= Extreme-order-statistic tolerance interval
{title2=$n\approx\vartheta/\varepsilon,\quad(1+\vartheta)e^{-\vartheta}=\delta$}
Let $X_1,\ldots,X_n$ be <independent and identically distributed> observations from a continuous <cumulative distribution function> $F$, with $n\geq2$. The interval from the sample minimum to maximum contains random population probability $C=F(X_{(n)})-F(X_{(1)})$. The <probability integral transform> identifies $C$ with the <uniformized sample range>, so
$$
\mathbb P(C<1-\varepsilon)=(1-\varepsilon)^{n-1}[1+(n-1)\varepsilon],\qquad0<\varepsilon<1.
$$
For small $\varepsilon$, setting $n\varepsilon\approx\vartheta$ makes this approximately $(1+\vartheta)e^{-\vartheta}$. Solve $(1+\vartheta)e^{-\vartheta}=\delta$ for the unique positive $\vartheta$ when $0<\delta<1$, then use the exact displayed inequality to choose an integer sample size.
This is a distribution-free tolerance interval for population content under the continuous iid model. It differs from a <confidence interval> for a fixed parameter. Atoms or dependent observations require their own treatment; uniformizing them does not automatically preserve the same ordinary sample-range calculation.
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