Finding probabilities for Weibull distributions
Probabilities for the Weibull distribution are usually found from the cumulative distribution function.
Cumulative distribution function
If \(X \sim \WeibullDistn(\alpha, \lambda)\) its cumulative distribution function is
\[ F(x) \;\;=\;\; P(X \le x) \;\;=\;\; 1 - e^{-(\lambda x)^{\alpha}} \]It is possible to integrate the Weibull distribution's pdf to find the cumulative distribution function, but we will prove the result by working in reverse — showing that \(F'(x) = f(x)\). This is done using the chain rule:
\[ \begin{align} F'(x) \;&=\; e^{-(\lambda x)^{\alpha}} \times \frac{d\;(\lambda x)^{\alpha}}{d\;x} \\ &=\; e^{-(\lambda x)^{\alpha}} \times \lambda^{\alpha} \times \alpha x^{\alpha - 1} \\ &=\; \alpha \lambda^{\alpha} x^{\alpha - 1} e^{-(\lambda x)^{\alpha}} \end{align} \]This is the pdf of the Weibull distribution.
Given values of \(x\), \(\alpha\) and \(\lambda\), these probabilities can be evaluated on a scientific calculator. Excel also has a function to evaluate cumulative Weibull probabilities, but its third parameter is the inverse of \(\lambda\), rather than \(\lambda\) itself. The cumulative probability could be found by typing into a spreadsheet cell
=WEIBULL.DIST( \(x\), \(\alpha\), \(1/\lambda\), true )
Relationship between \(E[X]\) and \(\lambda\)
Although the parameter \(\alpha\) has a meaningful interpretation since \(h(x) \propto x^{\alpha - 1}\), the value of the parameter \(\lambda\) is not easily interpreted. The mean lifetime of the items is an easier value to interpret than \(\lambda\) itself,
\[ E[X] \;=\; \frac 1 {\lambda} \Gamma\left(1 + \frac 1 {\alpha}\right) \]Example
If an item's hazard rate is proportional to the square root of its age, and its mean lifetime is 3 years, what is the probability that it will survive for longer than 10 years?
Since \(h(x) \;\;\propto\;\; x^{\alpha - 1}\),
\[ \alpha \;\;=\;\; 1.5 \]We now find the value of \(\lambda\).
\[ \lambda \;=\; \frac 1{E[X]} \Gamma\left(1 + \frac 1 {\alpha}\right) = \frac 1 3 \Gamma(1.66667) \]Using Excel to evaluate the gamma function gives \(\lambda = 0.3009\).
The probability of surviving longer than 10 years can be found in Excel with
=1 − WEIBULL.DIST( 10, 1.5, 1/ 0.3009, true )
giving a probability of 0.0054.
We now give an example in which the hazard rate decreases over time.
Example
If the item's hazard rate was inversely proportional to the square root of its age, and its mean lifetime is 3 years, what would be the corresponding probability of surviving for longer than 10 years? 40 years?
To make \(h(x) \;\;\propto\;\; x^{-0.5}\),
\[ \alpha \;\;=\;\; 0.5 \]The value of \(\lambda\) giving a mean lifetime of 3 years is now
\[ \lambda \;=\; \frac 1{E[X]} \Gamma\left(1 + \frac 1 {\alpha}\right) = \frac 1 3 \Gamma(3) \]Since \(\Gamma(3) = 2!\), \(\lambda = 0.6667\).
The probability of surviving longer than 10 years can be found in Excel with
=1 - WEIBULL.DIST( 10, 0.5, 1/0.6667, true )
giving a probability of 0.0756.
The probability of surviving longer than 40 years can be found in the same way to be 0.0057.
Check these probabilities against the shapes of the two Weibull distributions on the previous page. The probabilities when \(\alpha = 0.5\) may seem larger than expected in that diagram, but the right tail extends a long way beyond the right end of the axis.