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

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.

• A. Statement I is true; statement II is true; statement II is a correct explanation for statement I
• B. Statement I is true; statement II is true; statement II is not a correct explanation for statement I
• C. Statement I is true; statement II is false
• D. Statement I is false; statement II is true

Answer 1

Answer: (A)

Here,

$X$	$P(X)$	$XP(X)$	$X^{2}P(X)$
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, $\mu =\Sigma XP(X)=2.05$
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. $V(X)=1.4475$