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# Weibull Distribution

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: There is also a three-parameter version of the Weibull distribution.

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) = $alpha Gamma ! left( ! 1+frac{1}{beta} ! right)$.

Key statistical properties of the Weibull distribution are:

• Mean = $alpha Gamma ! left( ! 1+frac{1}{beta} ! right)$
• Median = $alpha ({ln 2})^{1/beta}$
• Mode (when β > 1) = $alpha ! left( ! frac{beta-1}{beta} ! right)^{1/beta}$
• Variance = $alpha^2 Gamma ! left( ! 1+frac{2}{beta} ! right)- mu^2$

Excel Function: Excel provides the following function in support of the Weibull distribution.

WEIBULL.DIST(x, βαcum) where α and β are the parameters in Definition 1 and cum = TRUE or FALSE

WEIBULL.DIST(x, βα, FALSE) = the value of the Weibull pdf f(x) at x

WEIBULL.DIST(x, βα, TRUE) = the value of  the Weibull cumulative distribution function F(x) at x

Versions of Excel prior to Excel 2010 use the WEIBULL function instead of the WEIBULL.DIST function.

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.DIST(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?

Thus

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

Eliminating α

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

and then we can calculate the probability that the component will last more than 2,000 hours using the WEIBULL.DIST function.

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, βα) = x such that WEIBULL.DIST(x, βα, TRUE) = p; i.e. the inverse of  WEIBULL.DIST(x, βα, TRUE)

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