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Instrument MI-07-191 · Statistics

Weibull Distribution Calculator

Does a part fail randomly, or does it wear out with age? The Weibull distribution's shape parameter is what lets a single model capture either behavior — or anything in between.

Instrument MI-07-191
Sheet 1 OF 1
Rev A
Verified
Type 07 — Probability Distributions SER. 2026-07191

Probability density f(x)

0.735759

f(x) = (k/lambda) x (x/lambda)^(k-1) x e^(-(x/lambda)^k)

0.632121 Cumulative probability F(x) = P(X ≤ x)
The working Every figure verified twice
  1. pdf = 2 ⁄ 1·pow(1 ⁄ 1, 2 − 1)·exp(0 − pow(1 ⁄ 1, 2)) = 0.735759
  2. cdf = 1 − exp(0 − pow(1 ⁄ 1, 2)) = 0.632121
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

The Weibull distribution is the workhorse of reliability engineering: it models the time until a component or system fails, and it does so far more flexibly than the exponential distribution because it adds a shape parameter, k, on top of a scale parameter, λ. That shape parameter is what makes it powerful — by adjusting k alone, the same formula can describe early-life 'infant mortality' failures, random constant-rate failures, or gradual wear-out failures that become more likely as a part ages.

The density function, f(x) = (k/λ) × (x/λ)^(k−1) × e^(−(x/λ)^k), and cumulative function, F(x) = 1 − e^(−(x/λ)^k), both reduce to simpler, well-known forms at specific values of k. When k = 1, the (x/λ)^(k−1) term becomes 1 and the formula collapses exactly to the exponential distribution with rate 1/λ — constant failure rate, no memory of age. When k = 2, it becomes exactly the Rayleigh distribution.

The value of k relative to 1 tells the reliability story directly: k less than 1 means the failure rate decreases over time (early failures are most likely — think manufacturing defects that show up fast, after which surviving units are increasingly robust); k equal to 1 means a constant failure rate (no aging effect, purely random failures); and k greater than 1 means the failure rate increases over time (true wear-out, like mechanical fatigue or corrosion, where age itself raises the risk). This single-parameter flexibility is why the Weibull distribution, rather than a fixed formula, is the standard choice in reliability and survival analysis.

f(x)=kλ(xλ)k1e(x/λ)kf(x) = \frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^{k}}F(x)=1e(x/λ)kF(x) = 1-e^{-(x/\lambda)^{k}}
k — shape parameter, governs whether failure risk decreases (k < 1), stays constant (k = 1), or increases (k > 1) with age, k > 0 · λ — scale parameter, the characteristic life, λ > 0 · x — the time or value being evaluated, x ≥ 0 · e — Euler's number, ≈ 2.71828.
  • Enter the failure-rate behavior into Shape parameter (k) — under 1 for decreasing risk, 1 for constant risk, over 1 for increasing risk with age.
  • Enter the characteristic life into Scale parameter (λ) — roughly the age by which about 63.2% of units have failed, in whatever time unit you're using.
  • Enter the time or value you want to evaluate into x (value, x ≥ 0).
  • Read Probability density f(x) and Cumulative probability F(x) = P(X ≤ x) — the density and the actual failure probability by time x.
  • Set k = 1 to reproduce exponential-distribution behavior, or k = 2 to reproduce Rayleigh-distribution behavior, using this same instrument.

Worked example — shape k = 2, scale λ = 1, evaluated at x = 1

Enter k = 2 into Shape parameter, λ = 1 into Scale parameter, and x = 1 into x — a component with an increasing failure rate (wear-out behavior, since k > 1), evaluated at x equal to the scale parameter itself. The instrument computes (x/λ)^k = 1² = 1, then e^(−1) ≈ 0.36787944.

Probability density f(x) reads 0.735758882343, computed as (2/1) × 1^1 × 0.36787944. Cumulative probability F(x) reads 0.632120558829, computed as 1 − 0.36787944. So there's about a 63.2% chance this component has already failed by x = 1 — and that same 63.2% figure at x = λ holds for every Weibull distribution regardless of k, since (λ/λ)^k always equals 1.

Questions

How is this related to the exponential distribution?

The exponential distribution is the special case of the Weibull distribution at k = 1 exactly, corresponding to a constant failure rate. This calculator's own second built-in check case (k = 1, λ = 2, x = 2) reduces to a PDF of 0.183939720586 and CDF of 0.632120558829 — identical to what this site's Exponential Distribution calculator gives for rate 1/λ = 0.5 at the same x. Use the exponential-specific calculator when you know the failure rate is truly constant with age; use this Weibull calculator when the failure rate might change over time.

How is this related to the Rayleigh distribution?

The Rayleigh distribution is the special case of the Weibull distribution at k = 2 exactly. This site's Rayleigh Distribution calculator uses a scale parameter σ that relates to this instrument's λ by λ = σ√2, since the two distributions write their exponents slightly differently. Use the Rayleigh-specific calculator when working with 2D-magnitude problems like wind speed or signal envelopes, where that σ parameterization and its mean formula are the natural fit.

What does the shape parameter k actually tell me?

It tells you how failure risk changes with age. k < 1 means the failure rate decreases over time — common in 'infant mortality' failures from manufacturing defects, where surviving units get progressively more reliable. k = 1 means a constant failure rate, with no memory of age at all. k > 1 means the failure rate increases over time — true wear-out, like fatigue cracking or bearing wear, where the risk of failure genuinely climbs the longer a part has been in service.

What does the scale parameter λ represent physically?

λ is the 'characteristic life' of the distribution — the point by which about 63.2% of units have failed, for any value of k. This comes directly from the CDF formula: F(λ) = 1 − e^(−(λ/λ)^k) = 1 − e^(−1) ≈ 0.632, regardless of k. It's a natural reference age for comparing different Weibull-distributed populations even when their k values differ.

Can k and λ be less than or equal to zero?

No — both k and λ must be strictly greater than zero. A shape parameter of zero or less has no meaningful interpretation in the formula (the exponent (x/λ)^(k−1) becomes undefined or nonsensical), and a scale parameter of zero or less would mean a characteristic life of zero or a negative timespan, neither of which makes physical sense for a failure-time model.

Where is the Weibull distribution used outside of reliability engineering?

Its flexible shape makes it a common choice for wind-speed modeling (estimating potential energy output at a wind farm site), survival analysis in medical research (time until an event like relapse or recovery), material strength and fatigue-life studies, and extreme-value modeling in fields like hydrology and insurance, wherever a flexible, positively-skewed, non-negative distribution with an adjustable hazard shape is needed.

References