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  • Consider the probability distribution of a random variable X: X 0 P(X) 0.1 Calculate (i) (ii) Variance of X.

Consider the probability distribution of a random variable X:
\begin{array}{|l|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & 0.1 & 0.25 & 0.3 & 0.2 & 0.15 \\ \hline \end{array}
Calculate (i) V\left ( \frac{X}{2} \right )  (ii) Variance of X.

Answers (1)

Given-

\begin{aligned} &\begin{array}{|l|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & 0.1 & 0.25 & 0.3 & 0.2 & 0.15 \\ \hline \mathrm{XP}(\mathrm{X}) & 0 & 0.25 & 0.6 & 0.6 & 0.60 \\ \hline \mathrm{X}^{2} \mathrm{P}(\mathrm{X}) & 0 & 0.25 & 1.2 & 1.8 & 2.40 \\ \hline \end{array}\\ &\operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right) \cdot-[\mathrm{E}(\mathrm{X})]^{2}\\ &\text { Where, }\\ &E(X)=\mu=\sum_{i=1}^{n} x_{i} P\left(x_{i}\right) \end{aligned}

\begin{aligned} &\mathrm{E}(\mathrm{X})^{2}=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^{2} \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)\\ &\therefore E(x)=0+0.25+0.6+0.6+0.60=2.05\\ &\text { And } E(X)^{2}=0+0.25+1.2+1.8+2.40=5.65\\ &\text { (i) } V\left[\frac{x}{2}\right]\\ &\text { It is known that } \operatorname{Var}(a x)=a^{2} \operatorname{yar}(x)\\ &\Rightarrow \mathrm{V}\left[\frac{\mathrm{X}}{2}\right]=\frac{1}{4} \mathrm{~V}(\mathrm{X})\\ &=\frac{1}{4}\left[5.65-(2.05)^{2}\right] \end{aligned}

\begin{aligned} &=\frac{1}{4}[5.65-4.2025]\\ &=\frac{1}{4} \times 1.4475\\ &=0.361875\\ &\text { (ii) } \mathrm{V}(\mathrm{X})\\ &\operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right)-[\mathrm{E}(\mathrm{X})]^{2}\\ &=5.65-(2.05)^{2}\\ &=5.65-4.2025\\ &=1.4475 \end{aligned}

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