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| | #REDIRECT [[Parameter_Estimation#Median_Ranks]] |
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| Median ranks are used to obtain an estimate of the unreliability, <math>Q({{T}_{j}}),</math> for each failure at a <math>50%</math> confidence level. In the case of grouped data, the ranks are estimated for each group of failures, instead of each failure.
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| For example, when using a group of 10 failures at 100 hours, 10 at 200 hours and 10 at 300 hours, Weibull++ estimates the median ranks (<math>Z</math> values) by solving the cumulative binomial equation with the appropriate values for order number and total number of test units.
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| For 10 failures at 100 hours, the median rank, <math>Z,</math> is estimated by using:
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| ::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
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| N \\
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| k \\
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| \end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}</math>
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| with:
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| ::<math>N=30,\text{ }J=10</math>
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| where one <math>Z</math> is obtained for the group, to represent the probability of 10 failures occurring out of 30.
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| For 10 failures at 200 hours, <math>Z</math> is estimated by using:
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| ::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
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| N \\
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| k \\
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| \end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}</math>
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| where:
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| ::<math>N=30,\text{ }J=20</math>
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| to represent the probability of 20 failures out of 30.
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| For 10 failures at 300 hours, <math>Z</math> is estimated by using:
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| ::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
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| N \\
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| k \\
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| \end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}</math>
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| where:
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| ::<math>N=30,\text{ }J=30</math>
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| to represent the probability of 30 failures out of 30.
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| =Alternative Computation=
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| Any rank can be computed by:
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| ::<math>M{{R}_{i}}=\frac{\frac{i}{N-i+1}}{{{F}_{1-\alpha ,2(N-i+1),2i}}+\frac{i}{N-i+1}}</math>
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| *where F is the F-distribution and
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| *<math>1-\alpha</math> is the confidence limit.
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| The Median Rank is obained by setting:
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| ::<math>1-\alpha=0.50</math>
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| or
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| ::<math>M{{R}_{i}}=\frac{\frac{i}{N-i+1}}{{{F}_{0.50 ,2(N-i+1),2i}}+\frac{i}{N-i+1}}</math>
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