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| ====Probability Plotting Example for the 1-parameter Exponential Distribution====
| | #REDIRECT [[Probability Plotting Example]] |
| Let's assume six identical units are reliability tested at the same application and operation
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| stress levels. All of these units fail during the test after operating for the following times (in hours), <math>{{T}_{i}}</math> : 96, 257, 498, 763, 1051 and 1744.
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| <br>
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| The steps for determining the parameters of the exponential <math>pdf</math> representing the
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| data, using probability plotting, are as follows:
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| <br>
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| Rank the times-to-failure in ascending order as shown next.
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| <br>
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| ::<math>\begin{matrix}
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| \text{Time-to-} & \text{Failure Order Number} \\
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| \text{failure, hr} & \text{out of a Sample Size of 6} \\
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| \text{96} & \text{1} \\
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| \text{257} & \text{2} \\
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| \text{498} & \text{3} \\
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| \text{763} & \text{4} \\
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| \text{1,051} & \text{5} \\
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| \text{1,744} & \text{6} \\
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| \end{matrix}</math>
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| <br>
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| Obtain their median rank plotting positions.
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| <br>
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| Median rank positions are used instead of other ranking methods because median ranks are at a
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| specific confidence level (50%).
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| <br>
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| The times-to-failure, with their corresponding median ranks, are shown next:
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| <br>
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| ::<math>\begin{matrix}
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| \text{Time-to-} & \text{Median} \\
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| \text{failure, hr} & \text{Rank, }% \\
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| \text{96} & \text{10}\text{.91} \\
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| \text{257} & \text{26}\text{.44} \\
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| \text{498} & \text{42}\text{.14} \\
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| \text{763} & \text{57}\text{.86} \\
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| \text{1,051} & \text{73}\text{.56} \\
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| \text{1,744} & \text{89}\text{.10} \\
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| \end{matrix}</math>
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| <br>
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| On an exponential probability paper, plot the times on the x-axis and their corresponding
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| rank value on the y-axis. Fig. 4 displays an example of an exponential probability paper. The
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| paper is simply a log-linear paper.
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| <br>
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| [[File:ALTA4.1.gif|center]]
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| <br>
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| <center>Fig. 4: Sample exponential probability paper.</center>
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| <br>
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| Draw the best possible straight line that goes through the <math>t=0</math> and <math>
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| (t)=100%</math> point and through the plotted points (as shown in Fig. 5).
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| <br>
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| At the <math>Q(t)=63.2%</math> or <math>R(t)=36.8%</math> ordinate point, draw a
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| straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hr which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%).</math>
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| <br>
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| [[File:ALTA4.2.gif|center]]
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| <br>
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| <center>Fig. 5: Probability plot for Example 1.</center>
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| <br>
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| <br>
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| Now any reliability value for any mission time <math>t</math> can be obtained. For example, the
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| reliability for a mission of 15 hr, or any other time, can now be obtained either from the plot or analytically (i.e. using the equations given in Section <math>5.1.1</math> ).
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| <br>
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| To obtain the value from the plot, draw a vertical line from the abscissa, at <math>t=15</math>
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| hr, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read
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| <math>R(t)</math> . In this case, <math>R(t=15)=98.15%</math> . This can also be obtained
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| analytically, from the exponential reliability function.
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