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Rayleigh Distribution with MLE Solution
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This example compares the results for a Rayleigh distribution (1P-Weibull with beta = 2).
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
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| S |
50
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| S |
50
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| S |
50
|
Result
- The model parameter [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++
- [math]\displaystyle{ \hat{\lambda} = \left (\frac{2}{\hat{\eta}} \right)^{2} = \left (\frac{2}{46.821851} \right)^{2} = 9.12289\times 10^{-4}\,\! }[/math]
- The standard deviation is:
