The probability distribution of a random variable $x$ is given as under

$P(X = x) = \left\{ {\begin{array}{llllllllllllllllllll}{k{x^2},x = 1,2,3}\\{2kx,x = 4,5,6}\\{0,{\rm{ otherwise }}}\end{array}} \right.$

where, $k$ is a constant. Calculate

(i) $E(X)$

(ii) $E\left( {3{X^2}} \right)$

(iii) $P(X \ge 4)$

We know that, $\Sigma {P_i} = 1$

$\Rightarrow$ $44k = 1 \Rightarrow k = \frac{1}{{44}}$

$\therefore$ $\Sigma XP(X) = k + 8k + 27k + 32k + 50k + 72k + 0$

$= 190k = 190 \times \frac{1}{{44}} = \frac{{95}}{{22}}$

(i) So, $\quad E(X) = \Sigma XP(X) = \frac{{95}}{{22}} = 4.32$

(ii) Also, $E\left( {{X^2}} \right) = \Sigma {X^2}P(X) = k + 16k + 81k + 128k + 250k + 432k$

$= 908{\rm{k}} = 908 \times \frac{1}{{44}}$

$= 20.636 = 20.64$

( approx)
$\therefore$ $E\left( {3{X^2}} \right) = 3E\left( {{X^2}} \right) = 3 \times 20.64 = 61.92 = 61.9$

(iii) $P(X \ge 4) = P(X = 4) + P(X = 5) + P(X = 6)$
$= 8k + 10k + 12k = 30k = 30 \cdot \frac{1}{{44}} = \frac{{15}}{{22}}$