Audited 31 Jul 2026·Last updated 28 Oct 2026·3 citations·Tier 3·0 uses

Weibull Distribution Calculator

Weibull CDF and density from shape k, scale λ and a value x: the share of a population failed by time x, used for component life, wind speed and strength.

Weibull Distribution Calculator

Cumulative probability
0.3717
F(x), the probability that a value is at or below x: the fraction of the population expected to have failed by x, from 0 to 1.
Probability density
0.0073
Model scope
Two-parameter Weibull with zero location; it does not estimate parameters, test fit, include censoring or competing risks, or justify reliability extrapolation.

Background.

Things rarely fail at a steady rate. Electronics that survive their first weeks often run for years, while bearings, belts and tyres last well and then wear out in a cluster. The Weibull distribution became the workhorse of reliability engineering because a single shape parameter can describe all of those patterns, and because it is simple enough to work out by hand.

Two numbers define it. The scale λ places the failures on the time axis, and it has a tidy meaning of its own: by x = λ, 63.2% of the population has failed, whatever the shape. The shape k governs how the risk of failure changes with age. Below 1 the risk falls over time, the signature of early defects being weeded out. At exactly 1 it is constant, and Weibull reduces to the exponential distribution, in which an old part is as good as a new one. Above 1 the risk climbs with age, which is wear-out.

Engineers in reliability are not the only users. Wind-energy studies commonly describe the spread of wind speeds at a site with a Weibull curve, and the strength of brittle materials such as ceramics is summarised by a Weibull modulus, the same k under another name: the higher it is, the more consistent the pieces.

You supply k, λ and a value x, and get back the cumulative probability F(x), the share expected to have failed, or to fall at or below x, together with the probability density there. This is the two-parameter form, which assumes failures can start at zero. It does not fit k and λ to your data, allow for units still running when a test ended, or judge whether a Weibull model suits your data in the first place.

What is weibull distribution calculator?

The Weibull distribution is a continuous probability distribution for positive quantities such as time to failure, wind speed or breaking strength. In the two-parameter form, the probability that a value is at most x is F(x) = 1 − exp[−(x/λ)^k], with shape k > 0 and scale λ > 0. Its complement, 1 − F(x), is the reliability or survival function, the share still working at x. The density f(x) measures how concentrated the probability is near x, per unit of x.

How to use this calculator.

  1. Enter the shape k from a fitted model, a supplier datasheet or a published study. Reliability software and Weibull probability plots report it, often under the symbol β.
  2. Enter the scale λ in the same unit as x, whether hours, cycles, kilometres or metres per second. Some sources call it the characteristic life and write it as η.
  3. Enter the value x you care about, such as a warranty period, a service interval or a design wind speed.
  4. Read the cumulative probability as the fraction failed, or at or below x, by that point, and subtract it from 1 for the fraction still running. The density is per unit of x and is mostly useful for plotting or for working out the hazard rate.

The formula.

F(x)=1-exp[-(x/λ)^k]

F(x) = 1 − exp[−(x/λ)^k] and f(x) = (k/λ)(x/λ)^(k−1) exp[−(x/λ)^k]. Dividing x by λ makes the argument dimensionless, so λ only stretches or squeezes the curve along the axis. Raising the ratio to the power k bends it: when x is below λ, a large k makes (x/λ)^k very small, so few failures happen early and the curve then rises steeply around λ. At x = λ the ratio is 1, so F = 1 − e⁻¹ ≈ 0.632 for every k. Dividing the density by the survival share gives the hazard rate, (k/λ)(x/λ)^(k−1), the instantaneous failure rate among units still running: it falls with x when k < 1, stays at 1/λ when k = 1 and rises when k > 1. Shape and scale must be positive and x cannot be negative. When k is below 1 the density grows without limit as x approaches zero, so it has no finite value at x = 0.

A worked example.

Example

Take a component whose life follows a Weibull curve with shape k = 1.5 and scale λ = 100 hours, and ask about x = 60 hours. The ratio x/λ is 0.6, and 0.6^1.5 = 0.4648. The survival share is exp(−0.4648) = 0.6283, so the cumulative probability is 1 − 0.6283 = 0.3717: about 37.2% of units are expected to fail within 60 hours. The density is (1.5 ÷ 100) × 0.6^0.5 × 0.6283 = 0.015 × 0.7746 × 0.6283 ≈ 0.0073 per hour. Because k is above 1, the risk is rising: density divided by survival gives a hazard of 0.0116 per hour at 60 hours. Had k been exactly 1, the same 60 hours would already have claimed 45.1% of units.

x Value60
scale Lambda100
shape K1.5

Frequently asked questions.

My datasheet gives β and η. Which is which?
In reliability notation β is the shape, k here, and η is the scale, λ here. Check the formula your source prints, though. Some texts write the scale as a rate that multiplies x, or fold it into x^k divided by a constant, and those values need converting before they mean the same as λ.
How do I find the time by which 10% of units will have failed?
Rearrange the formula: x = λ × (−ln(1 − p))^(1/k), where p is the failure fraction. With k = 1.5 and λ = 100 hours, the 10% point, often called the B10 life, is about 22.3 hours, and the median life, where p is one half, is about 78.3 hours.
Why is the mean life not equal to λ?
The mean is λ × Γ(1 + 1/k), where Γ is the gamma function. It equals λ only when k = 1. For k = 1.5 and λ = 100 hours, Γ(1.667) is about 0.9027, so the mean life is about 90.27 hours, a little below the characteristic life.
What if no failures can happen before a certain age?
That is the three-parameter Weibull, which adds a location γ, a failure-free period. Its curve is the two-parameter one shifted along the axis, so subtract γ from your x before entering it. If x is below γ, the probability of failure is zero.
Can I use this with test data where some units never failed?
Not to estimate the parameters. Units still running when a test stops are censored observations, and fitting k and λ properly needs maximum-likelihood software or a Weibull plot that accounts for them. Once you have trustworthy estimates, this page evaluates the resulting curve at any x.
What does a density of 0.0073 actually mean?
It is probability per unit of x. Multiplied by a short interval it approximates the chance of a value falling in that interval, so with x in hours, roughly 0.73% of all units are expected to fail during the hour around the 60-hour mark.

How this page was produced

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Quanta Calculator
Primary sources
3 cited below
Method
F(x)=1-exp[-(x/λ)^k]
Published
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