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

(i) Find the value of $k$.

(ii) Determine the mean of the distribution.

We have

(i) We know that, $\sum\limits_{i = 1}^n {{P_i}} = 1$, where ${P_i} \ge 0$

$\Rightarrow$ ${P_1} + {P_2} + {P_3} + {P_4} = 1$
$\Rightarrow$ $k + {k^2} + 2{k^2} + k = 1$

$\Rightarrow$ $3{k^2} + 2k - 1 = 0$

$\Rightarrow$ $3{k^2} + 3k - k - 1 = 0$

$\Rightarrow$ $3k(k + 1) - 1(k + 1) = 0$

$\Rightarrow$ $(3k - 1)(k + 1) = 0$

$\Rightarrow$ $k = 1/3 \Rightarrow k = - 1$

Since $k$ is $\ge 0 \Rightarrow k = 1/3$

(ii) Mean of the distribution $(\mu ) = E(X) = \sum\limits_{i = {1_i}}^n {{x_i}} {P_i}$

$= 0.5(k) + 1\left( {{k^2}} \right) + 1.5\left( {2{k^2}} \right) + 2(k) = 4{k^2} + 2.5k$

$= 4 \cdot \frac{1}{9} + 2.5 \cdot \frac{1}{3}$

$= \frac{{4 + 7.5}}{9} = \frac{{23}}{{18}}$