Question

A bag contains $30$ white balls and $10$ red balls. $16$ are drawn one by one randomly from the bag with replacement. If $X$ be the number of white balls drawn, then $\left(\frac{\text{mean of X}}{\text{standard deviation of X}}\right)$ is equal to :

• A. $3\sqrt{2}$
• B. $\frac{4\sqrt{3}}{3}$
• C. $4\sqrt{3}$
• D. $4$

Answer 1

The correct option is C $4\sqrt{3}$

$p=\frac{30}{40}=\frac{3}{4}q=\frac{10}{40}=\frac{1}{4}n=16$
$\therefore$ Mean, $\mu =np=16\times \frac{3}{4}=12$

Standard deviation, $\sigma =\sqrt{npq}=\sqrt{16\times \frac{3}{4}\times \frac{1}{4}}=\sqrt{3}$
Hence, $\left(\frac{\text{mean of X}}{\text{standard deviation of X}}\right)=\frac{12}{\sqrt{3}}=4\sqrt{3}$