Rayleigh Distribution with MLE Solution

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

Results in Weibull++

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

• The mean life is:

• The standard deviation is: