Question

For the following probability distribution, the standard deviation of the random variable $X$ is

(X)	$2$	$3$	$4$
P(X)	$0.2$	$0.5$	$0.3$

• A. $0.5$
• B. $0.6$
• C. $0.61$
• D. $0.7$

Answer 1

Answer: (D)

We have,

$X$	$P(X)$	$XP(X)$	$X^{2}P(X)$
$2$	$0.2$	$0.4$	$0.8$
$3$	$0.5$	$1.5$	$4.5$
$4$	$0.3$	$1.2$	$4.8$
		$\sum XP\left(X\right)=3.1$	$\sum X^{2}P\left(X\right)=10.1$

$\mu =\sum XP\left(X\right)=3.1$
var $X$, ${\sigma }^2=\sum X^{2}P\left(X\right)-{\mu }^2=10.1-{\left(3.1\right)}^2$
$=10.1-9.61=0.49$
$\therefore$ Standard deviation $=\sqrt{{\sigma }^2}=\sqrt{0.49}=0.7$