For the probability distribution of X given below $\begin{array}{|c|c|c|c|c|c|} \hline X=x & -2 & -1 & 0 & 1 & 2 \ \hline \mathrm{P}(\mathrm{X}=x) & 0 \cdot 2 & 0 \cdot 3 & 0 \cdot 1 & 0 \cdot 15 & 0 \cdot 25 \ \hline \end{array}$ The variance of $X$ is

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Aicharya · Accepted Answer

$\sum_{i=1}^{n} x_{i} P_{i}=E(X)$ $\begin{aligned} E(X) &=(-2)(0.2)+(-1)(0.3)+0(0.1)+1(0.15)+2(0.25) \ &=-0.4-0.3+0.15+0.50=-0.7+0.65=-0.05 \ E\left(X^{2}ight) &=\sum_{i=1}^{n} x_{i}^{2} P_{i} \ &=4(0.2)+1(0.3)+0(0.1)+1(0.15)+4(0.25) \ &=0.8+0.3+0.15+1=2.25 \ &=E\left(X^{2}ight)-[E(X)]^{2} \ &=2.25-0.0025=2.2475 \end{aligned}$

For the probability distribution of X given below $\begin{array}{|c|c|c|c|c|c|} \hline X=x & -2 & -1 & 0 & 1…

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