A discrete random variable X has the probability distribution giv… - Vidyadip

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 $(\mu)=\text{E}(\text{X})=\sum\limits_{\text{i=1}}^{\text{n}}\text{x}_{\text{i}}\text{P}_{\text{i}}$
$=0.5(\text{k})+1(\text{k}^2)+1.5(2\text{k}^2)+2(\text{k})$
$=4\text{k}^2+2.5\text{k}$
$=4\cdot\frac{1}{9}+2.5\cdot\frac{1}{3}\Big[\because\text{k}=\frac{1}{3}\Big]$
$=\frac{4+7.5}{9}=\frac{23}{18}$

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