Let $X$ be a discrete random variable whose probability distribution is defined as follows.

$P(X = x) = \left\{ {\begin{array}{llllllllllllllllllll}{k(x + 1),}&{{\rm{ for }}x = 1,2,3,4}\\{2kx,}&{{\rm{ for }}x = 5,6,7}\\{0,}&{{\rm{ otherwise }}}\end{array}} \right.$

where, $k$ is a constant. Calculate

(i) the value of $k$.

(ii) $E(X)$.

(iii) standard deviation of $X$.

$P(X = x) = \left\{ {\begin{array}{llllllllllllllllllll}{k(x + 1),}&{{\rm{ for }}x = 1,2,3,4}\\{2kx,}&{{\rm{ for }}x = 5,6,7}\\{0,}&{{\rm{ otherwise }}}\end{array}} \right.$

Thus, we have following table

(i) Since, $\Sigma {P_i} = 1$
$\Rightarrow$ $k(2 + 3 + 4 + 5 + 10 + 12 + 14) = 1 \Rightarrow k = \frac{1}{{50}}$

(ii)
$\therefore$ $E(X) = 2k + 6k + 12k + 20k + 50k + 72k + 98k + 0 = 260k$

$= 260 \times \frac{1}{{50}} = \frac{{26}}{5} = 5.2$

(iii) We know that,

${\mathop{\rm Var}\nolimits} (X) = \left[ {E\left( {{X^2}} \right)} \right] - {[E(X)]^2} = \Sigma {X^2}P(X) - {[\Sigma \{ XP(X)\} ]^2}$

$= [2k + 12k + 36k + 80k + 250k + 432k + 686k + 0] - {[5.2]^2}\quad$

[using Eq.(i)]

$= 29.96 - 27.04 = 2.92$

We know that, standard deviation of $X = \sqrt {{\mathop{\rm Var}\nolimits} (X)} = \sqrt {2.92} = 1.7088 = 1.7$ (approx)