Definition 1: The Weibull distribution has the probability density function (pdf)
for x ≥ 0. Here β > 0 is the shape parameter and α > 0 is the scale parameter.
The cumulative distribution function (cdf) is
The inverse cumulative distribution function is I(p) =
Observation: If x represents “time-to-failure”, the Weibull distribution is characterized by the fact that the failure rate is proportional to a power of time, namely β – 1. Thus β can be interpreted as follows:
- A value of β < 1 indicates that the failure rate decreases over time. This happens if there is significant “infant mortality”, or where defective items fail early with a failure rate decreasing over time as the defective items are weeded out of the population.
- A value of β = 1 indicates that the failure rate is constant over time. This might suggest random external events are causing mortality or failure.
- A value of β > 1 indicates that the failure rate increases with time. This happens if there is an “aging” process; e.g. if parts are more likely to wear out and/or fail as time goes on.
1/α can be viewed as the failure rate. The mean of the Weibull distribution is the mean time to failure (MTTF) or mean time between failures (MTBF) = .
Key statistical properties of the Weibull distribution are:
- Mean =
- Median =
- Mode (when β > 1) =
- Variance =
Excel Function: Excel provides the following function in support of the Weibull distribution.
WEIBULL(x, β, α, cum) where α and β are the parameters in Definition 1 and cum = TRUE or FALSE
WEIBULL(x, β, α, FALSE) = the value of the Weibull pdf f(x) at x
WEIBULL(x, β, α, TRUE) = the value of the Weibull cumulative distribution function F(x) at x
Excel 2010/2013/2016 also provide the additional function WEIBULL.DIST which is equivalent to WEIBULL.
Example 1: The time to failure of a very sensitive computer screen follows a Weibull distribution with α = 1,000 hours and β = .6. What is the probability that the screen will last more than 5,000 hours? What is the mean time to failure?
The probability that the screen will last no more than 5,000 hours
= WEIBULL(5000, .6, 1000, TRUE) = 0.92767.
Thus, the probability that the screen will last more than 5,000 hours = 1 – 0.92767 = 7.2%
MTTF = αΓ(1+1/β) = 1000Γ(1+1/.6) = 1000*EXP(GAMMALN(1 + 1/.6)) = 1,504.575 hours
Example 2: If the mean time to failure for a component which follows a Weibull distribution is 1,000 hours with a standard deviation of 400 hours, what is the probability that the component will last more than 2,000 hours?
We now solve these equations for α and β. First we simplify the second equation and then we take the natural log of both sides of both equations to get
which is equivalent to
The above equation takes the form h(β) = 0, which we solve using Excel’s Goal Seek capability by selecting Data > What If Analysis|Goal Seek and filling in the dialog box that appears as shown in Figure 2.
Once we obtain the value for β, we can calculate α using the equation
Figure 2 – Goal Seek initial guess
After clicking on the OK button, the result is shown in Figure 3.
Figure 3 – Goal Seek results
The values for α and β are shown in cells B5 and B3. The probability that the component will last more than 2,000 hours is 0.91% (cell B6).
Real Statistics Function: Since Excel doesn’t provide an inverse function, you can use the following function provided by the Real Statistics Resource Pack instead.
WEIBULL_INV(p, β, α) = x such that WEIBULL.DIST(x, β, α, TRUE) = p; i.e. the inverse of WEIBULL.DIST(x, β, α, TRUE)