A bag contains 30 white balls and 10 red ball | vedclass

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(B) The total number of balls is $30 + 10 = 40$. The probability of drawing a white ball is $p = \frac{30}{40} = \frac{3}{4}$. The probability of draw

Vedclass · Accepted Answer

$4\sqrt{3}$

Vedclass · Answer

(B) The total number of balls is $30 + 10 = 40$. The probability of drawing a white ball is $p = \frac{30}{40} = \frac{3}{4}$. The probability of drawing a red ball is $q = 1 - p = \frac{1}{4}$. The number of trials is $n = 16$. Since the balls are drawn with replacement,$X$ follows a binomial distribution $B(n, p)$. The mean of $X$ is $E(X) = np = 16 	imes \frac{3}{4} = 12$. The standard deviation of $X$ is $\sigma = \sqrt{npq} = \sqrt{16 	imes \frac{3}{4} 	imes \frac{1}{4}} = \sqrt{3}$. Therefore,the ratio is $\frac{	ext{mean}}{	ext{standard deviation}} = \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}$.

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

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