RD Sharma Class 12 Exercise VSA Mean and Variance of a Random variable Solutions Maths-Download PDF Free Online

RD Sharma Class 12 Exercise VSA Mean and Variance of a Random variable Solutions Maths-Download PDF Free Onli

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 04:37 PM IST

The Class 12 RD Sharma chapter 31 exercise FBQ solution is useful in solving questions from this chapter in career360. The RD Sharma class 12th exercise FBQ helps students to broaden their knowledge about the essential concepts and to understand them precisely which helps them to solve the questions with ease.

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RD Sharma Class 12 Solutions Chapter31VSA Mean and Variance of a Random Variable- Other Exercise

Mean and Variance of a Random Variable Excercise: 31 VSA

Mean and Variance of a Random Variable Exercise Fill in the Blanks Question 1

Answer:
$E\left ( X \right )=\sum_{i=1}^{n}x_{i}p_{i}$
Hints: -You must know the rule for finding mean from the Random variable.
Given: -
$X:x_{1}x_{2}-------x_{n}$
$P\left ( X \right ):p_{1}p_{2}-------p_{n}$
Solution:
Mean $\begin{aligned} &=E(X)=\sum_{i=1}^{n} X_{i} P\left(X=X_{i}\right) \\ \end{aligned}$
$\begin{aligned} &=x_{1} p_{1}+x_{2} p_{2}+x_{3} p_{3}+---+x_{n} p_{n} \\ &E(X)=\sum_{i=1}^{n} x_{i} p_{i} \end{aligned}$

Mean and Variance of a random Variable Exercise Fill in the Blanks Question 2

Answer: - $E\left(X^{2}\right)-[E(X)]^{2} \text { or } \sum_{i=1}^{n}\left(P_{i}\left(X_{i}\right)^{2}\right)-\left[\sum_{i=1}^{n}\left(P_{i} X_{i}\right)\right]^{2}$
Hint: - You must know the rules for finding variance of Random variables.
Given: -
$\begin{array}{ll} X: & x_{1} x_{2}----x_{n} \\ P(X): & p_{1} \quad p_{2}----p_{n} \end{array}$
Solution:
$\begin{aligned} \text { Variance } &=\operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2} \\\\ &=\sum_{i=1}^{n}\left(P_{i}\left(X_{i}\right)^{2}\right)-\left[\sum_{i=1}^{n}\left(P_{i} X_{i}\right)\right]^{2} \end{aligned}$
Where E(x) represents mean value for random variable x.

Mean and Variance of a random Variable Exercise Fill in the Blanks Question 3

Answer: $\frac{3}{10}$
Hint: - You must know the rules for finding values of random variables.
Given: -
$\begin{array}{|c|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(X) & c & 2 c & 2 c & 3 c & c^{2} & 2 c^{2} & 7 c^{2}+c \\ \hline \end{array}$
Solution: $P(X \leq 2)=P(X=1)+P(X=2)$
Therefore, we know that the sum of probabilities in a probability distribution is always 1.
$\begin{aligned} &\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6)+\mathrm{P}(\mathrm{X}=7)=1 \\ \end{aligned}$
$\begin{aligned} &c+2 c+2 c+3 c+c^{2}+2 c^{2}+7 c^{2}+c=1 \\\\ &10 c^{2}+9 c-1=0 \\ \end{aligned}$
$\begin{aligned} &10 c^{2}+10 c-c-1=0 \\\\ &10 c(c+1)-(c+1)=0 \\\\ &(10 c-1)(c+1)=0 \\\\ &c=\frac{1}{10}, c=-1 \end{aligned}$
Negative value does not support, we use c=1/10
$\begin{aligned} &P(X \leq 2)=P(X=1)+P(X=2) \\\\ &=\frac{1}{10} \times 1+\frac{1}{10} \times 2 \\\\ &=\frac{1}{10}+\frac{2}{10} \\\\ &=\frac{3}{10} \end{aligned}$

Mean and Variance of a random Exercise Fill in the Blanks Question 4

Answer: $\frac{1}{3}$
Hint: - You must know the rules for finding values of random variables.
Given: -
$\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline X & \frac{1}{2} & 1 & \frac{3}{2} & 2 \\ \hline P(X) & c & c^{2} & 2 c^{2} & c \\ \hline \end{array}\\ \end{aligned}$
Solution:
$\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline X & \frac{1}{2} & 1 & \frac{3}{2} & 2 \\ \hline P(X) & c & c^{2} & 2 c^{2} & c \\ \hline \end{array} \end{aligned}$
We know that the sum of probabilities in a probability distribution is always 1.
$\begin{aligned} &P\left(X=\frac{1}{2}\right)+P(X=1)+P\left(X=\frac{3}{2}\right)+P(2)=1 \\ \end{aligned}$
$\begin{aligned} &c+c^{2}+2 c^{2}+c=1 \\\\ &3 c^{2}+2 c-1=0 \\\\ &3 c^{2}+3 c-c-1=0 \\ \end{aligned}$
$\begin{aligned} &3 c(c+1)-1(c+1)=0 \\\\ &(3 c-1)(c+1)=0 \\\\ &3 c-1=0 \\\\ &c=\frac{1}{3} \end{aligned}$
Or $c+1=0$
$c=-1$
But value of (X) does not support the negative value.
$c=\frac{1}{3}$

Mean and Variance of a Random Variable Exercise Fill in the Blanks Question 5

Answer: $\frac{14}{5}$
Hint: - You must know the rules for solving problems of random variables.
Given: -
$\begin{array}{|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{1}{5} & \frac{3}{10} & \frac{2}{5} & \frac{1}{10} \\ \hline \end{array}$
Solution: - The probability distribution of x is given below: $\begin{aligned} &E\left(X^{2}\right)=\sum_{i=0}^{3} P_{i}\left(X_{i}\right)^{2} \\\\ &=(0)^{2} \times \frac{1}{5}+(1)^{2} \times \frac{3}{10}+(2)^{2} \times \frac{2}{5}+(3)^{2} \times \frac{1}{10} \\\\ &=0+\frac{3}{10}+\frac{8}{5}+\frac{9}{10} \\\\ &=\frac{14}{5} \end{aligned}$

The exercise in the book and the RD Sharma class 12th exercise FBQ is the fill-in blanks type questions, which consists of a total of 5 questions that covers all the essential concepts mentioned below-

Probability distribution
Standard deviation
Mean
Variance
Question based on Random Variables

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