Weibull Distribution: Parameterizations and Quantiles

Introduction

The Weibull distribution is a flexible model for nonnegative measurements, including lifetimes and failure times. Its two parameters have distinct roles: the shape parameter changes how probability is distributed across the positive axis, while the scale parameter sets the measurement scale. A reliable calculation begins by identifying the parameterization, because a quantity called a scale in one formula may appear as a power-rate coefficient in another.

This tutorial develops the shape-scale form, obtains its cumulative distribution function (CDF), density, and quantiles, and uses them to calculate interval probabilities. The emphasis is on translating formulas and using them correctly; moments and detailed comparisons of how shape changes the distribution are separate questions.

Definition and Basic Formulas

Definition: A random variable $X$ has a Weibull distribution with shape $k>0$ and scale $\lambda>0$, written $X\sim\operatorname{Weibull}(k,\lambda)$, if its CDF is $$ F(x)= \begin{cases} 0, & x<0,\\ 1-\exp\!\left[-(x/\lambda)^k\right], & x\geq 0. \end{cases} $$ Its support is $[0,\infty)$. Its density, for $x>0$, is $$ f(x)=\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1} \exp\!\left[-(x/\lambda)^k\right]. $$

The scale $\lambda$ has the same units as $X$, so $x/\lambda$ is dimensionless. The shape $k$ is dimensionless and must be positive. At $x=\lambda$, the CDF is $1-e^{-1}$, regardless of the value of $k$; this makes $\lambda$ a useful reference point, but not generally a quantile at a familiar probability such as one-half.

The survival function, the probability of exceeding $x$, follows by taking the complement of the CDF:

P(X>x)=1-F(x)=\exp\!\left[-(x/\lambda)^k\right],\qquad x\geq 0.

Because this distribution is continuous, $P(X=x)=0$ for every fixed $x$. Consequently, for $0\leq a\leq b$, it makes no difference whether either endpoint is included when calculating the probability between $a$ and $b$.

Deriving the Density and Checking Normalization

The density can be obtained by differentiating the CDF on the interior of its support. The calculation also verifies that the CDF rises rather than falls as $x$ increases.

Differentiate the exponent.
For $x>0$, $(x/\lambda)^k=x^k/\lambda^k$, so its derivative is $kx^{k-1}/\lambda^k=(k/\lambda)(x/\lambda)^{k-1}$.

Apply the chain rule.
Differentiating $1-\exp[-(x/\lambda)^k]$ gives the positive density stated above.

Check total probability.
Integrating the derivative from $0$ to $R$ gives $F(R)-F(0)=1-\exp[-(R/\lambda)^k]$. As $R\to\infty$, this tends to $1$, since $k,\lambda>0$. Thus the density integrates to $1$ over the support.

The density formula is stated for $x>0$. Depending on $k$, its limiting behavior as $x$ approaches zero can differ; a density's value at a single point does not affect probabilities or its integral. The CDF definition remains valid at $x=0$, where $F(0)=0$.

Parameterizations: Scale Form and Power-Rate Form

A common source of error is seeing the same Weibull law written with different-looking constants. In the shape-scale form used here, the exponent is $-(x/\lambda)^k$. Another common form writes that exponent as $-c x^k$, where $c>0$. Matching the exponents gives the conversion:

c=\lambda^{-k} \qquad\text{and}\qquad \lambda=c^{-1/k}.

Thus the power-rate form has CDF $1-\exp(-cx^k)$ for $x\geq0$. Its coefficient $c$ is not the scale $\lambda$: it has units of $X^{-k}$. When translating a formula, match the entire exponent, including the power, rather than identifying parameter names by sight.

Some references use different letters for shape and scale, and some software calls the shape parameter “scale” or uses a separate location parameter. The names alone do not determine the model. Check the formula for the CDF or survival function, identify the support, and then translate the constants into the shape-scale form before calculating.

Worked Example: Identifying Parameters and Finding Quantiles

Suppose $X$ has CDF $F(x)=1-\exp[-(x/3)^2]$ for $x\geq0$, and $F(x)=0$ for $x<0$. Identify its parameters, calculate $F(3)$, and find its median and 90th percentile.

Parameters. Comparing the exponent with $-(x/\lambda)^k$ gives shape $k=2$ and scale $\lambda=3$. The support is $[0,\infty)$.

At $x=3$, $(3/3)^2=1$, so $$ F(3)=1-e^{-1}\approx 1-0.3679=0.6321. $$

The quantile formula derived below gives $q_p=\lambda[-\log(1-p)]^{1/k}$. For the median, $p=0.5$: $$ q_{0.5}=3[-\log(0.5)]^{1/2} =3\sqrt{\log 2} \approx 3(0.8326)=2.498. $$

For the 90th percentile, $p=0.9$: $$ q_{0.9}=3[-\log(0.1)]^{1/2} =3\sqrt{\log 10} \approx 3(1.5174)=4.552. $$

The median is below $3$, consistent with $F(3)\approx0.6321>0.5$. The 90th percentile is above $3$, consistent with $F(3)<0.9$.

Quantile Function

For a continuous distribution whose CDF is strictly increasing on its support, the $p$-quantile for $0<p<1$ is the unique $x$ satisfying $F(x)=p$. The Weibull CDF is strictly increasing for $x>0$, because its density is positive there. Solving the CDF equation therefore gives the quantile directly.

Theorem: If $X\sim\operatorname{Weibull}(k,\lambda)$ with $k,\lambda>0$, its quantile function for $0<p<1$ is $$ q_p=F^{-1}(p)=\lambda[-\log(1-p)]^{1/k}. $$ With the endpoint convention $q_0=0$ and $q_1=+\infty$, this formula extends by limits to $p=0$ and $p=1$.

To verify the inverse, substitute $q_p$ into the CDF. Since $q_p/\lambda=[-\log(1-p)]^{1/k}$, raising to $k$ gives $-\log(1-p)$. Hence $$ F(q_p)=1-\exp\{\log(1-p)\}=p. $$ Strict increase of $F$ on the positive axis ensures there is no other solution. At the endpoints, $-\log(1-p)$ tends to $0$ as $p$ decreases to $0$, and tends to infinity as $p$ increases to $1$; these give the stated limiting quantiles.

An equivalent expression uses the survival probability $s=1-p$. The point exceeded with probability $s$, for $0<s<1$, is $\lambda[-\log(s)]^{1/k}$. This form is particularly convenient when a problem asks for an upper-tail probability or threshold.

Worked Example: Converting a Power-Rate Parameterization

A model is specified by $F(x)=1-\exp(-x^2/16)$ for $x\geq0$. Write it in shape-scale form, then calculate the probability that $X$ exceeds $6$ and the probability that $3<X\leq5$.

The exponent is $-c x^k$ with $k=2$ and $c=1/16$. Since $\lambda=c^{-1/k}$, the scale is $\lambda=(1/16)^{-1/2}=4$. Thus this is $\operatorname{Weibull}(2,4)$, because $(x/4)^2=x^2/16$.

For the upper-tail probability, $$ P(X>6)=\exp[-(6/4)^2] =\exp(-2.25)\approx0.1054. $$

For the interval, subtract the CDF at the lower endpoint from the CDF at the upper endpoint: $$ \begin{aligned} P(3<X\leq5) &=F(5)-F(3)\\ &=\left[1-\exp(-25/16)\right] -\left[1-\exp(-9/16)\right]\\ &=\exp(-9/16)-\exp(-25/16)\\ &\approx0.5698-0.2096=0.3602. \end{aligned} $$

The result is positive, as it must be for an interval probability, and it is less than $1$. The constants $1/16$ and $4$ describe the same model in different parameterizations; treating $1/16$ as the scale would produce an incorrect exponent.

Calculating Interval Probabilities

For any $0\leq a\leq b$, a direct CDF subtraction yields a useful general formula:

P(a<X\leq b)=F(b)-F(a) =\exp[-(a/\lambda)^k]-\exp[-(b/\lambda)^k].

This formula also covers $a=0$, because the first exponential is then $1$. For a one-sided upper-tail event, use $P(X>b)=\exp[-(b/\lambda)^k]$; for a lower-tail event $X\leq b$, use $1-\exp[-(b/\lambda)^k]$. Keeping the event and its complement distinct helps avoid subtracting in the wrong order.

Worked Example: Probability Between Two Measurements

Let $X\sim\operatorname{Weibull}(1.5,4)$, where the scale and the measured values use the same units. Find $P(2<X\leq5)$, and state the probability of exceeding $5$.

Apply the interval formula with $a=2$, $b=5$, $k=1.5$, and $\lambda=4$: $$ P(2<X\leq5) =\exp[-(2/4)^{1.5}]-\exp[-(5/4)^{1.5}]. $$

The two exponents are $$ (2/4)^{1.5}=0.5^{1.5}\approx0.3536, \qquad (5/4)^{1.5}=1.25^{1.5}\approx1.3975. $$

Therefore, $$ P(2<X\leq5)\approx e^{-0.3536}-e^{-1.3975} \approx0.7022-0.2472=0.4550. $$

The upper-tail probability at $5$ is the second survival probability: $$ P(X>5)=\exp[-(5/4)^{1.5}] \approx0.2472. $$

As a check, $P(X\leq5)=1-0.2472=0.7528$, while $P(X\leq2)=1-0.7022=0.2978$. Their difference is $0.7528-0.2978=0.4550$, agreeing with the interval calculation.

Common Parameterization and Probability Pitfalls

Interchanging shape and scale. In $(x/\lambda)^k$, $k$ is the exponent and $\lambda$ is the quantity dividing $x$. Identify their roles from the formula rather than from the order of arguments in a source.
Using the power-rate coefficient as a scale. If the exponent is $-cx^k$, then $\lambda=c^{-1/k}$, not $c$. Check by substituting the converted parameters back into the exponent.
Using the wrong tail. $F(x)$ is a lower-tail probability; the probability of exceeding $x$ is $1-F(x)=\exp[-(x/\lambda)^k]$.
Reversing the interval subtraction. For $a\leq b$, use $F(b)-F(a)$. Reversing the order gives a negative number, which cannot be a probability.
Applying the quantile formula at the wrong probability. A 90th percentile uses $p=0.9$, giving $-\log(0.1)$, not $-\log(0.9)$. The latter corresponds to a different tail probability.

These formulas make the Weibull distribution straightforward to use once its parameterization is explicit. Its CDF converts a threshold into a probability, its quantile function reverses that calculation, and CDF differences give probabilities for finite intervals. In every case, checking the exponent and whether the event is a lower tail, upper tail, or interval is a dependable final check.

Check Your Understanding

Try these questions using the shape-scale form and show the parameter conversions when needed.

For $X\sim\operatorname{Weibull}(3,2)$, write its CDF and density, including the support.
A Weibull model has CDF $1-\exp(-5x^2)$ for $x\geq0$. Find its shape and scale parameters.
Derive $q_p$ by solving $F(q_p)=p$, and explain why the solution is unique for $0<p<1$.
For $X\sim\operatorname{Weibull}(2,5)$, express $P(1<X\leq4)$ exactly using exponentials.
If $X\sim\operatorname{Weibull}(k,\lambda)$, write the threshold exceeded with probability $0.05$ as a function of $k$ and $\lambda$.