Q98 / 160 · TS EAMCET 2011 · Mathematics

The probability distribution of a random variable $X$ is given below. Figure
Then the variance of $X$ is

  • Correct answer
    A
    1
  • B
    2
  • C
    3
  • D
    4

Solution

$\begin{aligned} & \therefore \text { Mean, } m=\sum_{i=1}^4 p_i x_i \\ & =0 \times \frac{1}{10}+\frac{1 \times 2}{10}+\frac{2 \times 3}{10}+\frac{3 \times 4}{10} \\ & =0+\frac{2}{10}+\frac{6}{10}+\frac{12}{10} \\ & =\frac{20}{10}=2 \\ & \therefore \text { Variance }=\sum_{i=1}^4 p_i\left(x_i-m\right)^2\end{aligned}$ $\begin{aligned} & =\frac{1}{10}(0-2)^2+\frac{2}{10}(1-2)^2+\frac{3}{10}(2-2)^2 \\ & +\frac{4}{10}(3-2)^2 \\ & =\frac{4}{10}+\frac{2}{10}+0+\frac{4}{10} \\ & =\frac{10}{10}=1\end{aligned}$

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

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