Question

If $m$ and ${\sigma }^2$ are the mean and variance of the random variable $X$, whose distribution in given by

$X=x$	0	1	2	3
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

Then,

• A. $m={\sigma }^2=2$
• B. $m=1,{\sigma }^2=2$
• C. $m={\sigma }^2=1$
• D. $m=2,{\sigma }^2=1$

Answer 1

Answer: (C)

Given, distribution is

$X=x$	0	1	2	3
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

$\therefore$ Mean, $m=\sum _{i=1}^4{p}_i{x}_i$
$=0\times \frac{1}{3}+1\times \frac{1}{2}+2\times 0+3\times \frac{1}{6}=1$
Variance, ${\sigma }^2=\sum _{i=1}^4{p}_i{\left(x_i-m\right)}^2$
$=\frac{1}{3}(0-1{)}^2+\frac{1}{2}(1-1{)}^2+0(2-1{)}^2+\frac{1}{6}(3-1{)}^2=1$
$\therefore m={\sigma }^2=1$