If $\mu$ and $a^{2}$ are mean and variance of a random variable $X$ whose p. $m . f$. is given by $P(X=x)=\left(\begin{array}{l}6 \ x\end{array} ight)\left(\frac{1}{x} ight)^{x}\left(\frac{2}{x} ight)^{6-x}, x=0,1,2,3, \ldots \ldots 6$, then the value of $2 \mu+12 \sigma^{2}=$

Question

If $\mu$ and $a^{2}$ are mean and variance of a random variable $X$ whose p. $m . f$. is given by $P(X=x)=\left(\begin{array}{l}6 \ x\end{array}ight)\left(\frac{1}{x}ight)^{x}\left(\frac{2}{x}ight)^{6-x}, x=0,1,2,3, \ldots \ldots 6$, then the value of $2 \mu+12 \sigma^{2}=$

Aicharya · Accepted Answer

We have $P(X=x)={ }^{6} C_{x}\left(\frac{1}{3}ight)^{3}\left(\frac{2}{3}ight)^{6-x}$ $=\left(\frac{1}{3}+\frac{2}{3}ight)^{6}$ Thus, $\mathrm{n}=6, \mathrm{p}=\frac{1}{3}$ and $\mathrm{q}=\frac{2}{3}$ $	herefore$ Mean $=\mathrm{np}=(6)\left(\frac{1}{3}ight)=2$ and variance $=(6)\left(\frac{1}{3}ight)\left(\frac{2}{3}ight)=\frac{4}{3}$ $	herefore 2 \mu+12 \sigma^{2}=2(2)+12\left(\frac{4}{3}ight)=4+16=20$

If $\mu$ and $a^{2}$ are mean and variance of a random variable $X$ whose p. $m . f$. is given by…

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