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In Binomial distribution with parameters n=100n = 100 and p, Variance of distribution is maximum when p is equal to :

Solution

✅ Correct Option: 3

Variance of Binomial =npq=np(1−p)= npq = np(1-p).

For fixed nn, V(p)=np(1−p)V(p) = np(1-p). V′(p)=n(1−2p)=0⇒p=12V'(p) = n(1 - 2p) = 0 \Rightarrow p = \frac{1}{2}.

V′′(p)=−2n<0V''(p) = -2n < 0, so maximum at p=12p = \frac{1}{2}.

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