Weibull Distribution Examples | Calculator | Two Parameter

Weibull Distribution Calculator

Use this calculator to find the probability density and cumulative probabilities for two parameter Weibull distribution with parameter $\alpha$ and $\beta$.

How to Calculate Probabilities of Weibull distribution?

Step 1 - Enter the location parameter $\alpha$

Step 2 - Enter the scale parameter $\beta$

Step 2 - Enter the value of $x$

Step 4 - Click on "Calculate" button to get Weibull distribution probabilities

Step 5 - Gives the output probability at $x$ for Weibull distribution

Step 6 - Gives the output cumulative probabilities for Weibull distribution

Two-parameter Weibull distribution

Two parameter Weibull distribution can be defined by taking $\mu=0$. The pdf of two parameter Weibull distribution is given by

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

The distribution function of two-parameter Weibull distribution is

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

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. $P(X\leq 6)$

c. $P(1.8\leq X \leq 6)$

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 $X$

$$ \begin{aligned} E(X) &= \beta \Gamma (\dfrac{1}{\alpha}+1)\\ &=3\Gamma(\dfrac{1}{2}+1)\\ &=3\Gamma(3/2)\\ &=3\times\dfrac{1}{2}\Gamma(1/2)\\ &=\dfrac{3}{2}\times\sqrt{\pi}\\ &=\dfrac{3}{2}\times1.7725\\ &=2.6588 \end{aligned} $$

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

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. $P(1.8\leq X \leq 6)$

$$ \begin{aligned} P(1.8 \leq X\leq 6) &=F(6)-F(1.8)\\ &= \bigg[1-e^{-(6/3)^{2}}\bigg] -\bigg[1-e^{-(1.8/3)^{2}}\bigg]\\ &= e^{-(0.6)^{2}}-e^{-(2)^{2}}\\ &= e^{-(0.36)}-e^{-(4)}\\ &=0.6977-0.0183\\ &=0.6794 \end{aligned} $$
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 $X\sim W(\alpha = 300, \beta=0.5)$.

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, $P(X\geq 600)$

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

Conclusion

In this tutorial, you learned about how to calculate probabilities of Weibull distribution. You also learned about how to solve numerical problems based on Weibull distribution.

To read more about the step by step tutorial on Weibull distribution refer the link Weibull Distribution. This tutorial will help you to understand Weibull distribution and you will learn how to derive mean, variance, distribution function, median, mode, moment and other properties of Weibull distribution.

To learn more about other probability distributions, please refer to the following tutorial:

Probability distributions

Let me know in the comments if you have any questions on Weibull Distribution Examples and your thought on this article.