Rayleigh Distribution with MLE Solution

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Rayleigh Distribution with MLE Solution

This example validates the results for a Rayleigh distribution (1-parameter Weibull with beta = 2) in Weibull++ standard folios.

Reference Case

The data set is from Example 5.11 on page 283 in the book Reliability Engineering by Dr. Elsayed, Addison Wesley Longman, Inc, 1996.


State F/S Time to F/S
F 10
F 20
F 30
F 35
F 39
F 42
F 44
S 50
S 50
S 50


  • The model parameter is [math]\hat{\lambda} = \left (\frac{2}{\hat{\eta}} \right)^{2} = 9.12289\times 10^{-4}\,\![/math]
  • The Mean Life is 41.49
  • The Standard Deviation is 21.70

Results in Weibull++

  • The model parameter is:
[math]\hat{\lambda} = \left (\frac{2}{\hat{\eta}} \right)^{2} = \left (\frac{2}{46.821851} \right)^{2} = 9.12289\times 10^{-4}\,\![/math]

Rayleigh results.png

  • The mean life is:
Rayleigh meanlife.png

  • The standard deviation is:
Rayleigh std.png