Discussion
The failure time of a component is exponentially distributed with mean life 200 hours. The design of the component is modified after which the mean life has increased to 400 hours. What is the amount of increase in reliability at 800 hours ? (Let \(a = e^{-2}\).)
Answer: C
The increase in reliability is \(a(1 - a)\).
For an exponential life with mean \(\theta\), the reliability at time \(t\) is \(R(t) = e^{-t/\theta}\). At 800 hours:
\[ \begin{aligned} R_{\text{old}} &= e^{-800/200} = e^{-4} = a^{2} \\ R_{\text{new}} &= e^{-800/400} = e^{-2} = a \end{aligned} \]
The increase is \(a - a^{2} = a(1 - a) \approx 0.117\).
\(a\) and \(a^{2}\) are the new and old reliabilities themselves, and \((1 - a)/a\) is the relative increase \((a - a^{2})/a^{2}\), not the amount.
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