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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability5 Marks
Question
Let $X$ be a discrete random variable whose probability distribution is defined as follows:
$\text{P}(\text{X}=\text{x})=\begin{cases}\text{k}(\text{x}+1) & \text{for}\text{ x}= 1,2,3,4\\2\text{kx} & \text{for}\text{ x } =5,6,7\\0&\text{otherwise} \end{cases}$
where k is a constant. Calculate:
  1. The value of A if $E(X) = 2.94$
  2. Variance of $X.$
  3. Standard deviation of $X.$
✓

Answer

We have $\text{P}(\text{X}=\text{x})=\begin{cases}\text{k}(\text{x}+1) & \text{for}\text{ x}= 1,2,3,4\\2\text{kx} & \text{for}\text{ x } =5,6,7\\0&\text{otherwise} \end{cases}$
Thus, we have
$X$
$1$
$2$
$3$
$4$
$5$
$6$
$7$
Otherwise
$P(X)$
$2k$
$3k$
$4k$
$5k$
$10k$
$12k$
$14k$
$0$
$XP(X)$
$2k$
$6k$
$12k$
$20k$
$50k$
$72k$
$98k$
$0$
$X^2P(X)$
$2k$
$12k$
$36k$
$80k$
$250k$
$432k$
$686k$
$0$
Since, $\sum\text{P}_{\text{i}}=1$
$\Rightarrow\text{k}(2+3+4+5+10+12+14)=1$
$\Rightarrow\text{k}=\frac{1}{50}$
$\because\text{E}(\text{X})=\sum\text{XP}(\text{X})$
$\therefore\text{E}(\text{X})=2\text{k}+6\text{k}+12\text{k}+20\text{k}+50\text{k}+72\text{k}+98\text{k}+0$
$=260\text{k}=260\times\frac{1}{50}=\frac{26}{5}=5.2\ ......(\text{i})$
We know that,
$\text{Var}(\text{X})=\big[\text{E}(\text{X})^2\big]-\big[\text{E}(\text{X})\big]^2=\sum\text{X}^2\text{P}(\text{X})-\Big[\sum\left\{\text{XP}(\text{X})\right\}\Big]^2$
$=\big[2\text{k}+12\text{k}+36\text{k}+80\text{k}+250\text{k}+432\text{k}+686\text{k}+0\big]-\big[5.2\big]^2$
$=\big[1498\text{k}\big]-27.04$
$=\Big[1498\times\frac{1}{50}\Big]-27.04$
$=29.96-27.04=2.92$
We know that, standard deviation of $\text{X}=\sqrt{\text{Var}(\text{X})}=\sqrt{2.92}$
$=1.7088\cong1.7$

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