Interpretation of the Integral Used in the Weibull++ Comparison Wizard
The integral equation underlying the Comparison Wizard in Weibull++
can be intimidating. This article provides an explanation of the meaning of the equation using an analogy from
everyday life.
Joe, the reliability engineer, needs to make a recommendation to his boss about which of two different
designs is better from a reliability perspective. He tested 10 samples of each design and found that part
of the time, Design A is more reliable than Design B and part of the time, Design B is more reliable than
Design A. In order to decide which design to recommend, Joe decides to determine the probability that
Design B would outlast Design A over the entire life of the product.
Joe, an avid basketball fan, realizes that he could relate the problem he needs to solve with his
favorite sport. He thinks about the two best players in the league this year: the power forward Clyde and
the elusive point guard Bonnie. He knows their free throw statistics by heart: Clyde is a 75% free
throw shooter, and Bonnie shoots 90% from the line. So if these two basketball greats had a free throw
contest, Joe’s intuition tells him that Bonnie would win more often than Clyde. But how often would Bonnie
make more free throws than Clyde?
Joe decides to compute the percentage of times Bonnie would win if each player attempted 5 free throw shots.
He knows that the probability that a player will make exactly S shots is described by the binomial
distribution, given by:
Where N is the number of shots attempted and FTP is the free throw percentage. For example, the probability
that Clyde will make exactly 4 out of 5 shots is:
Using the binomial distribution, Joe creates the following table showing the probabilities of Bonnie and
Clyde making exactly S shots in 5 attempts:
Figure 1: Probability of making exactly S shots.
From the table above, Joe can see that in a 5 shot contest, Bonnie will most likely
make all 5 shots and Clyde will most likely make 4 out of 5. Since the probabilities
are independent of each other, Joe computes the joint probability of this outcome as:
Similarly, Joe computes the joint probabilities of all 36 outcomes as shown in the following table.
Figure 2: Joint probabilities of all possible outcomes
All of the outcomes in the shaded diagonal row are ties, i.e. Bonnie and Clyde make the same number of shots.
Adding up the probabilities on the diagonal, Joe sees that the probability of the contest ending in a tie
is 29%. All of the outcomes above the diagonal represent the probabilities of Clyde making more shots
than Bonnie, so Joe expects Clyde to have a 13% chance of winning the contest. All of the outcomes below
the diagonal represent the probabilities of Bonnie making more shots than Clyde. Thus, Joe expects a 58%
chance that Bonnie will win.
Joe now knows that Bonnie will make more free throws than Clyde, but the joint probability table method
will be too cumbersome if he wants to figure out the probability of Bonnie winning
a contest of, say, 100 free throws. So he thinks about what exactly would constitute a win for Bonnie. If Clyde
makes S shots, then Bonnie wins if she makes at least S+1 shots. For example, if Clyde makes
exactly 2 shots, then Bonnie wins if she makes 3, 4, or 5 shots. Thus, the probability of Bonnie
winning the contest, given that Clyde makes exactly 2 shots, is:
Joe realizes that a) the probability that a random variable will be exactly a certain value is given by its
probability density function, f(t), and b) the probability that a random variable will be greater than a
certain value is given by its reliability, R(t). So he rewrites the above equation as follows:
Joe now needs to perform only 6 calculations to
determine the probability of Bonnie winning the
contest instead of the 36 required using the table of
joint probabilities. He also comes up with a general
formula for all possible outcomes for the number of shots Clyde makes (which is simply the sum of the above equation):
Relating the above equation with the problem of the two engineering designs, Joe sees one difference between his method of
determining the winner of the free throw contest and his method of determining if Design B will outlast Design A:
in the case of the free throw contest, he used a discrete distribution to describe the possible outcomes of the
contest; however, in his work with the engineering designs, he used continuous distributions to fit his
times-to-failure data. With this new insight, Joe changes his equation from a summation to an integral:
Since the integral is complicated to solve, Joe uses the Comparison Wizard in
Weibull++ to compute the probability that Design B
will outlast Design A. Joe creates a Folio called "New Data" and enters the test results in
two data sheets; one called "Design A" and another called
"Design B." He calculates
the parameters for each data set as shown in Figures 3a and 3b. In addition, he plots the
reliability versus time curve for both designs as shown in Figure 4.
Figure 3a: Input data and calculated parameters for Design A
Figure 3b: Input data and calculated parameters for Design B
Figure 4: Reliability multiplot for Design A and Design B
From the reliability plot, Joe can see that Design B will outlast Design A most of the time, but in
order to quantify the probability that B will outlast A, Joe uses the Comparison Wizard as shown
in Figure 5.
Figure 5: Computation of probability that Design B will outlast Design A
The 81% probability that Design B will outlast Design A is roughly equivalent to 4 to 1
odds that Design B will survive longer. Joe knows that a 50% probability will make it impossible
to tell which design will last longer, and that a probability of 80% or greater is recommended in
order to draw any firm conclusions. Since the Comparison Wizard returned a value slightly greater
than 80%, Joe recommends Design B to management.
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