:2. Solution. We switch to a 0-1 box with two tickets marked 1
and eight marked 0. (The number of 5's in 100 draws from the
original box is like the sum of 100 draws from the 0-1 box.) The
box average is .2 and the box SD is the square root of .2 x .8 =
.16, or .4, so the EV of the sum of the draws is .2 x 100 = 20
and the SE of the sum of the draws is .4 x 10 = 4. Since
(12-20)/4 = -8/4 = -2, and since (24-20)/4 = 4/4 = 1, the chance
is approximately equal to the area under the normal curve between
-2 and 1. The area between -2 and 0 is one-half of 95%, or
47.5%. The area between 0 and 1 is one-half of 68%, or 34%. So
the chance that the number of 5's is between 12 and 24 is 47.5 +
34 = 81.5%, which is closest to 82%.