If X is a random variable with probability distribution as given … - Vidyadip

Question

If X is a random variable with probability distribution as given below:

X = x_i	0	1	2	3
P(X = X_i)	k	3k	3k	k

The value of k and its variance are:

$\frac{1}{8},\frac{22}{27}$
$\frac{1}{8},\frac{23}{27}$
$\frac{1}{8},\frac{24}{27}$
$\frac{1}{8},\frac{3}{4}$

✓

Answer

$\frac{1}{8},\frac{3}{4}$

Solution:

$\sum\limits_0^3\text{P}(\text{x})=1$

$\text{k}+3\text{k}+3\text{k}+\text{k}=1$

$\text{k}=\frac{1}{8}$

$\text{x}$	$\text{P}(\text{x})$	$\text{x}\text{P}(\text{x})$	$\text{x}^2\text{P}(\text{x})$
$0$	$\frac{1}{8}$	$0$	$0$
$1$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{3}{8}$
$2$	$\frac{3}{8}$	$\frac{6}{8}$	$\frac{12}{8}$
$3$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{9}{8}$
$\text{Total}$		$\text{E(x)}=\frac{12}{8}=1.5$	$\text{E}(\text{x}^2)=3$

$\text{V(x)}=\text{E}(\text{x}^2)-[\text{E}(\text{x})^2]$

$\text{V(x)}=3-(1.5)^2$

$\text{V(x)}=0.75=\frac{3}{4}$

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