Probability — Class 12 Maths Solution

We have,

${{\rm{X}}^{\rm{2}}}{\rm{P}}\left( {\rm{X}} \right)$

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

where, $E(X) = \mu = \sum\limits_{i = 1}^n {{x_i}} {P_i}(xi)$

and $E\left( {{X^2}} \right) = \sum\limits_{i = 1}^n {x_i^2} P\left( {{x_i}} \right)$

$\therefore$ $E(X) = 0 + 0.25 + 0.6 + 0.6 + 0.60 = 2.05$

$E\left( {{X^2}} \right) = 0 + 0.25 + 1.2 + 1.8 + 2.40 = 5.65$

(i) $V\left( {\frac{X}{2}} \right) = \frac{1}{4}V(X) = \frac{1}{4}\left[ {5.65 - {{(2.05)}^2}} \right]$

$= \frac{1}{4}[5.65 - 4.2025] = \frac{1}{4} \times 1.4475 = 0.361875$

(ii) $V(X) = 1.4475$