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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Aicharya · Accepted Answer

Given, $p=0.3, q=1-p=0.7$ Also, Mean $(\mu)=3 	imes$ Standard deviation $(\sigma)$. $\Rightarrow \quad \mu=3(\sigma)$...(i) $\begin{aligned} & 	ext { Now, } \mu=n p=0.3 n \ & 	ext { and } \sigma=\sqrt{n p q}=\sqrt{n(0 \cdot 3)(0 \cdot 7)}=\sqrt{0.21 n} \ & 	ext { From Eq. (i), }(0.3) n=3 \sqrt{(0.21) n}\end{aligned}$ $\begin{array}{ll}\Rightarrow & (0.1) n=\sqrt{(0.21) n} \ \Rightarrow & \left(\frac{n}{10}ight)^2=\left(\frac{21 n}{100}ight) \ \Rightarrow & n^2-21 n=0 \Rightarrow n(n-21)=0 \ \Rightarrow & n=0,21\end{array}$ But $n 
eq 0$ Thus, $n=21$.

A random variable $X$ follows the binomial distribution and $X \sim B(n, 0.3)$. If the mean of $X$ is three…

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