%I #14 Dec 19 2012 13:32:18
%S 1,9,5,9,9,6,3,9,8,4,5,4,0,0,5,4,2,3,5,5,2,4,5,9,4,4,3,0,5,2,0,5,5,1,
%T 5,2,7,9,5,5,5,5,0,0,7,7,8,6,9,5,4,8,3,9,8,4,7,6,9,5,2,6,4,6,3,6,1,6,
%U 3,5,2,7,4,1,4,4,8,8,2,6,6,7,7,9,8,2,5,4,7,0,9,4,9,2,8,1,4,2,0,6,0,1,7,9,9
%N Decimal expansion of the standard normal deviate for a 95% confidence interval.
%H Eric Weisstein's MathWorld, <a href="http://mathworld.wolfram.com/ConfidenceInterval.html">Confidence Interval</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/1.96">1.96</a>
%F Solution to (2/sqrt(2*pi))*integral_{t..infinity} exp(-x^2/2) dx = 5/100.
%e 1.95996398454005423552459443052055152795555...
%t RealDigits[N[Sqrt[2]*InverseErf[95/100], 105]][[1]]
%K cons,nonn
%O 1,2
%A _Jean-François Alcover_, Dec 17 2012