Question

If the sum of mean and Variance of a binomial distribution for $14$ pairs is $\frac{48}{7},$ then the variance is :

• A. $18/7$
• B. $19/7$
• C. $20/7$
• D. $21/7$

Answer 1

The correct option is C $20/7$

$np+npq=\frac{48}{7}$

$p+pq=\frac{48}{7n}=\frac{48}{7\times 14}=\frac{24}{49}$

$p(1+1-p)=\frac{24}{49}$

$p(2-p)=\frac{24}{49}$

$\Rightarrow 2p-p^2=\frac{24}{49}\Rightarrow p^2-2p=\frac{-24}{49}$

$\Rightarrow p=\frac{2}{7}$ or $\frac{12}{7}$

$\because$ $p\le 1$ $\therefore$ $p=\frac{2}{7}$

$\therefore$ $q=\frac{5}{7}$ $\therefore$ variance $=npq=14\times \frac{2}{7}\times \frac{5}{7}=\frac{20}{7}$ [C]