Correct Answer - Option 3 : μ'
(2) + μ'
(1)
Explanation
The moment about origin of binomial distribution is given by
μ'2 = E(X2) = ∑x2(n/x)pxqn – x
⇒ ∑{x(x – x) + x}[n(n – 1)/x(x – 1)](n – 2/x – 2)pxq3 – x
⇒ n(n – 1)p2[∑nx= 2(n – 2/x – 2)px – 2qn – x] + np
⇒ n(n -1)p2 × ( q + p)n – 2 + np
⇒ n(n – 1)p2 + np
⇒ n2p2 – np2 + np
Factorial moment of binomial distribution
μ’(r) = E(x(r)) = ∑nx = 0 × x2p(x)
⇒ ∑nx = 0 × x(r)[n!/x!(n – x)!] × px – rqn – r
⇒ n(r)pr∑nx = r [(n – r)!/(x – r)(n – x)!] × px – rqn – x
⇒ n(r)pr(q+ p)n – r
⇒ n(r)p(r) [ p + q] = 1
⇒ μ’(1) = E(x1)= np = mean
⇒ μ’(2) = E(x2) = n2p2 =n(n – 1)p2
⇒ n2p2 – np2
μ’(2) + μ’(1) = n2p2 – np2 + np
∴ μ’(2)+ μ’(1) = μ’2