If a discrete random variable X has probability distribution as follows $\begin{array}{|c|c|c|c|c|} \hline \mathrm{X}=x & 0 & 1 & 2 & 3 \ \hline \mathrm{P}[\mathrm{X}=x] & \mathrm{k} & 3 \mathrm{k} & 3 \mathrm{k} & \mathrm{k} \ \hline \end{array}$ Then $\operatorname{var}(\mathrm{X})=$

Question

Aicharya · Accepted Answer

Here $\mathrm{k}+3 \mathrm{k}+3 \mathrm{k}+\mathrm{k}=8 \mathrm{k}=1 \Rightarrow \mathrm{k}=\frac{1}{8}$ $\begin{aligned} \sum \mathrm{P}_{\left(\mathrm{x}_{1}ight)} \mathrm{x}_{\mathrm{i}} &=0 	imes \frac{1}{8}+1 	imes \frac{3}{8}+2 	imes \frac{3}{8}+3 	imes \frac{1}{8}=\frac{12}{8}=\frac{3}{2} 	ext { and } \ \sum \mathrm{P}_{\left(\mathrm{x}_{i}ight)} \mathrm{x}_{\mathrm{i}}^{2} &=\left(0 	imes \frac{1}{8}ight)+\left(1 	imes \frac{3}{8}ight)+\left(4 	imes \frac{3}{8}ight)+\left(9 	imes \frac{1}{8}ight)=\frac{24}{8}=3 \ 	ext { Variance } &=\mathrm{V}(\mathrm{X}) \ &=\sum \mathrm{P}_{\left(\mathrm{x}_{i}ight)} \mathrm{x}_{\mathrm{i}}^{2}-\left[\sum \mathrm{P}_{\left(\mathrm{x}_{i}ight)} \mathrm{x}_{\mathrm{i}}ight]^{2} \ &=3-\left(\frac{3}{2}ight)^{2}=3-\frac{9}{4}=\frac{3}{4} \end{aligned}$

If a discrete random variable X has probability distribution as follows $\begin{array}{|c|c|c|c|c|} \hline…

Solution