Question

Some of the eggs at a market are sold in boxes of six. The number, X, of broken eggs in a box has the probability distribution given in the following table.

x 0 1 2 3 4 5 6
P(X=x) 0.8 0.14 0.03 0.02 0.01 0 0


(Find the expectation and variance of the number of unbroken eggs in a box)

Answer 1

E(x) = Summation xp(x) = 0(0.8) + 1(0.14) + 2(0.03) + 3(0.02) + 4(0.01) + 5(0) + 6(0) = 0.14 + 0.06 + 0.06 + 0.04 = 0.3

Variance = [(0 - 0.3)^2 * 0.8 + (1 - 0.3)^2 * 0.14 + (2 - 0.3)^2 * 0.03 + (3 - 0.3)^2 * 0.02 + (4 - 0.3)^2 * 0.01 + (5 - 0.3)^2 * 0 + (6 - 0.3)^2 * 0] = 0.072 + 0.0686 + 0.0867 + 0.1458 + 0.1369 = 0.51