Rayleigh Distribution with MLE Solution

This example compares the results for a Rayleigh distribution (1-parameter Weibull with beta = 2).

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


 * 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


 * 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: