Open access peer-reviewed chapter

Weibull Distribution

Written By

Jaroslav Menčík

Submitted: January 8th, 2016 Reviewed: February 3rd, 2016 Published: April 13th, 2016

DOI: 10.5772/62375

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Weibull distribution is very flexible in fitting empirical data, such as strength or time to failure. Several methods for the determination of parameters are described, including direct fitting using solvers available in universal programs. Also finding of parameters of exponential distribution is described. The use of Weibull distribution is illustrated on examples.


  • Probability
  • reliability
  • Weibull distribution
  • exponential distribution
  • determination of parameters
  • least squares method
  • solver

A special position in reliability assessment pertains to Weibull distribution, which offers great flexibility in fitting empirical data. The distribution function (Fig. 1a) is

F(t) = 1  exp{[(tt0)/a]b} ,E1

with parameters a, b, and t0. The scale parameter ais related to the values of tand ensures that the distribution is independent of the units of t(e.g. minutes or hours). The constant bis shape parameter. Depending on its value, Weibull distribution can approximate various, even very different shapes (Fig. 5 in Chapter 2). It is suitable for the characterization of time to failure as well as strength or load; therefore, it became popular in reliability assessment. The constant t0 is the threshold value that corresponds to the minimum possible value and characterizes the position of the distribution on the t-axis. (tis the usual symbol for time; for other quantities, other symbols may be used.)

Figure 1.

Weibull distribution functionF(t): (a) original coordinate system, (b) transformed coordinates (Weibull probabilistic paper).

1. Determination of parameters in a two-parameter distribution

The strength or time to failure cannot attain negative values, so that the threshold parameter is often assumed zero, t0 = 0. The distribution function (1) will thus have only two parameters:

F(t) = 1  exp[ (t/a)b] .E2

Parameters aand bcan be found easily, as the transformed data can be fitted by a straight line. Double logarithmic transformation and rearrangement change Equation (2) to

lnt= lna+ (1/b) ln{ln[1/(1 F)]} ,E3

which corresponds to the equation of straight line (Fig. 1b)


where Y= ln t, X= ln{ln[1/(1 – F)]}, A= ln a, B= 1/b.

The method of linearization was very popular in the past, and it is still often used for the determination of parameters from the operation data via a special diagram, called Weibull paper (Fig. 1b). For its construction, the individual measured values tj and the corresponding values Fj of the empirical distribution function are needed. The tj values are obtained by rankordering the ndata from operation (e.g. times to failure) from the minimal value (j= 1) to maximal (j= n). The corresponding values of distribution function are calculated as

Fj=j/ (n+ 1);E5

jis the rank number and nis the total number of measured values. The explanation of formula (5), common for order statistics, is simple. If we have, say, 100 values and order them from the minimal to maximal, then the probability Fthat twill be smaller or equal to the lowest of 100 values, t1, is 1:100. The probability of tt2 is 2/100, etc.; generally, Fj = j/n. In Equation (5), 1 was added to the denominator because of mathematical correctness; the probability Fthat twill be smaller or equal tn must be smaller than 1, simply because if more measurements would be done, values higher than tn could appear. Also other formulas exist for the calculation of empirical Fj values [e.g. Fj = (j– ½)/n], but none can be recommended unequivocally, especially when considering the fact that bigger errors in the determination of distribution parameters can arise due to the small amount of data than due to the formula used for Fj.

The regression constants A, Bcan be obtained by fitting the empirical data by a straight line (using Weibull paper or a program for curve fitting, such as “Insert Trendline” in Excel). Then, the constants in the distribution function (2) are obtained from Aand Bby inverse transformation:

b= 1/B,a= exp (A) .E6

Plotting the empirical data into the coordinate system X= ln{ln[1/(1 – F)]}, Y= ln t, enables a good visual check. In the ideal case, if Equation (2) is valid, the data lie on a straight line.


2. Determination of parameters in a three-parameter distribution

A two-parameter distribution is not always suitable. Sometimes, the transformed data do not lie on a straight line, or it is obvious that the distribution should have a threshold value t0 higher than zero. In such case, the use of a two-parameter distribution as a base for dimensioning could lead to uneconomical design, and a three-parameter function (1) would be better.

The parameters in this distribution can be found by the procedure for a two-parameter function if tin Equation (2) is replaced by the expression tt0; the constant t0 must be chosen in advance. For various t0 values, the shape of empirical distribution varies. The best t0 value is such for which the transformed data best resemble a straight line. However, a more straightforward procedure exists.

Direct determination of parameters

The constants a, b, and t0 can also be obtained in a simpler way without any transformation. The solution of Equation (1) for tgives the formula for quantiles:

t=t0+a{ln[1/(1 F)]1/b} .E7

This equation and the least-squares method are used in search for such values of a, b, and t0, which minimize the sum of squared differences between the measured and the calculated values of t,

(tj,meastj,calc)2= min !E8

If a suitable solver is available for such minimization (one is present also in Excel), it is then sufficient to prepare one series of measured data, tj,meas, and another series of the tj,calc values, calculated via Equation (7) for the same values of Fj using the parameters a, b, and t0. Solver’s command to minimize the expression (8) by changing a, b, and t0 will do the job. An example is shown at the end of this chapter.

Remark: Formula (7) is also suitable for the determination of a ”minimum guaranteed value“(e.g. strength or time to failure) for acceptably low probability F.

In addition to flexibility, Weibull distribution has one more advantage. The shape parameter bin Equation (1) or (2) is related to the character of failures. This is well visible at the bathtub curve (Fig. 1 in Chapter 4). The values b< 1 are typical of decreasing failure rate λand may thus indicate the period of early failures. On the contrary, b>1 corresponds to increasing failure rate λand is typical of the period of aging or wear out. The value b= 1 corresponds to the constant failure rate λ= const, with failures from many various reasons (see Chapter 4). The exponent bthus can inform generally about the possible kind of failures and about the period in the life of an object even if the amount of data is not large. However, caution is necessary. If the data from a long period are fitted by Weibull distribution, failures from various reasons and stages can be mixed, and the relation of bto the kind of failures is not unambiguous.

Remark: Weibull distribution was proposed in 1939 by the Swedish engineer Waloddi Weibull, who studied the strength of materials, life endurance of ball bearings, and fatigue life of mechanical components and other quantities. Later, it appeared that this very useful distribution belongs to the family of extreme value distributions [1, 2]. More on Weibull distribution and its applications can be found, for example, in [3 - 5].


3. Exponential distribution

Let us now look at a special and very important case. With the shape parameter b= 1, Weibull distribution simplifies to exponential distribution

F(t)=1exp[(t/a)],orF(t) = 1exp[ (tt0)/a] .E9

The probability density and distribution function are depicted in Fig. 5. The parameters aand t0 can be determined similarly as described above. If t0 = 0, the remaining parameter ais usually calculated from the mean time to failure, as it will be shown in Chapter 20. Typical of exponential distribution is that the standard deviation has the same or similar value as the mean.

The determination of parameters and use of Weibull and exponential distribution will be demonstrated in the following examples.

Example 1

The strength (S) of a new alloy was measured on seven specimens, with the following results: 203, 223, 248, 265, 290, 313, and 342 MPa. Solve the following three problems:

A. Determine the parameters of Weibull distribution for this alloy using:

  1. Two-parameter distribution and linearized data;

  2. Two-parameter distribution, applying Solver on the original data without transformation); and

  3. Three-parameter distribution, applying Solver on the nontransformed data.

B. Calculate (for each case) the probability that the strength will be lower than 120 MPa.

C. Calculate (for each distribution) the “minimum guaranteed” strength such that the probability of the actual strength being lower equals: 0.05 – 0.01 – 0.001.


Task A. Determination of distribution parameters

  1. Linearized two-parameter Weibull distribution. The strength values, ordered from minimum to maximum, are given in Table 1 together with the values of distribution function, calculated as Fj = j/(n+ 1), with n= 7; see also Fig. 2. The distribution function F(t) = 1 – exp[– (t/a)b] was transformed to linear form; see Equation (4) and the following formulas. The transformed values are in the columns Xj and Yj. Note: The values of distribution function are fixed (deterministic), as they correspond to the number of measured values, whereas the strengths exhibit random variations. Therefore, Fis the independent variable and tis the dependent variable.


Table 1.

Measured values S(Fj) and those calculated using three methods.

Subscript c means calculated; lin2 – linearized, two parameters; sol2 – nonlinearized, Solver, two parameters; sol3 – nonlinearized, Solver, three parameters

The transformed values were fitted by linear function (4); see columns Xj and Yj in Table 2. The regression constants were A= 5.673642 and B= 0.194844. The inverse transformation has given a= exp A= 291.0928 and b= 1/B= 5.132311, so that the two-parameter distribution function is F(t) = 1 – exp[– (S/291.0928)5.13231]. The corresponding calculated values Sj are in column Sj,c,lin2 and depicted by a curve in Fig. 2.

  1. Two-parameter distribution, application of Solver on untransformed data. In this case, the strength values tj,calc were calculated for the individual Fj values using Equation (7), with t0 = 0. (See column tj,calc,2p.) Now, a quantity for characterization of the quality of the fit was defined: sum of the squared differences of the measured and calculated strengths, ∑(tmeastcalc)2. The Solver then changes the constants aand bof the distribution function (7) automatically until the sum of squared differences attains a minimum. In the investigated example, the “optimum” constants were a= 291.7807 and b= 4.964505, near to the results of the linearized problem. The calculated values are in column Sj,c,sol2.

  2. Three-parameter distribution, application of Solver on untransformed data. The difference from the previous case is the full form of distribution function (7). Also here the sum of the squared differences of the measured and calculated strength, ∑(tmeastcalc)2, was minimized. The resultant constants were a= 154.9796, b= 2.4156, and t0 = 133.7975. The calculated values are in column Sj,c,sol2. The calculated distribution function is plotted by the thick curve in Fig. 2. Also the distribution function of a two-parameter distribution function is shown (thin curve). The curves for cases (a) and (b) were very close to each other.

Task B. Determination of probability S ≤ 120 MPa

The probabilities are as follows:

  1. 0.010533, (b) 0.012069, and (c) 0; the minimum possible value is t0 = 133.8 MPa.

Task C. Determination of guaranteed strength

The results are in the following table.

Probability of lower strengthGuaranteed strength (MPa) with the constants from the method:

Note the big difference between the two- and three-parameter distributions for very low failure probabilities (cf. also Fig. 2). According to the three-parameter model, the minimum (threshold) strength is 133.8 MPa.

Figure 2.

Measured values of strength (S) and approximate distribution functions (F) for various approximations in Example 1. Thick curve – case c, three-parameter function; thin curves – cases a, b, two-parameter curves.

Example 2

Eight components (n= 8) were tested until failure. The failures occurred at the following times tj: 65, 75, 90, 120, 250, 510, 520, and 760 h. Calculate the mean time to failure and failure rate. Calculate also the standard deviation, so that you can assess whether exponential distribution may be used for the time to failure.


MTTF= ∑tj/n= (65+75+90+120+250+510+520+760)/8 = 298.750 h.

The sample standard deviation [Equation (4) in Chapter 2] is σMTTF = 264.288 h. This is reasonably close to the sample mean, and an exponential distribution may be assumed. For this case, failure rate λ= 1/MTTF= 1/298.75 = 0.003347 h–1. The determination of confidence interval for λwill be demonstrated in a similar case in Chapter 20.


  1. 1. Rao S S. Reliability-Based Design. New York: McGraw-Hill; 1992. 569 p.
  2. 2. Gumbel J E. Statistics of Extremes. New York: Columbia University Press; 1958. 375 p.
  3. 3. Bentley J P. Introduction to Reliability and Quality Engineering. Harlow, England: Addison-Wesley; 1999. 202 p.
  4. 4. Abernethy R B. The New Weibull Handbook. 4th ed. North Palm Beach, Florida: Robert B Abernethy; 2001. ISBN 0-9653062-1-6.
  5. 5. VDA 3. Zuverlässigkeitssicherung bei Automobilherstellern und Lieferanten. Frankfurt am Main: Verband der Automobilindistrie e.V; 1984.

Written By

Jaroslav Menčík

Submitted: January 8th, 2016 Reviewed: February 3rd, 2016 Published: April 13th, 2016