Chapter 9: Problem 25
We say that \(\zeta\) is a \(p\) -percentile of the distribution \(F\) if \(F(\zeta)=p\). Show that if \(\zeta\) is a \(p\) -percentile of the IFRA distribution \(F\), then $$ \begin{array}{ll} \bar{F}(x) \leqslant e^{-\theta x}, & x \geqslant \zeta \\ \bar{F}(x) \geqslant e^{-\theta x}, & x \leqslant \zeta \end{array} $$ where $$ \theta=\frac{-\log (1-p)}{\zeta} $$
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