If $\mathrm{P}(\mathrm{X}=2)=0.3, \mathrm{P}(\mathrm{X}=3)=0.4, \mathrm{P}(\mathrm{X}=4)=0.3$, then the…
- 1.6
- 6.6
- 3.6
- 0.6
Solution

$\begin{aligned} & \mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \mathrm{P}\left(x_{\mathrm{i}}\right) \\ & \quad=2(0.3)+3(0.4)+4(0.3) \\ & \begin{aligned} \mathrm{E}(\mathrm{X}) & =3 \\ \text { Variance } & =\mathrm{E}\left(x^2\right)-[\mathrm{E}(x)]^2 \\ & =2^2(0.3)+3^2(0.4)+4^2(0.3)-(3)^2 \\ & =0.6\end{aligned}\end{aligned}$
Asked in: MHT CET 2024 (04 May Shift 2)