Originally Posted by

**TKHunny** 50 balls! This is a nicely large sample to suggest the use of the Normal Distribution as a reasonable approximation to whatever the given distribution is.

Distribution for 1 ball

X = Payoff Random Variable

E[X] = (1/2)(1) + (1/4)*(2) + (1/4)(0) = 1

Mean(X) = 1

E[X^2] = (1/2)(1)^2 + (1/4)*(2)^2 + (1/4)(0)^2 = 1.5

Var(X) = E[X^2] - (E[X])^2 = 1.50 - 1 = 0.5

Then we have:

Mean(50*X) = 50*Mean(X) = 50

Var(50*X) = (50^2)*Var(X) = 1250

Standard Deviation (50*X) = sqrt(Var(50*X)) = 35.355

Now what?