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  • For the following probability distribution determine standard deviation of the random variable X.

For the following probability distribution determine standard deviation of the random variable X.
\begin{array}{|l|l|l|l|} \hline \mathrm{X} & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & 0.2 & 0.5 & 0.3 \\ \hline \end{array}

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Given-

\begin{aligned} &\begin{array}{|l|l|l|l|} \hline \mathrm{X} & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & 0.2 & 0.5 & 0.3 \\ \hline \mathrm{XP}(\mathrm{X}) & 0.4 & 1.5 & 1.2 \\ \hline \mathrm{X}^{2} \mathrm{P}(\mathrm{X}) & (4 \times 0.2) & (9 \times 0.5) & 16 \times 0.3 \\ & =0.8 & =4.5 & =4.8 \\ \hline \end{array}\\ &\text { It is known that standard deviation of }\\ &\mathrm{X}=\sqrt{\operatorname{VarX}}\\ &\text { Where, Var }=E\left(X^{2}\right)-[E(X)]^{2}\\ &=\sum_{i=1}^{n} x_{i}^{2} P\left(x_{i}\right)-\left[\sum_{i=1}^{n} x_{i} P_{i}\right]^{2} \end{aligned}

\begin{aligned} &\therefore \operatorname{Var} \mathrm{X}=[0.8+4.5+4.8]-[0.5+1.5+1.2]^{2}\\ &=10.1-(3.1)^{2}\\ &=10.1-9.61\\ &=0.49\\ &\text { Hence, standard deviation of }\\ &\mathrm{X}=\sqrt{\operatorname{Var} \mathrm{X}}=\sqrt{0.49}=0.7 \end{aligned}

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