Q53 / 90 · JEE Main 2019 (11 Jan Shift 2) · Mathematics

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:

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

Solution

$P($ white ball $)=\frac{30}{40}=\frac{3}{4}, Q$ (red ball) $=\frac{10}{40}=\frac{1}{4}, n=16$ $\frac{\text { Mean of } X}{\text { standard deviation of } X}=\frac{n P}{\sqrt{n P Q}}=\frac{\sqrt{n P}}{\sqrt{Q}}$ $=\sqrt{\frac{16 \times \frac{3}{4}}{\frac{1}{4}}}=\sqrt{48}=4 \sqrt{3}$

Hence, the correct answer is option: 'B'.

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