Median Ranks: Difference between revisions
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For 10 failures at 100 hours, the median rank, <math>Z,</math> is estimated by using: | For 10 failures at 100 hours, the median rank, <math>Z,</math> is estimated by using: | ||
<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix} | ::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix} | ||
N \\ | N \\ | ||
k \\ | k \\ | ||
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with: | with: | ||
<math>N=30,\text{ }J=10</math> | ::<math>N=30,\text{ }J=10</math> | ||
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For 10 failures at 200 hours, <math>Z</math> is estimated by using: | For 10 failures at 200 hours, <math>Z</math> is estimated by using: | ||
<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix} | ::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix} | ||
N \\ | N \\ | ||
k \\ | k \\ | ||
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where: | where: | ||
<math>N=30,\text{ }J=20</math> | ::<math>N=30,\text{ }J=20</math> | ||
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For 10 failures at 300 hours, <math>Z</math> is estimated by using: | For 10 failures at 300 hours, <math>Z</math> is estimated by using: | ||
<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix} | ::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix} | ||
N \\ | N \\ | ||
k \\ | k \\ | ||
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where: | where: | ||
<math>N=30,\text{ }J=30</math> | ::<math>N=30,\text{ }J=30</math> | ||
to represent the probability of 30 failures out of 30. | to represent the probability of 30 failures out of 30. |
Revision as of 16:08, 29 June 2011
Median ranks are used to obtain an estimate of the unreliability, [math]\displaystyle{ Q({{T}_{j}}), }[/math] for each failure at a [math]\displaystyle{ 50% }[/math] confidence level. In the case of grouped data, the ranks are estimated for each group of failures, instead of each failure. 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]\displaystyle{ Z }[/math] values) by solving the cumulative binomial equation with the appropriate values for order number and total number of test units. For 10 failures at 100 hours, the median rank, [math]\displaystyle{ Z, }[/math] is estimated by using:
- [math]\displaystyle{ 0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix} N \\ k \\ \end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}} }[/math]
with:
- [math]\displaystyle{ N=30,\text{ }J=10 }[/math]
where one [math]\displaystyle{ Z }[/math] is obtained for the group, to represent the probability of 10 failures occurring out of 30.
For 10 failures at 200 hours, [math]\displaystyle{ Z }[/math] is estimated by using:
- [math]\displaystyle{ 0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix} N \\ k \\ \end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}} }[/math]
where:
- [math]\displaystyle{ N=30,\text{ }J=20 }[/math]
to represent the probability of 20 failures out of 30.
For 10 failures at 300 hours, [math]\displaystyle{ Z }[/math] is estimated by using:
- [math]\displaystyle{ 0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix} N \\ k \\ \end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}} }[/math]
where:
- [math]\displaystyle{ N=30,\text{ }J=30 }[/math]
to represent the probability of 30 failures out of 30.