The discrete random variables X and Y are ind | vedclass

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(A) For a binomial distribution $X \sim B(n, p)$,the variance is given by $Var(X) = n 	imes p 	imes q$,where $q = 1 - p$. Given $X \sim B(n_1, 0.5)$

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$7$

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(A) For a binomial distribution $X \sim B(n, p)$,the variance is given by $Var(X) = n 	imes p 	imes q$,where $q = 1 - p$. Given $X \sim B(n_1, 0.5)$,so $p_1 = 0.5$ and $q_1 = 1 - 0.5 = 0.5$. The variance is $n_1 	imes 0.5 	imes 0.5 = 6$. $n_1 	imes 0.25 = 6 \Rightarrow n_1 = \frac{6}{0.25} = 24$. Given $Y \sim B(n_2, 0.4)$,so $p_2 = 0.4$ and $q_2 = 1 - 0.4 = 0.6$. The variance is $n_2 	imes 0.4 	imes 0.6 = 6$. $n_2 	imes 0.24 = 6 \Rightarrow n_2 = \frac{6}{0.24} = 25$. Therefore,$\sqrt{n_1 + n_2} = \sqrt{24 + 25} = \sqrt{49} = 7$.

The discrete random variables $X$ and $Y$ are independent from one another and are defined as $X \sim B(n_1, 0.5)$ and $Y \sim B(n_2, 0.4)$. If the variance of both $X$ and $Y$ is $6$,then $\sqrt{n_1+n_2}=$

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