Question

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

$\begin{array}{cccccccc}X& 1& 2& 3& 4& 2A& 3A& 5A\\ P\left(X\right)& \frac{1}{2}& \frac{1}{2}& \frac{1}{5}& \frac{3}{25}& \frac{1}{10}& \frac{1}{25}& \frac{1}{25}\end{array}$

Calculate

(i) the value of A, if E (X) = 2.94.

(ii) variance of X.

Answer 1

(i) We have, $\sum XP\left(X\right)=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$

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

Since, E(X) = $\sum$ XP(X)

$\Rightarrow 2.94=\frac{69+26A}{50}$

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

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

(ii) We know that

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

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

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

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

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

$=\frac{953}{50}-[2.94{]}^2$ [$\because$ E(X)=2.94]

= 19.0600 - 8.6436 = 10.4164