A random variable $X$ has the following probability distribution The mean and variance of X are respectively

A random variable $X$ has the following probability distribution
The mean and variance of X are respectively
  1. $2 \cdot 3$ and $6 \cdot 1$
  2. 2.3 and 0.81
  3. 2.3 and 0.1
  4. $2 \cdot 3$ and $0 \cdot 9$

Solution

$\begin{aligned} \text { Mean } & =\mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right) \\ & =1(0.2)+2(0.4)+3(0.3)+4(0.1) \\ & =2.3 \\ \operatorname{Var}(\mathrm{X}) & =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ & =1^2(0.2)+2^2(0.4)+3^2(0.3)+4^2(0.1) \\ & =0.2+1.6+2.7+1.6-5.29 \\ & =0.81\end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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