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

\[ F(x) \;\;=\;\; P(X \le x) \;\;=\;\; 1 - e^{-(\lambda x)^{\alpha}} \]

(Proved in full version)

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 )

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,

\[ E[X] \;=\; \frac 1 {\lambda} \Gamma\left(1 + \frac 1 {\alpha}\right) \]

Question

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?

(Solved in full version)

We now give an example in which the hazard rate decreases over time.

Question

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?

(Solved in full version)