Question

Consider the probability distribution of a random variable X:

X	0	1	2	3	4
P(X)	0.1	0.25	0.3	0.2	0.15

Calculate ${\displaystyle \text{V}\left(\frac{\text{X}}{2}\right)}$

Answer 1

Here, we have

X	0	1	2	3	4
P(X)	0.1	0.25	0.3	0.2	0.15

We know that: Var(X) = E(X²) – [E(X)]²

Where E(X) = ${\displaystyle \sum _{\text{i}=1}^{\text{n}}{x}_{\text{i}}{\text{p}}_{\text{i}}}$ and E(X²) = ${\displaystyle \sum _{\text{i}=1}^{\text{n}}{\text{p}}_{\text{i}}x{\text{i}}^2}$

∴ E(X) = 0 × 0.1 + 1 × 0.25 + 2 × 0.3 + 3 × 0.2 + 4 × 0.15

= 0 + 0.25 + 0.6 + 0.6 + 0.6

= 2.05

E(X²) = 0 × 0.1 + 1 × 0.25 + 4 × 0.3 + 9 × 0.2 + 16 × 0.15

= 0 + 0.25 + 1.2 + 1.8 + 2.40

= 5.65

 ${\displaystyle \text{V}\left(\frac{\text{X}}{2}\right)=\frac{1}{4}\text{V}\left(\text{X}\right)}$

= ${\displaystyle \frac{1}{4}[5.65-{\left(2.05\right)}^2]}$

= ${\displaystyle \frac{1}{4}[5.65-4.2025]}$

= ${\displaystyle \frac{1}{4}\times 1.4475}$

= 0.361875   .......${\displaystyle [\because \text{V}\left(\frac{\text{X}}{2}\right)=\frac{1}{2}\text{V}\left(\text{X}\right)]}$