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Mathematics | Weibull Distribution Model

  • Last Updated : 19 Feb, 2021

Introduction :

Suppose an event can occur several times within a given unit of time. When the total number of occurrences of the event is unknown, we can think of it as a random variable X. When this random variable X follows Weibull Distribution Model (closely related to the exponential distribution) then its probability density function is given as follows.

f(x) = \alpha\beta x^{\beta-1}e^{-\alpha x^{\beta}}
only when 
x > 0, α >0, β > 0.  
f(x) = 0 , Otherwise

The cumulative distribution function of Weibull Distribution is obtained as follows.

F(x) = \int_{0}^{x}\alpha\beta w^{\beta-1}e^{-\alpha w^{\beta}} dw
Putting y = w^b   , 

We will get the following expression.

F(x) = \int^{x^b}_{0} \alpha e^{-\alpha y} dy = 1-e^{-\alpha x^\beta}

This means, when X has the Weibull distribution then Y = X^β    has an exponential distribution.



Expected Value :

The Expected Value of the Beta distribution can be found by summing up products of Values with their respective probabilities.

\mu = E(X) = \int^{\infty}_{-\infty} x.f(x) dx
\mu = \int^{\infty}_{0} x. \alpha\beta x^{\beta-1}e^{-\alpha x^{\beta}} dx
Putting u = \alpha x^\beta  , 

We will get the following expression as follows.

\mu = \alpha^{-1/\beta} \int^{\infty}_0 u^{1/\beta} e^{-u} du

Using the definition of the gamma function, we will get the following expression as follows.

\mu = \alpha^{-1/\beta} \Gamma(1+\frac{1}{\beta})

Variance and Standard Deviation :

The Variance of the Beta distribution can be found using the Variance Formula.

σ^2 = E( X − μ )^2 = E( X^2 ) − μ^2
E(X^2) = \int^{\infty}_{-\infty} x^2.f(x) dx
E(X^2) = \int^{\infty}_{0} x^2. \alpha\beta x^{\beta-1}e^{-\alpha x^{\beta}} dx
Putting u = \alpha x^\beta   , 

We will get the following expression as follows.

E(X^2) = \alpha^{-2/\beta} \int^{\infty}_0 u^{2/\beta} e^{-u} du

Using the definition of the gamma function, we will get the following.



E(X^2) = \alpha^{-2/\beta} \Gamma(1+\frac{2}{\beta})
So, Var(X) = E(X^2) - \mu^2
Var(X) = \sigma^2 = \alpha^{-2/\beta} [\Gamma(1+\frac{2}{\beta})- \{\Gamma (1+\frac{1}{\beta})\}^2]

Standard Deviation is given as follows.

\sigma = \alpha^{-1/\beta} \sqrt{[\Gamma(1+\frac{2}{\beta})- \{\Gamma (1+\frac{1}{\beta})\}^2]}

Example – 

Suppose that the lifetime of a certain kind of emergency backup battery (in hours) is a random variable X having the Weibull distribution with α = 0.1 and β = 0.5. Find

  1. The mean lifetime of these batteries;
  2. The probability that such a battery will last more than 300 hours.

Solution –

1. Substituting the value of α and β in the formula of mean we will get the following.

\mu = (0.1)^{-2}\Gamma(3) = 200    hours

2. The probability of the battery lasting for more than 300 hours is given by the following.

\int_{300}^{\infty} (0.05)x^{-0.5} e^{-0.1x^{0.5}} dx = e^{-0.1(300)^{0.5}} = 0.177

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