The Variance of the following probability distribution of X is X 0 1 2 3 P(X) q^3 3q^2 p 3qp^2 p^3 where 0 < p < 1, q = 1 - p

Step-by-step solution

The given probability distribution represents a binomial distribution with parameters n = 3 and probability of success p , since P(X = x) = ³C_ x p^x q^ 3-x for x = 0, 1, 2, 3 . For a binomial distribution, the variance is given by npq . Substituting n = 3 , the variance is 3pq . Alternatively, calculating the variance using expectations: E(X) = x P(x) = 0(q^3) + 1(3q^2 p) + 2(3qp^2) + 3(p^3) E(X) = 3p(q^2 + 2pq + p^2) = 3p(q+p)^2 = 3p E(X^2) = x^2 P(x) = 0(q^3) + 1(3q^2 p) + 4(3qp^2) + 9(p^3) E(X^2) = 3p(q^2 + 4pq