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

RD Sharma Class 12 Solutions Chapter31 FBQ Mean and Variance of a Random Variable- Other Exercise

Mean and Variance of a Random Variable Excercise: 31 FBQ

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 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

RD Sharma class 12 solutions chapter 31 exercise FBQ are the highly recommended study material because of the following reasons :-

The RD Sharma class 12th exercise FBQ is helpful in solving questions as the material provides you with solved question samples that make it easier for students to take reference from it for better understanding.
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The questions are prepared by experts from around the country and provide tips to solve questions in a way that might not be taught in school.
The RD Sharma class 12th exercise FBQ can easily be downloaded from the careers360 website.

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

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Mean and Variance of a Random Variable Excercise: 31 FBQ

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