For the probability distribution
Then the $\operatorname{Var}(\mathrm{X})$ is (Given : $\left.(0.25)^2=0.0625,(0.35)^2=0 \cdot 1225,(0.45)^2=0 \cdot 2025\right)$

Question

For the probability distribution Then the $\operatorname{Var}(\mathrm{X})$ is (Given : $\left.(0.25)^2=0.0625,(0.35)^2=0 \cdot 1225,(0.45)^2=0 \cdot 2025\right)$

Aicharya · Accepted Answer

$\begin{aligned} E(X)= & (-2)(0.1)+(-1)(0.2)+0(0.2)+(1)(0.3) \ & +2(0.15)+3(0.05) \ = & -0.2-0.2+0+0.3+0.3+0.15 \ = & 0.35\end{aligned}$ $\begin{aligned} 	herefore \quad \operatorname{Var}(\mathrm{X}) & =\mathrm{E}\left(\mathrm{X}^2ight)-[\mathrm{E}(\mathrm{X})]^2 \ & =(-2)^2(0.1)+(-1)^2(0.2)+0^2(0.2) \ & +1^2(0.3)+2^2(0.15)+3^2(0.05)-(0.35)^2 \ & =0.4+0.2+0+0.3+0.6\end{aligned}$ $\begin{aligned} & =1.95-(0.35)^2 \ & =1.95-0.1225 \ & =1.8275\end{aligned}$

For the probability distribution Then the $\operatorname{Var}(\mathrm{X})$ is (Given : $\left.(0.25)^2=0…

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