Question

If for a binomial distribution mean is $\frac{10}{3}$ and sum of mean and variance is $\frac{40}{9}$. Then the parameters are

• A. $\frac{2}{3},10$
• B. $\frac{2}{3},20$
• C. $5,\frac{2}{3}$
• D. $4,\frac{2}{3}$

Answer 1

The correct option is C $5,\frac{2}{3}$

Mean $=np=\frac{10}{3}$
Variance $=npq=\frac{40}{9}$
The sum of Mean and variance,
$\Rightarrow np+np(1-p)=\frac{40}{9}$

$\Rightarrow np(2-p)=\frac{40}{9}$

$\Rightarrow \frac{10}{3}(2-p)=\frac{40}{9}$

$\Rightarrow 2-p=\frac{4}{3}$
$\therefore p=\frac{2}{3}$
Hence, $n=5$