This Weibull distribution has special meaning to reliability experts; however, it can be used to model other phenomena as well. As illustrated in the figure, the distribution is defined only for nonnegative variables and is skewed to the right. It has two parameters and .

The parameter affects the form of the distribution. For a given , the parameter affects the location of the distribution. The cases of the figure have adjusted so that each distribution has the mean of 1. When is 1, the distribution is an exponential. As increases and the mean is held constant, the variance decreases and the distribution becomes more symmetric.

Conveniently, the cumulative distribution has a closed form expression.

Reliability

In reliability modeling, the random variable is the time to failure of a component. The hazard function is defined at the left as a function of the density and cumulative distribution. The hazard function can be viewed as the failure rate as a function of time.

For =1, the hazard function is constant, and we say that the component has a constant failure rate. The distribution for time to failure is the exponential distribution. This is often the assumption used for electronic components.

For = 2, the hazard rate is increasing linearly with time. The probability of failure in a small interval of time, given the component has not failed previously, is growing with time. This is the characteristic of wear out. As the component gets older, it begins to wear out and the likelihood of failure increases. For larger values of the hazard function increases at a greater than linear rate, indicating accelerating wear out.

Alternatively, for < 1, the hazard function is decreasing with time. This models the characteristic of infant mortality. The component has a high failure rate during its early life but if it survives that period, it becomes less likely to fail.

Example

Example: A truck tire has a mean life of 20,000 miles. A conservative owner decides to always replace a tire at 15,000 miles rather than risk failure. What is the probability that the tire will fail before it is replaced? Assume the life of the tire has a Weibull distribution with parameter = 2.

For this problem we are given the mean and the value of . We compute for the distribution by manipulating the equation for the mean. The computation yields the value shown at the left.

The owner decides to replace the tire at 15,000 miles rather than risk failure. We ask for the probability that the tire will fail before it is replaced. The answer is computed as 0.357.

This result looks a little risky for our conservative driver. He asks, how soon must the tire be discarded so the chance of failure is less than 0.1? To answer this question, we solve for the value of x that yields 0.1 as the value of the cumulative. This is accomplished by the inverse probability function, computed in T16. The tire must be replaced at about 7,325 miles.