Life of Incandescent Light Bulbs: Difference between revisions
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'''Do the following:''' | '''Do the following:''' | ||
#Plot the data on a Weibull probability plot and obtain the Weibull model parameters. | #Plot the data on a Weibull probability plot and obtain the Weibull model parameters. | ||
#Compute the B10 life of the bulbs. | #Compute the B10 life of the bulbs. | ||
<br> | |||
The median ranks for the the <math>{{j}^{th}}</math> failure out of N units is obtained by solving the cumulative binomial equation for <math>Z</math> . This however requires numerical solution. Tables of median ranks can be used in lieu of the solution. | |||
represents the rank, or unreliability estimate, for the failure[15; 16] in the following equation for the cumulative binomial: | |||
<math>P=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix} | <math>P=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix} | ||
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k \\ | k \\ | ||
\end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}</math> | \end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}</math> | ||
[[Category:Weibull_Examples]] | [[Category:Weibull_Examples]] |
Revision as of 03:31, 25 June 2011
This example uses time-to-failure data from a life test done on incandescent light bulbs. The observed times-to-failure are given in the next table.
Order Number | Hours-to-failure |
---|---|
1 | 361 |
2 | 680 |
3 | 721 |
4 | 905 |
5 | 1010 |
6 | 1090 |
7 | 1157 |
8 | 1330 |
9 | 1400 |
10 | 1695 |
Do the following:
- Plot the data on a Weibull probability plot and obtain the Weibull model parameters.
- Compute the B10 life of the bulbs.
The median ranks for the the [math]\displaystyle{ {{j}^{th}} }[/math] failure out of N units is obtained by solving the cumulative binomial equation for [math]\displaystyle{ Z }[/math] . This however requires numerical solution. Tables of median ranks can be used in lieu of the solution.
represents the rank, or unreliability estimate, for the failure[15; 16] in the following equation for the cumulative binomial:
[math]\displaystyle{ P=\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 }[/math] is the sample size and [math]\displaystyle{ j }[/math] the order number. The median rank is obtained by solving this equation for [math]\displaystyle{ Z }[/math] at [math]\displaystyle{ P=0.50, }[/math]
[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]