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The probability distribution of a discrete random variable X is given as under:
\begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{X} & 1 & 2 & 4 & 2 \mathrm{~A} & 3 \mathrm{~A} & 5 \mathrm{~A} \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{2} & \frac{1}{5} & \frac{3}{25} & \frac{1}{10} & \frac{1}{25} & \frac{1}{25} \\ \hline \end{array}

Calculate :
(i) The value of A if E(X) = 2.94
(ii) Variance of X.

Answers (1)

i ) Given-

E(X) = 2.94

It is known to us that μ = E(X)

\\ \because E(X)=\sum_{i=1}^{n} x_{i} p_{i} \\ =1 \times \frac{1}{2}+2 \times \frac{1}{5}+4 \times \frac{3}{25}+2 A \times \frac{1}{10}+3 A \times \frac{1}{25}+5 A \times \frac{1}{25} \\ =\frac{1}{2}+\frac{2}{5}+\frac{12}{25}+\frac{A}{5}+\frac{3 A}{25}+\frac{A}{5} \\ =\frac{25+20+24+10 A+6 A+10 A}{50} \\ =\frac{69+26 A}{50} \\ =2.94=\frac{69+26 A}{50} \quad \text { [given: } \left.E(X)=2.94\right] \\ \Rightarrow 2.94 \times 50=69+26 A \\ \Rightarrow 147-69=26 A \\ \Rightarrow \quad 78=26 A \\ \Rightarrow A=\frac{78}{26} \\x1

\begin{aligned} &\Rightarrow A=3\\ &\text { (ii) As we know that, }\\ &\operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}\\ &=\Sigma X^{2} P(X)-[\Sigma\{X P(X)\}]^{2}\\ &=\Sigma X^{2} P(X)-(2.94)^{2}\\ &\begin{array}{l} \text { We first find } \Sigma X^{2} P(X) \\ =1^{2} \times \frac{1}{2}+2^{2} \times \frac{1}{5}+4^{2} \times \frac{3}{25}+(2 A)^{2} \times \frac{1}{10}+(3 A)^{2} \times \frac{1}{25}+(5 A)^{2} \times \frac{1}{25} \end{array}\\ &=\frac{1}{2}+\frac{4}{5}+\frac{48}{25}+\frac{36}{10}+\frac{81}{25}+\frac{225}{25}\\ &=\frac{25+40+96+180+162+450}{50}\\ &=\frac{953}{50}\\ &=19.06\\ &\operatorname{Var}(X)=19.06-(2.94)^{2}\\ &=19.06-8.6436\\ &=10.4164 \end{aligned}

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