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A219330 Number of random selections (with replacement) needed from a normal population to assure a greater than one-half chance that the selected group contains the top 10th percentile individual, top 1st percentile individual, the 0.1 percentile, 0.01 percentile etc... 0

%I #21 May 31 2022 16:29:01

%S 7,69,693,6932,69315,693147,6931472,69314718,693147181,6931471806,

%T 69314718056,693147180560,6931471805600,69314718055995,

%U 693147180559945,6931471805599453,69314718055994531,693147180559945310,6931471805599453094,69314718055994530942

%N Number of random selections (with replacement) needed from a normal population to assure a greater than one-half chance that the selected group contains the top 10th percentile individual, top 1st percentile individual, the 0.1 percentile, 0.01 percentile etc...

%C a(n) is the least number of random selections (with replacement) required that assures a group having at least a 0.5 probability of containing the top 100/(10^n)th percentile individual. Converges toward log(2)*10^n with increasing n.

%D F. Reif, Fundamentals of Statistical and Thermal Physics, McGraw-Hill, 1965

%F a(n) = ceiling((log(2))/(log(10^n/(10^n-1)))).

%e To assure a greater than 1/2 chance of finding an individual with, say, an IQ in the top one out of a thousand or 1/(10^3), you must select at least (with replacement) a(3) = 693 individuals.

%o (PARI) a(n)=ceil((log(2))/(log(10^n/(10^n-1)))); \\ _Joerg Arndt_, Apr 15 2013

%Y Cf. A002162, A014088, A033810, A050255.

%K nonn

%O 1,1

%A _Zacariaz Martinez_, Apr 11 2013

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