Let X be a discrete random variable whose probability distribution is defined as follows: P(X=x)= k(x+1) & for x= 1,2,3,4 2kx & for x =5,6,7 0&otherwise 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.

We have P(X=x)= k(x+1) & for x= 1,2,3,4 2kx & for x =5,6,7 0&otherwise 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, _ i =1 (2+3+4+5+10+12+14)=1 =1 50 (X)= (X) (X)=2k+6k+12k+20k+50k+72k+98k+0 =260k=260 1 50 =26 5 =5.2 ......(i) We know that, Var(X)= [E(X)^2 ]- [E(X) ]^2= ^2P(X)- [ XP(X) ]^2 = [2k+12k+36k+80k+250k+432k+686k+0 ]- [5.2 ]^2 = [1498k ]-27.04 = [1498 1 50 ]-27.04 =29.96-27.04=2.92 We know that, standard deviation of X= Var(X) =2.92Z =1.7088 1.7

Let X be a discrete random variable whose probability distributio…

Download App

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:

The value of $A$ if $E(X) = 2.94$
Variance of $X$.
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 $\text{X}=\sqrt{\text{Var}(\text{X})}=\sqrt{2.92}$Z
$=1.7088\cong1.7$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Probability→Probability — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Probability — MATHS STD 12 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions