
A statistical distribution describes how likely each value of a variable is. In Six Sigma the choice matters because capability, control limits, sample sizes, and hypothesis tests all assume one. Use the normal for most continuous measurements, the binomial for counts of defective units, and the Poisson for counts of defects.
For times and life data, use the exponential, Weibull, or lognormal; for tests and confidence intervals, use the t, chi-square, and F sampling distributions.
Which distribution fits your data?
| Distribution | Type | Typical Six Sigma use | Parameters | Mean / variance |
|---|---|---|---|---|
| Normal | Continuous | Dimensions, weights, many process outputs; basis of Cp/Cpk and X̄ charts | μ, σ | μ / σ² |
| Binomial | Discrete | Number of defective units in a sample of n; p and np charts | n, p | np / np(1 − p) |
| Poisson | Discrete | Defects per unit, per area, or per time; c and u charts | λ | λ / λ |
| Hypergeometric | Discrete | Sampling without replacement from a small finite lot | N, K, n | nK/N / — |
| Uniform | Continuous | Every value in a range equally likely; rounding error, random start points | a, b | (a + b)/2 / (b − a)²/12 |
| Exponential | Continuous | Time between random events; constant failure rate | λ (or mean θ = 1/λ) | 1/λ / 1/λ² |
| Lognormal | Continuous | Right-skewed positive data, such as repair or cycle times | μ, σ of ln(x) | — |
| Weibull | Continuous | Reliability and life data; shape < 1 infant mortality, = 1 random, > 1 wear-out | shape β, scale η | — |
| t, chi-square, F | Sampling | Tests and intervals for means, variances, and ratios of variances | degrees of freedom | — |
Look-up tables for the sampling distributions are on this site: z, t, chi-square, F, and Poisson.
Worked examples
- Binomial. A process makes 5% defectives. In a sample of 20, P(no defectives) = 0.9520 = 0.358, and P(at most one) = 0.736. Seeing zero defectives in 20 is weak proof of a good process.
- Poisson. A panel averages 2 paint defects. P(a panel has none) = e−2 = 0.135, and P(4 or more) = 0.143.
- Hypergeometric. A lot of 50 contains 3 defectives, and you inspect 5 without replacement. P(the sample shows none) = 0.724, so most samples would miss them.
- Normal. If a characteristic is normal and the upper limit is 2.5σ above the mean, the fraction above the limit is P(Z > 2.5) = 0.0062, or about 6,210 ppm. See process capability.
- Exponential. With a mean time between failures of 500 hours, P(running 100 hours without failure) = e−100/500 = 0.819.
- Weibull. With shape 1.5 and scale 2,000 hours, reliability at 1,000 hours = e−(1000/2000)1.5 = 0.702.
Distribution profiles
Normal distribution
Symmetric and bell-shaped, defined by its mean μ and standard deviation σ. About 68.27% of values fall within ±1σ, 95.45% within ±2σ, and 99.73% within ±3σ. Any normal value converts to a standard score z = (x − μ)/σ, which you look up in a z-table. It underlies standard capability indices, X̄ and individuals charts, and t-tests.
Binomial distribution
The number of defective units x in n independent trials, each with the same probability p: P(x) = C(n, x) px(1 − p)n − x. Mean np, variance np(1 − p). Use it for pass/fail data with a fixed sample size, as on p and np charts and in acceptance sampling with replacement or large lots. With n = 20 and p = 0.05, P(0) = 0.358.
Poisson distribution
The number of events in a fixed interval of time, area, or units when events occur independently at a constant average rate λ: P(x) = e−λλx/x!. Mean and variance both equal λ. Use it for defects per unit (c and u charts), arrivals, or calls per hour. Probabilities are tabulated on the Poisson table and cumulative Poisson table.
Hypergeometric distribution
The number of defectives x in a sample of n drawn without replacement from a lot of N items that contains K defectives: P(x) = C(K, x) C(N − K, n − x) / C(N, n). Mean nK/N. Use it for small, finite lots where each draw changes the odds. In the lot of 50 with 3 defectives, a sample of 5 expects 0.3 defectives and shows none 72.4% of the time.
Uniform distribution
Every value between a and b is equally likely, with density 1/(b − a). Mean (a + b)/2, variance (b − a)²/12. For a uniform between 2 and 8, the mean is 5, the standard deviation is 1.73, and P(x < 3.5) = 1.5/6 = 0.25. It models rounding error, random start times, and random-number generation for simulation and sampling.
Exponential distribution
The time between independent random events with constant rate λ: f(t) = λe−λt, reliability R(t) = e−λt, mean 1/λ, variance 1/λ². It is memoryless: a unit that has run 400 hours is no more likely to fail in the next hour than a new one. With a 500-hour mean, R(100) = 0.819 and the median life is 500 × ln 2 ≈ 347 hours. It is the Weibull with shape 1.
Lognormal distribution
A positive, right-skewed variable whose natural logarithm is normal. If ln(x) has mean 2.0 and standard deviation 0.5, the median is e2.0 ≈ 7.39, the mean is e2.0 + 0.5²/2 ≈ 8.37, and P(x > 15) ≈ 0.078. Repair times, cycle times, particle sizes, and some fatigue lives behave this way. To analyze it, take logs, use normal methods, and transform back.
Weibull distribution
A flexible life distribution with shape β and scale η: R(t) = e−(t/η)β. β < 1 means a falling failure rate (infant mortality), β = 1 a constant rate (the exponential), and β > 1 a rising rate (wear-out). The scale η is the characteristic life, the age by which 63.2% have failed. A three-parameter form adds a location γ, a failure-free period before which no failures occur: R(t) = e−((t − γ)/η)β for t > γ. A one-parameter form fixes β from experience so that small data sets can still estimate η. With β = 1.5 and η = 2,000 hours, R(1,000) = 0.702 and the B10 life (10% failed) is about 446 hours.
Student’s t distribution
The distribution of (x̄ − μ)/(s/√n) when σ is estimated from the sample. It is symmetric like the normal but has heavier tails, set by the degrees of freedom (n − 1 for one sample). The two-sided 95% critical value is 2.776 at 4 df, 2.228 at 10, and 2.042 at 30, approaching the normal’s 1.960. Use it for confidence intervals and t-tests on means; values are on the t-table. The chi-square and F distributions play the same role for variances.
The central limit theorem
The central limit theorem says that the distribution of sample averages approaches a normal distribution as the sample size n grows, whatever the shape of the individual values, provided they are independent and have a finite variance. The average of the means equals the population mean μ, and their standard deviation, the standard error, is σ/√n.
| Sample size n | Standard error (σ = 1) | Skewness of the averages |
|---|---|---|
| 1 (individual exponential values) | 1.000 | 2.00 (strongly skewed) |
| 5 | 0.447 | 0.89 |
| 30 | 0.183 | 0.37 (close to symmetric) |
The table uses a strongly skewed exponential population (skewness 2). The skewness of an average falls as 2/√n; a simulation of 100,000 averages of five values gave a standard error of 0.447 and a skewness of 0.91, matching theory. Six Sigma relies on this in three places: X̄ charts with subgroups of 4–5 behave close to normal; t-tests and confidence intervals on means are robust to moderate skew in larger samples; and sample-size formulas for means use σ/√n. The theorem does not make individual values normal, so capability and ppm estimates for a skewed characteristic still need the right distribution or a transformation.
Checking the shape before you analyze
- Plot the data first with a histogram and a normal probability plot.
- Run a normality test such as Anderson-Darling. With very large samples, tiny departures become significant, so judge the plot as well.
- Ask whether the data are actually a mixture, for example two machines or shifts. A bimodal shape usually means stratify, not transform.
- If the data are truly skewed, fit a better distribution (lognormal, Weibull) or transform (Box-Cox), or use nonparametric tests.
Frequently asked questions
What is the difference between the binomial and Poisson distributions?
The binomial counts defective units out of a fixed number n, where each unit is either good or bad. The Poisson counts defects, where one unit can have several and there is no fixed upper limit. The Poisson also approximates the binomial when n is large and p is small.
When should I use the hypergeometric instead of the binomial?
Use it when you sample without replacement from a small, finite lot, so each draw changes the odds for the next. When the sample is less than about 10% of the lot, the binomial is a close approximation.
Does Six Sigma require normal data?
No. Many tools assume normality, such as standard capability indices and t-tests on small samples, but non-normal data can be analyzed with the right distribution, a transformation, or nonparametric methods. Check the shape before choosing the tool.
Why is the Weibull distribution used in reliability?
Its shape parameter lets one model describe decreasing, constant, or increasing failure rates, so a fitted Weibull tells you whether failures are early-life, random, or wear-out, which points to different fixes.
Related guides
Last updated: · Published by Lean Six Sigma Training Guide (sixsigmatrainingfree.com)