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

Calculate
(i) the value of $A$, if $E(X) = 2.94$.
(ii) variance of $X$.

(i) We have, $\Sigma XP(X) = \frac{1}{2} + \frac{2}{5} + \frac{{12}}{{25}} + \frac{{2A}}{{10}} + \frac{{3A}}{{25}} + \frac{{5A}}{{25}}$

$= \frac{{25 + 20 + 24 + 10A + 6A + 10A}}{{50}} = \frac{{69 + 26A}}{{50}}$

Since, $E(X) = \Sigma XP(X)$
$\Rightarrow$ $2.94 = \frac{{69 + 26A}}{{50}}$

$\Rightarrow$ $\quad 26A = 50 \times 2.94 - 69$

$\Rightarrow$ $A = \frac{{147 - 69}}{{26}} = \frac{{78}}{{26}} = 3$

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

$= \Sigma {X^2}P(X) - {\left[ {\sum X P(X)} \right]^2}$

$= \frac{1}{2} + \frac{4}{5} + \frac{{48}}{{25}} + \frac{{4{A^2}}}{{10}} + \frac{{9{A^2}}}{{25}} + \frac{{25{A^2}}}{{25}} - {[E(X)]^2}$

$= \frac{{25 + 40 + 96 + 20{A^2} + 18{A^2} + 50{A^2}}}{{50}} - {[E(X)]^2}$

$= \frac{{161 + 88{A^2}}}{{50}} - {[E(X)]^2} = \frac{{161 + 88 \times {{(3)}^2}}}{{50}} - {[E(X)]^2}$

$= \frac{{953}}{{50}} - {[2.94]^2}$

$= 19.0600 - 8.6436 = 10.4164$