Question

In a binomial distribution, the mean is $15$ and variance of $10$. Then, parameter $n$ is

• A. $28$
• B. $16$
• C. $45$
• D. $25$

Answer 1

Answer: (C)

$45$

Given, mean, $np=15$
and variance, $npq=10$
$\Rightarrow np(1-p)=10$
$\Rightarrow 15(1-p)=10$
$\Rightarrow 1-p=\frac{2}{3}$
$\Rightarrow p=1-\frac{2}{3}\Rightarrow p=\frac{1}{3}$
$\therefore n\times \left(\frac{1}{3}\right)=15\Rightarrow n=45$.