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| '''Standard Actuarial Example'''
| | #REDIRECT [[Non-Parametric_Life_Data_Analysis]] |
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| Find reliability estimates for the data in Example 10 using the standard actuarial method.
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| '''Solution'''
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| The solution to this example is similar to that of Example 10, with the exception of the inclusion of the <math>n_{i}^{\prime }</math> term, which is used in Eqn. (standact). Applying this equation to the data, we can generate the following table:
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| <center><math>\begin{matrix}
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| Start & End & Number of & Number of & Adjusted & {} & {} \\
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| Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, n_{i}^{\prime } & 1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} & \mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} \\
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| 0 & 50 & 2 & 4 & 53 & 0.962 & 0.962 \\
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| 50 & 100 & 0 & 5 & 46.5 & 1.000 & 0.962 \\
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| 100 & 150 & 2 & 2 & 43 & 0.953 & 0.918 \\
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| 150 & 200 & 3 & 5 & 37.5 & 0.920 & 0.844 \\
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| 200 & 250 & 2 & 1 & 31.5 & 0.937 & 0.791 \\
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| 250 & 300 & 1 & 2 & 28 & 0.964 & 0.762 \\
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| 300 & 350 & 2 & 1 & 25.5 & 0.922 & 0.702 \\
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| 350 & 400 & 3 & 3 & 21.5 & 0.860 & 0.604 \\
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| 400 & 450 & 3 & 4 & 15 & 0.800 & 0.484 \\
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| 450 & 500 & 1 & 2 & 9 & 0.889 & 0.430 \\
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| 500 & 550 & 2 & 1 & 6.5 & 0.692 & 0.298 \\
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| 550 & 600 & 1 & 0 & 4 & 0.750 & 0.223 \\
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| 600 & 650 & 2 & 1 & 2.5 & 0.200 & 0.045 \\
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| \end{matrix}</math></center>
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| As can be determined from the preceding table, the reliability estimates for the failure times are:
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| <center><math>\begin{matrix}
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| Failure Period & Reliability \\
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| End Time & Estimate \\
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| 50 & 96.2% \\
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| 150 & 91.8% \\
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| 200 & 84.4% \\
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| 250 & 79.1% \\
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| 300 & 76.2% \\
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| 350 & 70.2% \\
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| 400 & 60.4% \\
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| 450 & 48.4% \\
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| 500 & 43.0% \\
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| 550 & 29.8% \\
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| 600 & 22.3% \\
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| 650 & 4.5% \\
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| \end{matrix}</math></center>
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