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A random variable $X$ follows the binomial distribution and $X \sim B(n, 0.3)$. If the mean of $X$ is three times as large as the standard deviation of $X$, then $n=$
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The correct answer is:
$21$
Given, $p=0.3, q=1-p=0.7$
Also, Mean $(\mu)=3 \times$ Standard deviation $(\sigma)$.
$\Rightarrow \quad \mu=3(\sigma)$...(i)
$\begin{aligned} & \text { Now, } \mu=n p=0.3 n \\ & \text { and } \sigma=\sqrt{n p q}=\sqrt{n(0 \cdot 3)(0 \cdot 7)}=\sqrt{0.21 n} \\ & \text { From Eq. (i), }(0.3) n=3 \sqrt{(0.21) n}\end{aligned}$
$\begin{array}{ll}\Rightarrow & (0.1) n=\sqrt{(0.21) n} \\ \Rightarrow & \left(\frac{n}{10}\right)^2=\left(\frac{21 n}{100}\right) \\ \Rightarrow & n^2-21 n=0 \Rightarrow n(n-21)=0 \\ \Rightarrow & n=0,21\end{array}$
But $n \neq 0$
Thus, $n=21$.
Also, Mean $(\mu)=3 \times$ Standard deviation $(\sigma)$.
$\Rightarrow \quad \mu=3(\sigma)$...(i)
$\begin{aligned} & \text { Now, } \mu=n p=0.3 n \\ & \text { and } \sigma=\sqrt{n p q}=\sqrt{n(0 \cdot 3)(0 \cdot 7)}=\sqrt{0.21 n} \\ & \text { From Eq. (i), }(0.3) n=3 \sqrt{(0.21) n}\end{aligned}$
$\begin{array}{ll}\Rightarrow & (0.1) n=\sqrt{(0.21) n} \\ \Rightarrow & \left(\frac{n}{10}\right)^2=\left(\frac{21 n}{100}\right) \\ \Rightarrow & n^2-21 n=0 \Rightarrow n(n-21)=0 \\ \Rightarrow & n=0,21\end{array}$
But $n \neq 0$
Thus, $n=21$.
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