A discrete random variable X has the probability distribution given as below: X 0.5 1 1.5 2 P(X) k k^2 2k^2 k 1. Find the value of k. 2. Determine the mean of the distribution.

We have, X 0.5 1 1.5 2 P(X) k k^2 2k^2 k 1. We know that _i=1^ n P_ i =1, where P_ i 0 P_1+ P_2 + P_3 + P_4= 1 k + k^2 + 2k^2 + k = 1 3k^2 + 2k - 1 = 0 3k^2 + 3k - k - 1 = 0 3k(k + 1) - 1(k + 1) = 0 (3k - 1)(k + 1) = 0 =1 3 and k = -1 Since, K 0, we take k=1 3 2. Mean of the distribution ( )=E(X)= _ i=1 ^ n x_ i P_ i =0.5(k)+1(k^2)+1.5(2k^2)+2(k) =4k^2+2.5k =4 1 9 +2.5 1 3 [ =1 3 ] =4+7.5 9 =23 18

A discrete random variable X has the probability distribution giv…

Download App

Question

A discrete random variable $X$ has the probability distribution given as below:

$X$	$0.5$	$1$	$1.5$	$2$
$P(X)$	$k$	$k^2$	$2k^2$	$k$

Find the value of $k.$
Determine the mean of the distribution.

✓

Answer

We have,

$X$	$0.5$	$1$	$1.5$	$2$
$P(X)$	$k$	$k^2$	$2k^2$	$k$

We know that $\sum\limits_\text{i=1}^{\text{n}}\text{P}_{\text{i}}=1,$ where $\text{P}_{\text{i}}\geq0$

$\Rightarrow P_1+ P_2 + P_3 + P_4= 1$
$\Rightarrow k + k^2 + 2k^2 + k = 1$
$\Rightarrow 3k^2 + 2k - 1 = 0$
$\Rightarrow 3k^2 + 3k - k - 1 = 0$
$\Rightarrow 3k(k + 1) - 1(k + 1) = 0$
$\Rightarrow (3k - 1)(k + 1) = 0$
$\Rightarrow\text{k}=\frac{1}{3}$ and $k = -1$
Since, $\text{K}\geq0,$ we take $\text{k}=\frac{1}{3}$

Mean of the distribution $(\mu)=\text{E}(\text{X})=\sum\limits_{\text{i=1}}^{\text{n}}\text{x}_{\text{i}}\text{P}_{\text{i}}$

$=0.5(\text{k})+1(\text{k}^2)+1.5(2\text{k}^2)+2(\text{k})$
$=4\text{k}^2+2.5\text{k}$
$=4\cdot\frac{1}{9}+2.5\cdot\frac{1}{3}\Big[\because\text{k}=\frac{1}{3}\Big]$
$=\frac{4+7.5}{9}=\frac{23}{18}$