In Binomial distribution with parameters n = 12 and \(p=\displaystyle\frac{1}{3}\), value of \(E(X^2)+E(X)\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{68}{3}\)
For a binomial distribution, the mean is np and the variance is npq, with q = 1 − p = 2/3:
\( \displaystyle E(X) = 12 \times \frac{1}{3} = 4 \)
\( \displaystyle \mathrm{Var}(X) = 12 \times \frac{1}{3} \times \frac{2}{3} = \frac{8}{3} \)
Since \(\mathrm{Var}(X) = E(X^2) - [E(X)]^2\):
\( \displaystyle E(X^2) = \frac{8}{3} + 16 = \frac{56}{3} \)
Hence
\( \displaystyle E(X^2) + E(X) = \frac{56}{3} + 4 \)
\( \displaystyle = \frac{68}{3} \)