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Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability5 Marks
Question
The probability distribution of a discrete random variable X is given as under:
$\text{X}$ $1$ $2$ $4$ $2\text{A}$ $3\text{A}$ $5\text{A}$
$\text{P}(\text{X})$ $\frac{1}{2}$ $\frac{1}{5}$ $\frac{3}{25}$ $\frac{1}{10}$ $\frac{1}{25}$ $\frac{1}{25}$
Calculate:
  1. The value of A if E(X) = 2.94
  2. Variance of X.
✓

Answer

We have
$\sum\text{X}\text{P}(\text{X})=\frac{1}{2}+\frac{2}{5}+\frac{2}{25}+\frac{2\text{A}}{10}+\frac{3\text{A}}{25}+\frac{5\text{A}}{25}$
$=\frac{25+20+24+10\text{A}+6\text{A}+10\text{A}}{50}$
$=\frac{69+26\text{A}}{50}$
Since, $\text{E}(\text{X})=\sum\text{X}\text{P}(\text{X})$
$\Rightarrow2.94=\frac{69+26\text{A}}{50}$
$\Rightarrow26\text{A}=50\times2.94-69$
$\Rightarrow\text{A}=\frac{147-69}{26}$
$=\frac{78}{26}=3$
We know that,
$\text{Var}(\text{X})=\text{E}(\text{X}^2)-[\text{E}(\text{X})]^2$
$=\sum\text{X}^2\text{P}(\text{X})-[\text{E}(\text{X})]^2$
$=\frac{1}{2}+\frac{4}{5}+\frac{48}{25}+\frac{4\text{A}^2}{10}+\frac{9\text{A}^2}{25}+\frac{25\text{A}^2}{25}-[\text{E}(\text{X})]^2$
$=\frac{25+40+96+20\text{A}^2+18\text{A}^2+50\text{A}^2}{50}-[\text{E}(\text{X})]^2$
$=\frac{161+88\text{A}^2}{50}-[\text{E}(\text{X})]^2$
$=\frac{161+88\times(3)^2}{50}-[\text{E}(\text{X})]^2$
$=\frac{953}{50}-[2.94]^2$ $[\because\text{E}(\text{X})=2.94]$
$=19.06-8.6436$
$=10.4164$

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