A ceiling the largest of ten rarely exceeds

Question

Draw 10 independent numbers uniformly from [0,1] and let M be the largest. Find the level t that the maximum stays below exactly 95\% of the time (the 95th percentile of M).

Accepted Answer

The maximum is below t when all ten numbers are below t, so P(M \le t) = t^{10}. Set this equal to 0.95: t^{10} = 0.95 \quad\Rightarrow\quad t = 0.95^{1/10} \approx 0.9949. The 95th percentile of the maximum of ten uniforms sits at about 0.995, extremely close to the top. With ten draws it is nearly certain that at least one lands very near 1. Takeaway. To find a percentile of the maximum, invert t^n: the p-th quantile is p^{1/n}.

A ceiling the largest of ten rarely exceeds

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