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 which of the following is correct.

• 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

The given data:

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

The mean can be calculated by, $m=\sum _{i=0}^3{x}_i{P}_i$

$$\begin{gathered} \Rightarrow m=\left(0\times \frac{1}{3}\right)+\left(1\times \frac{1}{2}\right)+\left(2\times 0\right)+\left(3\times \frac{1}{6}\right) \\ \Rightarrow m=0+\frac{1}{2}+0+\frac{1}{2} \\ \Rightarrow m=1 \end{gathered}$$

The variance can be calculated by, ${\sigma }^2=\sum _{i=0}^3{P}_i{\left(x_i-m\right)}^2$

$$\begin{gathered} \Rightarrow {\sigma }^2=\left[\frac{1}{3}\times {\left(0-1\right)}^2\right]+\left[\frac{1}{2}\times {\left(1-1\right)}^2\right]+\left[0\times {\left(2-1\right)}^2\right]+\left[\frac{1}{6}\times {\left(3-1\right)}^2\right] \\ \Rightarrow {\sigma }^2=\left[\frac{1}{3}\times 1\right]+\left[\frac{1}{2}\times 0\right]+0+\left[\frac{1}{6}\times 4\right] \\ \Rightarrow {\sigma }^2=\frac{1}{3}+0+0+\frac{2}{3} \\ \Rightarrow {\sigma }^2=1 \end{gathered}$$

Therefore, $m={\sigma }^2=1$.

Hence, option(C) is the correct option.