Probability Plotting for the Weibull Distribution

One method of calculating the parameters of the Weibull distribution is by using probability

plotting. To better illustrate this procedure, consider the following example from Kececioglu

[18].

Example 1

Assume that six identical units are being reliability tested at the same application and

operation stress levels. All of these units fail during the test after operating the following

number of hours, : 93, 34, 16, 120, 53 and 75. Estimate the values of the parameters for a

two-parameter Weibull distribution and determine the reliability of the units at a time of 15

hours.

Solution to Example 1

The steps for determining the parameters of the Weibull pdf representing the data, using

probability plotting, are outlined in the following instructions.

First, rank the times-to-failure in ascending order as shown next.

Obtain their median rank plotting positions. Median rank positions are used instead of other

ranking methods because median ranks are at a specific confidence level (50%). Median

ranks can be found tabulated in many reliability books. They can also be estimated using the

following equation,

where i is the failure order number and N is the total sample size.

The exact median ranks are found in Weibull++ by solving,

for MR, where N is the sample size and i the order number.

The times-to-failure, with their corresponding median ranks, are shown next.

On a Weibull probability paper, plot the times and their corresponding ranks. A sample of a

Weibull probability paper is given in Figure 7, and the plot of the data in the example in Figure

8.

Fig. 7: Example of Weibull probability plotting paper

Draw the best possible straight line through these points, as shown below, then obtain the

slope of this line by drawing a line, parallel to the one just obtained, through the slope indicator.

This value is the estimate of the shape parameter , in this case = 1.4.

Fig. 8: Probability plot of data in Probability Plotting Example

At the Q(t) = 63.2% ordinate point, draw a straight horizontal line until this line intersects the

fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa.

The value at the intersection of the abscissa is the estimate of . For this case = 76 hours.

(This is always at 63.2% since Q(T) = 1 - = 1- = 0.632 = 63.2%.)

Now any reliability value for any mission time t can be obtained. For example the reliability for

a mission of 15 hours, or any other time, can now be obtained either from the plot or

analytically.

To obtain the value from the plot, draw a vertical line from the abscissa, at t = 15 hours, to the

fitted line. Draw a horizontal line from this intersection to the ordinate and read Q(t), in this

case Q(t) = 9.8%. Thus, R(t) = 1 - Q(t) = 90.2%.

This can also be obtained analytically, from the Weibull reliability function, since the estimates

of both of the parameters are known or,

Probability Plotting for the location parameter,

The third parameter of the Weibull distribution is utilized when the data do not fall on straight

line, but fall on either a concave up or down curve. The following statements can be made

regarding the value of :

Case 1: If the curve for MR versus is concave down and the curve for MR versus ( - )

is concave up, then there exists a such that 0 < < , or has a positive value.

Case 2: If the curves for MR versus and MR versus ( - ) are both concave up, then

there exists a negative , which will straighten out the curve of MR versus .

Case 3: If neither one of the previous two cases prevail, then either reject the Weibull pdf as

one capable of representing the data, or proceed with the multiple population (mixed Weibull)

analysis.

To obtain the location parameter, :

?/ΦΟΝΤ> Subtract the same arbitrary value, , from all the times to failure, and replot the

data.

?/ΦΟΝΤ> If the initial curve is concave up subtract a negative from each failure time.

?/ΦΟΝΤ> If the initial curve is concave down subtract a positive from each failure time.

?/ΦΟΝΤ> Repeat until the data plots on an acceptable straight line.

?/ΦΟΝΤ> The value of is the subtracted (positive or negative) value that places the points

in an acceptable straight line.

The other two parameters are then obtained using the techniques previously described. Also,

it is important to note that we used the term subtract a positive or negative gamma, where

subtracting a negative gamma is equivalent to adding it. Note that when adjusting for gamma,

the x-axis scale for the straight line becomes (T - ).

Example 2

Six identical units are reliability tested under the same stresses and conditions. All units are

tested to failure, and the following times-to-failure are recorded: 48, 66, 85, 107, 125 and 152

hours. Find the parameters of the three-parameter Weibull distribution using probability

plotting.

Solution to Example 2

The following figure shows the results. Note that since the original data set was concave down,

17.26 was subtracted from all the times-to-failure and re-plotted, resulting in a straight line,

thus = 17.26. (We used Weibull++ to get the results. To perform this by hand, one would

attempt different values of , using a trial and error methodology, until an acceptable straight

line is found. When performed manually, you do not expect decimal accuracy.)

Fig. 9: Probability Plot of data in Probability Plotting for the Location Parameter

example

Weibull Rank Regression on Y

Performing rank regression on Y requires that a straight line mathematically be fitted to a set

of data points such that the sum of the squares of the vertical deviations from the points to the

line is minimized. This is in essence the same methodology as the probability plotting method,

except that we use the principle of least squares to determine the line through the points, as

opposed to just 밻 yeballing?it.

The first step is to bring our function into a linear form. For the two-parameter Weibull

distribution, the cdf (cumulative density function) is,

(8)

Taking the natural logarithm of both sides of the equation yields,

or,

(9)

Now let,

(10)

(11)

and,

(12)

which results in the linear equation of,

The least squares parameter estimation method (also known as regression analysis) was

discussed in the Statistical Background chapter and the following equations for regression on

Y were derived in the Parameter Estimation chapter.

(13)

and,

(14)

In this case the equations for and are,

and,

The F s are estimated from the median ranks.

Once and are obtained, then and can easily be obtained from Eqns. (11) and (12).

The Correlation Coefficient

The correlation coefficient is defined as follows,

where, = covariance of x and y, = standard deviation of x and = standard deviation

of y.

The estimator of is the sample correlation coefficient, , given by,

(15)

Example 3

Consider the data in Example 1, where six units were tested to failure and the following failure

times were recorded: 16, 34, 53, 75, 93 and 120 hours. Estimate the parameters and the

correlation coefficient using rank regression on Y, assuming that the data follow the two-

parameter Weibull distribution.

Solution to Example 3

Construct a table as shown below.

Table 1- Least Squares Analysis

Utilizing the values from Table 1, calculate and using Eqns. (13) and (14),

or,

and,

or,

Therefore, from Eqn. (12),

and from Eqn (11)

or,

The correlation coefficient can be estimated using Eqn. (15),

The above example can be repeated using Weibull++ 6. The steps for using the application

are as follows:

?/ΦΟΝΤ> Start Weibull++ and create a new Data Folio.

?/ΦΟΝΤ> Select the Times to Failure option.

?/ΦΟΝΤ> Enter the times-to-failure in the spreadsheet (ignore the Subset ID column). The

times-to-failure need not be sorted, Weibull++ will automatically sort the data.

?/ΦΟΝΤ> Select the desired method of analysis. Note that we are assuming that the

underlying distribution is the Weibull, so make sure that the Weibull distribution is selected.

(Just click Weibull. It will turn red when selected.)

Also, so that you get the same results as this example, switch to the Set Analysis page

and make sure you are using the Rank Regression on Y calculation method with this

example.

Note that this can also be done from the Main page by clicking the left bottom box under

the Results area. Each time you click that box you will see the method switch between

MLE, RRX, and RRY.

?/ΦΟΝΤ> Under Parameters/Type on the main page, choose 2. Click the Calculate icon or

select Calculate Parameters from the Data menu. The results should appear in the Data

Folio's Results Area. The next figure shows the results for this example.

You can now plot the results by clicking the Plot icon or by selecting Plot Probability from

the Data menu.

The Weibull probability plot for these data is shown next.

The confidence bounds, as determined from the Fisher matrix, can also be plotted.

If desired, the Weibull pdf representing these data can be written as,

or,

You can also plot the Weibull pdf by selecting Pdf Plot from the Special Plot Type option on

the control panel to the right of the plot area.

From this point on, different results, reports and plots can be obtained.

Weibull Rank Regression on X

Performing a rank regression on X is similar to the process for rank regression on Y, with the

difference being that the horizontal deviations from the points to the line are minimized, rather

than the vertical.

Again the first task is to bring our cdf function, Eqn. (8), into a linear form. This step is exactly

the same as in the regression on Y analysis and Eqns. (9), (10), (11) and (12) apply in this

case too. The derivation from the previous analysis begins on the least squares fit part, where

in this case we treat x as the dependent variable and y as the independent variable.

The best-fitting straight line to the data, for regression on X (see the Statistical Background

chapter), is the straight line,

(16)

The corresponding equations for and are,

and,

where,

and,

and the F( ) values are again obtained from the median ranks.

Once and are obtained, solve Eqn. (16) for y, which corresponds to,

Solving for the parameters from Eqns. (11) and (12) we get,

(17)

and,

(18)

The correlation coefficient is evaluated as before using Eqn. (15).

Example 4

Repeat Example 1 using rank regression on X.

Solution to Example 4

Table 1, constructed in Example 3, can also be applied to this example. Using the values from

this table we get,

or,

and,

or,

Therefore, from Eqn. (18),

and from Eqn. (17),

The correlation coefficient is found using Eqn. (15):

The results and the associated graph using Weibull++ 6 are given next. Note that the slight

variation in the results is due to the number of significant figures used in the estimation of the

median ranks. Weibull++ by default uses double precision accuracy when computing the

median ranks.

Three-Parameter Weibull Regression

When the MR versus points plotted on the Weibull probability paper do not fall on a satisfactory straight line and the points fall on a curve, then a location parameter, , might

exist which may straighten out these points.

The goal in this case is to fit a curve, instead of a line, through the data points, using nonlinear

regression The Gauss-Newton method can be used to solve for the parameters, , and ,

by performing a Taylor series expansion on F( ; , , ). Then the nonlinear model is

approximated with linear terms and ordinary least squares are employed to estimate the

parameters. This procedure is iterated until a satisfactory solution is reached.

Weibull++ 6 calculates the value of by utilizing an optimized Nelder-Mead algorithm, and

adjusts the points by this value of such that they fall on a straight line, and then plots both

the adjusted and the original unadjusted points. To draw a curve through the original

unadjusted points, if so desired, choose the Weibull 3P Line Unadjusted for Gamma option

from the Show Plot Line submenu under the Plot Options menu. The returned estimations of

the parameters are the same when selecting RRX or RRY.

The results and the associated graph for the previous example using the three-parameter

Weibull case are shown next:

Maximum Likelihood Estimation for the Weibull Distribution

As it was outlined in the Statistical Background chapter, maximum likelihood estimation works

by developing a likelihood function based on the available data and finding the values of the

parameter estimates that maximize the likelihood function. This can be achieved by using

iterative methods to determine the parameter estimate values that maximize the likelihood

function, but this can be rather difficult and time-consuming, particularly when dealing with the

three-parameter distribution. Another method of finding the parameter estimates involves

taking the partial derivatives of the likelihood function with respect to the parameters, setting

the resulting equations equal to zero and solving simultaneously to determine the values of the

parameter estimates. The log-likelihood functions and associated partial derivatives used to

determine maximum likelihood estimates for the Weibull distribution are covered in the

Distribution Log-Likelihood Equations chapter.

Example 5

Repeat Example 1 using maximum likelihood estimation.

Solution to Example 5

In this case we have non-grouped data with no suspensions or intervals, i.e. complete data.

The equations for the partial derivatives of the log-likelihood function are derived in the

Distribution Log-Likelihood Equations chapter and given next:

and,

Solving the above equations simultaneously we get,

The variance/covariance matrix is found to be,

The results and the associated graph using Weibull++ (MLE) are shown next.

You can view the variance/covariance matrix directly from the Quick Calculation Pad (QCP) of

Weibull++ by clicking the Fisher Matrix button.

Note that the decimal accuracy displayed and used is based on your individual User Setup.