Question

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

$X$	$0$	$1$	$2$	$3$
$P\left(X\right)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

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

The correct option is C

$m={\sigma }^2=1$

Explanation for the correct option:

Step 1: Solve for the mean

For the given probability distribution, the mean is given as

$m=\sum _{i=0}^3{P}_i{x}_i$

$\Rightarrow$$m=0\times \frac{1}{3}+1\times \frac{1}{2}+2\times 0+3\times \frac{1}{6}$

$\Rightarrow$$m=1$

Step 2: Solve for the variance

The variance is given as

${\sigma }^2=\sum _{i=0}^3{P}_i{\left(x_i-m\right)}^2$

$\Rightarrow$${\sigma }^2=\frac{1}{3}{\left(0-1\right)}^2+\frac{1}{2}{\left(1-1\right)}^2+0{\left(2-1\right)}^2+\frac{1}{6}{\left(3-1\right)}^2$

$\Rightarrow$${\sigma }^2=\frac{1}{3}+\frac{2}{3}$

$\Rightarrow$${\sigma }^2=1$

Thus the value of $m={\sigma }^2=1$.

Hence option(C) is the correct answer.