In Binomial distribution with parameters n = 100 and p, Variance of distribution is maximum when p is equal to :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{2}\)
The variance of a binomial distribution is npq, with q = 1 − p:
\(V = 100\,p(1-p) = 100\left(p - p^{2}\right)\)
Differentiating with respect to p:
\( \displaystyle \frac{dV}{dp} = 100(1 - 2p) = 0 \)
\( \displaystyle p = \frac{1}{2} \)
Since \(\displaystyle\frac{d^{2}V}{dp^{2}} = -200 \lt 0\), this gives the maximum variance, 100 × \(\frac{1}{2}\) × \(\frac{1}{2}\) = 25.