Rayleigh Distribution with MLE Solution: Difference between revisions
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Kate Racaza (talk | contribs) (Created page with '{{Reference Example}} This example compares the results for a Rayleigh distribution (1P-Weibull with beta = 2). {{Reference_Example_Heading1}} The data set is from Example 5.…') |
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{{Reference Example}} | {{Reference Example}} | ||
This example | This example validates the results for a Rayleigh distribution (1-parameter Weibull with beta = 2) in Weibull++ standard folios. | ||
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{{Reference_Example_Heading2}} | {{Reference_Example_Heading2}} | ||
{| {{table}} | {| {{table|15%}} | ||
! State F/S | !State F/S | ||
!Time to F/S | !Time to F/S | ||
|- | |- | ||
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{{Reference_Example_Heading3}} | {{Reference_Example_Heading3}} | ||
* The model parameter <math>\hat{\lambda} = \left (\frac{2}{\hat{\eta}} \right)^{2} = 9.12289\times 10^{-4}\,\!</math> | * 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 Mean Life is 41.49 | ||
* The Standard Deviation is 21.70 | * The Standard Deviation is 21.70 | ||
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{{Reference_Example_Heading4}} | {{Reference_Example_Heading4}} | ||
* The model parameter | * 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> | ::<math>\hat{\lambda} = \left (\frac{2}{\hat{\eta}} \right)^{2} = \left (\frac{2}{46.821851} \right)^{2} = 9.12289\times 10^{-4}\,\!</math> | ||
Latest revision as of 16:14, 28 September 2015
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