A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with…
A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If $X$ be the number of white balls drawn, then
$\left(\frac{\text { mean of } \mathrm{X}}{\text { standard deviation of } \mathrm{X}}\right)$ is equal to:
4
$4 \sqrt{3}$
$3 \sqrt{2}$
$\frac{4 \sqrt{3}}{3}$
Solution
$P($ white ball $)=\frac{30}{40}=\frac{3}{4}, Q$ (red ball) $=\frac{10}{40}=\frac{1}{4}, n=16$
$\frac{\text { Mean of } X}{\text { standard deviation of } X}=\frac{n P}{\sqrt{n P Q}}=\frac{\sqrt{n P}}{\sqrt{Q}}$ $=\sqrt{\frac{16 \times \frac{3}{4}}{\frac{1}{4}}}=\sqrt{48}=4 \sqrt{3}$