Problem 112 Solution

The mean number of reds is 1,000,000 * 0.4 = 400,000.

The standard deviation of the number of reds is sqr(1,000,000 * 0.4 * (1-.4)) =~ 489.9

Let l denote the lower bound of the range, let u denote the upper bound, and let x denote the number of reds drawn:

Pr (l <= x <= u) = .95

Pr (l-400,000 <= x-400,000 <= u-400,000) = .95

Pr ((l-.5-400,000)/489.9 <= Z <= (u+.5-400,000)/489.9) = .95 where Z denotes a random variable distributed according to the standard normal distribution.

Next we want the probability that the number of reds will fall on either side of this range to be .025 per side.

Pr( Z <= (u+.5-400,000)/489.9) = .975

(u+.5-400,000)/489.9 = 1.96

u-399,999.5 = 960.2

u =~ 400,960

Pr( Z <= (l-400,000.5)/489.9) = .025

(l-400,000.5)/489.9 = -1.96

l = 399,040

So the range is 399,040 to 400,960

Michael Shackleford, A.S.A.