Weibull Distribution

In this tutorial we will discuss about the Weibull distribution.Weibull distribution is a continuous probability distribution. Weibull distribution is one of the most widely used probability distribution in reliability engineering.

This tutorial help you to understand how to calculate probabilities related to Weibull distribution.

Three parameter Weibull Distribution

A continuous random variable $X$ is said to have a Weibull distribution with three parameters $\mu$, $\alpha$ and $\beta$ if the probability density function of Weibull random variable $X$ is

$$ \begin{equation*} f(x;\alpha, \beta)=\left\{ \begin{array}{ll} \frac{\alpha}{\beta} \big(\frac{x-\mu}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}, & \hbox{$x>\mu$, $\alpha, \beta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

  • $\alpha$ is the shape parameter
  • $\beta$ is the scale parameter
  • $\mu$ is the location parameter.

Two-parameter Weibull Distribution

Let $\mu=0$. Then the pdf of two parameter Weibull distribution is given by $$ \begin{equation*} f(x;\alpha, \beta)=\left\{ \begin{array}{ll} \frac{\alpha}{\beta} \big(\frac{x}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x}{\beta}\big)^\alpha}, & \hbox{$x>0$, $\alpha, \beta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

Standard Weibull Distribution

If we let $\mu=0$ and $\beta =1$, then the distribution of $X$ is called standard Weibull distribution. Then the pdf of standard Weibull distribution is

$$ \begin{equation*} f(x;\beta)=\left\{ \begin{array}{ll} \alpha x^{\alpha-1}e^{-x^\alpha}, & \hbox{$x>0$, $\beta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

Mean of Two-parameter Weibull Distribution

The mean of Two-parameter Weibull distribution is $E(X) = \beta \Gamma (\dfrac{1}{\alpha}+1)$.

Variance of Two-parameter Weibull Distribution

The variance of Two-parameter Weibull distribution is $V(X) = \beta^2 \bigg(\Gamma (\dfrac{2}{\alpha}+1) -\bigg(\Gamma (\dfrac{1}{\alpha}+1) \bigg)^2\bigg)$.

Example 1

The lifetime $X$ (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters $\alpha = 2$ and $\beta = 3$. Compute the following:

a. $E(X)$ and $V(X)$

b. $P(X\leq 5)$

c. $P(1.8\leq X \leq 5)$

d. $P(X\geq 3)$.

Solution

Let $X$ denote the lifetime (in hundreds of hours) of vaccume tube. Given that $X\sim W(\alpha,\beta)$, where $\alpha =2$ and $\beta=3$.

The probability density function of $X$ is

$$ \begin{aligned} f(x;\alpha, \beta)&=\frac{\alpha}{\beta} \big(\dfrac{x}{\beta}\big)^{\alpha-1}e^{-\big(\dfrac{x}{\beta}\big)^\alpha};\; x>0,\alpha,\beta>0. \end{aligned} $$

The distribution function of $X$ is

$$ \begin{aligned} F(x) &= 1- e^{-\big(x/\beta\big)^\alpha}. \end{aligned} $$

a. Mean and variance of $X$

$$ \begin{aligned} E(X) &= \beta \Gamma (\dfrac{1}{\alpha}+1)\\ &=3\Gamma(\dfrac{1}{2}+1)\\ &=3\Gamma(3/2)\\ &=3\times\dfrac{1}{2}\Gamma(1/2)\\ &=\dfrac{3}{2}\times\sqrt{\pi}\\ &=\dfrac{3}{2}\times1.7725\\ &=2.6587 \end{aligned} $$

$$ \begin{aligned} V(X) &= \beta^2 \bigg[\Gamma (\dfrac{2}{\alpha}+1) -\bigg(\Gamma (\dfrac{1}{\alpha}+1) \bigg)^2\bigg]\\ &=3^2 \bigg[\Gamma (\dfrac{2}{2}+1) -\bigg(\Gamma (\dfrac{1}{2}+1) \bigg)^2\bigg]\\ &=9\bigg[\Gamma(2)-\big(\Gamma(3/2)\big)^2\bigg]\\ &=9\bigg[1-\bigg(\frac{1}{2}\Gamma(1/2)\bigg)^2\bigg]\\ &=9\bigg[1-\bigg(\frac{\sqrt{\pi}}{2}\bigg)^2\bigg]\\ &=9\bigg[1-\bigg(\frac{\sqrt{3.1416}}{2}\bigg)^2\bigg]\\ &=1.931846 \end{aligned} $$

b. $P(X\leq 6)$

$$ \begin{aligned} P(X\leq 6) &=F(6)\\ &= 1-e^{-(6/3)^{2}}\\ &= 1-e^{-(2)^{2}}\\ &= 1-e^{-(4)}\\ &=1-0.0183\\ &=0.9817 \end{aligned} $$

c. $P(1.8\leq X \leq 5)$

$$ \begin{aligned} P(1.8 \leq X\leq 6) &=F(6)-F(1.8)\\ &= \bigg[1-e^{-(6/3)^{2}}\bigg] -\bigg[1-e^{-(1.8/3)^{2}}\bigg]\\ &= e^{-(0.6)^{2}}-e^{-(2)^{2}}\\ &= e^{-(0.36)}-e^{-(4)}\\ &=0.6977-0.0183\\ &=0.6794 \end{aligned} $$

d. $P(X\geq 3)$

$$ \begin{aligned} P(X\geq 3) &=1-P(X< 3)\\ &= 1-F(3)\\ &= 1-\bigg[1-e^{-(3/3)^{2}}\bigg]\\ &= e^{-(1)^{2}}\\ &=0.3679 \end{aligned} $$

Example 2

Assume that the life of a packaged magnetic disk exposed to corrosive gases has a Weibull distribution with $\alpha = 300$ hours and $\beta = 0.5$.

Calculate the probability that

a. a disk lasts at least 600 hours,

b. a disk fails before 500 hours.

Solution

Let $X$ denote the life of a packaged magnetic disk exposed to corrosive gases in hours.

Given that $X\sim W(\alpha = 300, \beta=0.5)$.

The probability density function of $X$ is

$$ \begin{aligned} f(x;\alpha, \beta)&=\frac{\alpha}{\beta} \big(\dfrac{x}{\beta}\big)^{\alpha-1}e^{-\big(\dfrac{x}{\beta}\big)^\alpha};\; x>0,\alpha,\beta>0. \end{aligned} $$

The distribution function of $X$ is

$$ \begin{aligned} F(x) &= 1- e^{-\big(x/\beta\big)^\alpha}. \end{aligned} $$

a. The probability that a disk fails before 500 hours is

$$ \begin{aligned} P(X\leq 500) &=F(500)\\ &= 1-e^{-(500/300)^{0.5}}\\ &= 1-e^{-(1.6667)^{0.5}}\\ &= 1-e^{-(1.291)}\\ &=1-0.275\\ &=0.725 \end{aligned} $$

b. The probability that a disk lasts at least 600 hours, $P(X\geq 600)$

$$ \begin{aligned} P(X\geq 600) &=1-P(X< 600)\\ &= 1-F(600)\\ &= 1-\bigg[1-e^{-(600/300)^{0.5}}\bigg]\\ &= e^{-(2)^{0.5}}\\ &=0.2431 \end{aligned} $$

Hope this article helps you understand how to solve the numerical problems on Weibull distribution.

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