Question

In a binomial distribution, the mean is $4$ and variance is $3$. Then its mode is :

• A. 5
• B. 6
• C. 4
• D. None of these

Answer 1

Answer: (C)

4

Answer 2

In Binomial distribution, Mean $= np$,
Variance $= npq$ and the mode is $r$
if for $x = r$, the probability function $p(x)$ is maximum.
Given $np = 4$ and $npq = 3$
$\therefore q = \frac{3}{4} p = 1 -q = 1 - \frac{3}{4} = \frac{1}{4}$
Also, $n = \frac{4}{p} = \frac{4}{1/4} = 16$
Now , $\left(n+1\right)p = \left(16 +1\right) \frac{1}{4} = \frac{17}{4} = 4 + \frac{1}{4}$
$\Rightarrow$ The distribution will have unique mode (unimodal) & the mode $= 4$