Q80 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics

The variance of the random variable $X$ having the following distribution Figure

  • A
    $\frac{1}{3}$
  • B
    $\frac{2}{3}$
  • C
    $\frac{4}{3}$
  • Correct answer
    D
    $\frac{5}{3}$

Solution

$\begin{aligned} & \because \operatorname{var}(x)=E\left(x^2\right)-[E(x)]^2 \\ & \begin{aligned} E(x) & =\Sigma P_i x_i \\ & =\frac{1}{6} \times(-2)+\frac{1}{6}(-1)+\frac{1}{3} \times 0+\frac{1}{6}(1)+\frac{1}{6}(2) \\ & =-\frac{2}{6}-\frac{1}{6}+0+\frac{1}{6}+\frac{2}{6}=0 \\ E\left(x^2\right) & =\Sigma P_i x_i^2 \\ & =\frac{1}{6}(-2)^2+\frac{1}{6}(-1)^2+\frac{1}{3} \times(0)^2+\frac{1}{6}(1)^2+\frac{1}{6}(2)^2 \\ & =\frac{4}{6}+\frac{1}{6}+0+\frac{1}{6}+\frac{4}{6}=\frac{10}{6}=\frac{5}{3} \\ \therefore \operatorname{var}(x) & =\frac{5}{3}-(0)^2=\frac{5}{3} .\end{aligned}\end{aligned}$

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

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