Question

If the variance of a random variable $X$ is ${\sigma }^2$ and variance of random variable $2X-5$ is $m{\sigma }^2+n.$ then $\left(\frac{m+n}{2}\right)=$

Answer 1

We know that, $V(ax\pm b)=a^{2}V\left(x\right)$
$\Rightarrow V(2X-5)=2^2\cdot {\sigma }^2+0$
$\Rightarrow m=4,n=0\Rightarrow \frac{m+n}{2}=2$