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

• A

  $m=\sigma^2=2$
• B

  $m=1, \sigma^2=2$
• Correct answer
  C

  $m=\sigma^2=1$
• D

  $m=2, \sigma^2=1$

Solution

Given, distribution is
$\begin{aligned}\therefore Mean, m & =\sum_{i=1}^4 p_i x_i \\ & =0 \times \frac{1}{3}+1 \times \frac{1}{2}+2 \times 0+3 \times \frac{1}{6}\end{aligned}$ $=0+\frac{1}{2}+0+\frac{1}{2}=1$ Variance, $\sigma^2=\sum_{i=1}^4 p_i\left(x_i-m\right)^2$ $\begin{aligned} & =\frac{1}{3}(0-1)^2+\frac{1}{2}(1-1)^2 \\ & \quad+0(2-1)^2+\frac{1}{6}(3-1)^2 \\ & =\frac{1}{3}+0+0+\frac{2}{3}=1 \\ & m=\sigma^2=1 \end{aligned}$ $\therefore \quad m=\sigma^2=1$

Hence, the correct answer is option: 'C'.