Q.
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
Statement I The value of $V\left(\frac{X}{2}\right)$ is $0.3618$.
Statement II The variance of $X$ is $1.4475$.
Choose the correct option.
| $X$ | 0 | 1 | 2 | 3 | 4 |
| $P(X)$ | 0.1 | 0.25 | 0.3 | 0.2 | 0.15 |
Probability - Part 2
Solution:
Here,
$X$
$P(X)$
$XP(X)$
$X^2P(X)$
0
0.10
0
0
1
0.25
0.25
0.25
2
0.30
0.60
1.20
3
0.20
0.60
1.80
4
0.15
0.60
2.40
Now, $ \mu=\Sigma X P(X)=2.05$
and $ \text{var}(X)=\Sigma X^2 P(X)-\mu^2 \left[\because \mu=2.05, X^2 P(X)=5.65\right]$
$=5.65-(2.05)^2$
$=5.65-42025$
$=1.4475$
I. $V\left(\frac{X}{2}\right)=\frac{1}{4} \times V(X)$
$=\frac{1}{4} \times 1.4475$
$=0.3618$
II. $V(X)=1.4475$
| $X$ | $P(X)$ | $XP(X)$ | $X^2P(X)$ |
|---|---|---|---|
| 0 | 0.10 | 0 | 0 | 1 | 0.25 | 0.25 | 0.25 |
| 2 | 0.30 | 0.60 | 1.20 |
| 3 | 0.20 | 0.60 | 1.80 |
| 4 | 0.15 | 0.60 | 2.40 |