Q80 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics

If $\mathrm{X}$ is a random variable such that $P(X=-2)=P(X=-1)=P(X=2)=P(X=1)=\frac{1}{6}$ and $P(X=0)=\frac{1}{3}$, then the mean of $X$ is

  • A
    $\frac{5}{3}$
  • B
    1
  • Correct answer
    C
    0
  • D
    $\frac{3}{5}$

Solution

$p(X=-2)=p(X=-1)=p(X=2)=p(X=1)=\frac{1}{6}$ $\begin{aligned} & \& p(X=0)=\frac{1}{3} \\ & \text { Mean }=(-2) \cdot p(X=-2)+(-1) p(X=-1)+0 \cdot p(X=0)+ \\ & \text { 1. } p(X=1)+2 \cdot p(X=2) \\ & =-2 \times \frac{1}{6}-\frac{1}{6}+0+\frac{1}{6}+2 \times \frac{1}{6} \Rightarrow \text { Mean }=0\end{aligned}$

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

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