Determine MTBF Given a Weibull Distribution

Determine MTBF Given a Weibull Distribution

Gary A. K. Reliable & regal 1000-block Nelson.
Gary A. K.
Reliable & regal 1000-block Nelson.

First off, not sure why anyone would want to do this, yet one of the issues I’ve heard concerning abandoning the use of MTBF is client ask for MTBF. If they will not accept reliability probabilities at specific durations, and insist on using MTBF, you probably should provide a value to them.

Let’s say you have a Weibull distribution model that described the time to failure distribution of your product. You’ve done the testing, modeling, and many field data analysis and know for the requestor’s application this is the best estimate of reliability performance. You can, quite easily calculate the MTBF value.

As you know, if theβ parameter is equal to one then the characteristic life, η, is equal to MTBF. If β is less than or greater than one, then use the following formula to determine the mean value, MTBF, for the distribution.

$latex \displaystyle&s=4 \mu =\eta \Gamma \left( 1+\frac{1}{\beta } \right)$

You’ll need the Gamma function and the Weibull parameters. The further β is from one, the bigger the difference between η and MTBF.

You can find a little more information and background at the article Calculate the Mean and Variance on the site.

Author: Fred Schenkelberg

I am an experienced reliability engineering and management consultant with my firm FMS Reliability. My passion is working with teams to create cost-effective reliability programs that solve problems, create durable and reliable products, increase customer satisfaction, and reduce warranty costs.

6 thoughts on “Determine MTBF Given a Weibull Distribution”

  1. Hi,

    I’m doing my best to better understand how to move away from the misunderstandings caused by using MTTF and/or MTBF when generating/testing specifications in new product development. However, I find it confusing that, in a previous post about the use of MTBF in a fan data sheet, you say that MTBF assumes a constant failure rate, whilst above you generalise its calculation to any Weibull distribution. Does the definition/use of MTBF only apply to beta=1, or is that again part of the whole confusion?

    1. Hi Nicola,

      Yes, it is part of the confusion. One can calculate MTBF from any set of data. The failure pattern might be increasing, decreasing or whatever, and you can tally total time and divide by the number of failures.

      The hard part is when only given MTBF to represent reliability; you do not have any information on the changing nature of the failure rate over time. Thus must are left to assume a constant hazard rate, which, is rarely if ever true.

      I find MTBF oversimplifies failure data to the point of making the summary without value.



      1. Thanks a lot!
        There is much confusion out there and your work is helping me a lot to shed some light on the whole reliability and MTTF testing/analysis affair.

    1. Sorry for the late reply.
      I haven’t got a lot of examples to share, it’s mostly a story of a hard to demonstrate reliability goal. I am working on an electro-mechanical device and I reviewed some old reliability testing. Since the device is quite expensive and requires almost continuous attention to run, the shortcut of assuming beta=1 was taken in the past to reduce the no. of samples and test hours. Subsequently someone realised that and fit a Weibull distribution, although over only 1 failure and 9 right censored data… Again, an unreliable result.
      My approach is now to use competitive failure modes and analyse reliability of the critical components of the device independently, using a semi-quantitative assessment as we cannot do a comprehensive statistical analysis.


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