Weibull Distribution
In this tutorial we will discuss about the Weibull distribution and examples. 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 and step by step guide on Weibuill Distribution Examples for different numerical problems.
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 $V(X) = \beta^2 \bigg(\Gamma (\dfrac{2}{\alpha}+1) -\bigg(\Gamma (\dfrac{1}{\alpha}+1) \bigg)^2\bigg)$.
Lets solve few of the Weibull distribution examples with detailed guide to compute probbility and variance for different numerical problems.
Weibull Distribution 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 $\beta=3$.
Using above formula of Two parameter Weibull distribution example can be solved as below
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. $P(X\geq 3)$
$$ \begin{aligned} P(X\geq 3) &=1-P(X< 3)\\ &= 1-F(3)\\ &= 1-\bigg[1-e^{-(3/3)^{2}}\bigg]\\ &= e^{-(1)^{2}}\\ &=0.3679 \end{aligned} $$
Weibull Distribution 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 $X\sim W(\alpha = 300, \beta=0.5)$.
Using above formula of Two parameter Weibull distribution example can be solved as below:
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, $P(X\geq 600)$
$$ \begin{aligned} P(X\geq 600) &=1-P(X< 600)\\ &= 1-F(600)\\ &= 1-\bigg[1-e^{-(600/300)^{0.5}}\bigg]\\ &= e^{-(2)^{0.5}}\\ &=0.2431 \end{aligned} $$
I Hope above article with step by step guide on Weibull Distribution Examples helps you understand how to solve the numerical problems on Weibull distribution.
