Rayleigh Distribution with MLE Solution: Difference between revisions

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{{Reference Example}}
{{Reference Example}}


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




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{| {{table}}
{| {{table|15%}}
! State F/S
!State F/S
!Time to F/S
!Time to F/S
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Latest revision as of 16:14, 28 September 2015

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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.


Data

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


Result

  • The model parameter is [math]\displaystyle{ \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]\displaystyle{ \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
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