Weibull distribution calculator
This Weibull distribution calculator evaluates the two-parameter Weibull distribution for a value x and gives the usual reliability figures, from failure probability to B10 life.
Example: Shape parameter k (β) 2 · Scale parameter λ (η) 10 · Value x (e.g. time or load) 5 · Probability for the quantile (%) 10 → Result: F(5) = 0.221199, f(5) = 0.0778801. Source: NIST/SEMATECH e-Handbook of Statistical Methods, 1.3.6.6.8 Weibull Distribution: pdf, cdf, hazard function, percent point function, mean Γ((γ+1)/γ), median ln(2)^(1/γ), mode and standard deviation (read 14.10.2026). Updated: .
How it is calculated
The Weibull distribution has the CDF F(x) = 1 − e^(−(x/λ)^k) and the density f(x) = (k/λ)·(x/λ)^(k−1)·e^(−(x/λ)^k). With k = 2, λ = 10 and x = 5 the probability of failing by x is 0.221199, the density 0.0778801.
Formulas
- Reliability (survival): R(x) = 1 − F(x) = e^(−(x/λ)^k)
- Hazard rate: h(x) = f(x)/R(x) = (k/λ)·(x/λ)^(k−1)
- Mean: λ·Γ(1 + 1/k), median: λ·(ln 2)^(1/k)
- Mode: λ·((k − 1)/k)^(1/k) for k > 1, otherwise 0
- Standard deviation: λ·√(Γ(1 + 2/k) − Γ(1 + 1/k)²)
- Quantile: x_p = λ·(−ln(1 − p))^(1/k); for p = 10 % this is the B10 life.
At x = λ the CDF is always 1 − e^(−1) ≈ 0.632121, whatever k is – that is why λ is called the characteristic life. With k = 1 the Weibull distribution is the exponential distribution, with k = 2 the Rayleigh distribution.
Effect of the shape parameter (λ = 10, x = 5)
| k | F(5) | Mean | Median | B10 |
|---|---|---|---|---|
| 0.5 | 0.506931 | 20 | 4.80453 | 0.111008 |
| 1 | 0.393469 | 10 | 6.931472 | 1.053605 |
| 2 | 0.221199 | 8.862269 | 8.325546 | 3.245928 |
| 3.5 | 0.084595 | 8.997472 | 9.005785 | 5.257328 |
A small k spreads the failures widely: with k = 0.5 half of all units fail before 4.80453 although the mean is 20. A large k concentrates them around λ. Related: normal distribution and Poisson distribution.
Limits
The tool uses the two-parameter form (location μ = 0). It evaluates the distribution for known parameters; it does not fit k and λ to failure data.
Frequently asked questions
What do k and λ mean in the Weibull distribution?
k is the shape: below 1 the failure rate falls, at 1 it is constant, above 1 it rises. λ is the scale or characteristic life: by x = λ, 63.2 % have failed.
How do I calculate the Weibull CDF?
F(x) = 1 − e^(−(x/λ)^k). For k = 2, λ = 10, x = 5: 1 − e^(−0.25) ≈ 0.221199.
What is the mean of a Weibull distribution?
λ·Γ(1 + 1/k). For k = 2 and λ = 10 that is 10·Γ(1.5) ≈ 8.862269.
What is the B10 life?
The time by which 10 % of the units have failed: λ·(−ln 0.9)^(1/k). For k = 2, λ = 10 it is about 3.245928.
Is the exponential distribution a Weibull distribution?
Yes, with k = 1 the Weibull distribution becomes the exponential distribution with mean λ.
Sources and legal basis
- NIST/SEMATECH e-Handbook of Statistical Methods, 1.3.6.6.8 Weibull Distribution: pdf, cdf, hazard function, percent point function, mean Γ((γ+1)/γ), median ln(2)^(1/γ), mode and standard deviation (read 14.10.2026)
- NIST Digital Library of Mathematical Functions §5.2: gamma function Γ(z) used for mean and standard deviation (read 14.10.2026)
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