For a Binomial distribution, \(B(n, p)\), where \(p + q = 1\), the sum and product of mean and variance are 8 and 12 respectively, then the value of \(n\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 9
For a binomial distribution, the mean is \(np\) and the variance is \(npq\). Their sum and product are 8 and 12, so they are the roots of
\(t^2 - 8t + 12 = 0\)
\((t - 6)(t - 2) = 0\)
\(t = 6 \text{ or } 2\)
Since \(q \lt 1\), the variance is less than the mean, so
\(np = 6,\quad npq = 2\)
\( \displaystyle q = \frac{npq}{np} = \frac{2}{6} = \frac{1}{3},\quad p = \frac{2}{3} \)
\( \displaystyle n = \frac{6}{p} = 6 \times \frac{3}{2} = 9 \)