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

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RD Sharma Class 12 Exercise MCQ Mean and Variance of a Random variable Solutions Maths

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

The RD Sharma class 12 solution of Mean and Variance of a Random Variable exercise MCQ is one of the exciting chapters of mathematics Class 12, and students once understand the basic logic of solving. RD Sharma solutions They can perform well in this chapter. Students can make the best use of the RD Sharma class 12th exercise MCQ while solving exercise problems of class 12 RD Sharma. The Class 12 RD Sharma chapter 31 exercise MCQ solution is given in a stepwise manner using simple language for understanding the students.

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

Mean and Variance of a Random Variable Excercise: 31 MCQ

Mean and Variance of a Random Variable exercise multiple choice questions question 1

Answer:

$a=\frac{1}{81}$

Hint:

To solve this we add the probability of all the variable, $\sum_{i=0}^{n} P\left(X_{i}\right)=1$

Given:

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P\left ( x \right )$	$a$	$3a$	$5a$	$7a$	$9a$	$11a$	$13a$	$15a$	$17a$

Solution:

The sum of all probabilities is equal to 1,

$\sum_{i=0}^{n} P\left(X_{i}\right)=1$

Hence, $a + 3 a + 5 a + 7 a + 9 a + 11 a + 13 a + 15 a + 17 a =1$

$\therefore a=\frac{1}{81}$

Mean and Variance of a Random Variable exercise multiple choice questions question 2

Answer:

$0.77$

Hint:

To solve this we use $P\left ( E\cup F \right )=P\left ( E \right )+P\left ( F \right )-P\left ( E\cap F \right )$

Given:

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P\left ( X \right )$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

Solution:

$\begin{aligned} &P(E \cup F)=P(E)+P(F)-P(E \cap F) \\ &P(E)=P(X=2)+P(X=3)+P(X=5)+P(X=7) \ldots\{\text { prime numbers from } 1 \text { to } 8\} \\ &=0.23+0.12+0.2+0.07 \\ &=0.62 \end{aligned}$

$\begin{aligned} &P(F)=P(X=1)+P(X=2)+P(X=3) \ldots\{X<4\} \\ &=0.15+0.23+0.12 \\ &=0.50 \\ \end{aligned}$

$\begin{aligned} &P(E \cap F)=P(\text { Xisaprimenumber }<4) \\ &=P(X=2)+P(X=3) \\ &=0.23+0.12 \\ &=0.35 \\ \end{aligned}$

Mean and Variance of a Random Variable exercise multiple choice questions question 3

Answer:
0.4
Hint:
To solve this we use mean formula
Given:
$P\left ( X=3 \right )=2P\left ( X=1 \right ) and P\left ( x=2 \right )=0.3$ then $P\left ( X=0 \right )$
Solution:
$\begin{aligned} &\text { Mean }=\sum_{i=0}^{3} X_{i} P\left(X_{i}\right)\\ &1.3=X_{0} P(X=0)+X_{1} P(X=1)+X_{2} P(X=2)+X_{3} P(X=3)\\ &1.3=0 \times P(X=0)+1 P(X=1)+2 \times 0.3+3 \times 2 \boldsymbol{P}(\boldsymbol{X}=\mathbf{1})\\ &1.3=0.6+7 P(X=1)\\ &1.3-0.6=7 P(X=1)\\ &0.7=7 P(X=1)\\ &P(X=1)=0.1 \end{aligned}$

$\begin{aligned} As we know, \sum_{i=0}^{n} P\left(X_{i}\right)=1 \end{aligned}$
$\begin{aligned} &P(X=0)+P(X=1)+P(X=2)+P(X=3)=1 \\ &P(X=0)+0.1+0.3+0.2=1 \\ &P(X=0)=1-0.6 \\ &P(X=0)=0.4 \end{aligned}$

Mean and Variance of a Random Variable exercise multiple choice questions question 4

Answer:
$\frac{1}{10}$
Hint:
To solve this we add all the variables
Given:

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P\left ( X \right )$	$0$	$2p$	$2p$	$3p$	$p^{2}$	$2p^{2}$	$7p^{2}$	$2p$

Solution:
$\begin{aligned} &\text { As we know, } \sum_{i=0}^{n} P\left(X_{i}\right)=1 \\ &0+2 p+2 p+3 p+p^{2}+2 p^{2}+7 p^{2}+2 p=1 \\ &10 p^{2}+9 p-1=0 \\ &10 p^{2}+10 p-p-1=0 \\ &10 p(p+1)-1(p+1)=0 \\ &(10 p-1)(p+1)=0 \\ &p=\frac{1}{10} \text { or }-1 \end{aligned}$
Probability cannot be negative. So value of $p=\frac{1}{10}$

Mean and Variance of a Random Variable exercise multiple choice questions question 5

Answer:
$\frac{3}{4}$
Hint:
To solve this we add all variables
Given:
$\begin{array}{ccccc} X & 0 & 1 & 2 & 3 \\ P(X) & k & 3 k & 3 k & k \end{array}$
Solution:
$\begin{aligned} As we know, \sum_{i=0}^{n} P\left(X_{i}\right)=1 \end{aligned}$
$\begin{aligned} &P(X=0)+P(X=1)+P(X=2)+P(X=3)=1 \\ &k+3 k+3 k+k=1 \\ &8 k=1 \\ &k=\frac{1}{8} \end{aligned}$
As we know variance, Var= $\sum X^{2}P\left ( X \right )-\left ( \sum XP\left ( X \right ) \right )^{2}$

$X$	$P\left ( X \right )$	$XP\left ( X \right )$	$X^{2}P\left ( X \right )$
$0$	$\frac{1}{8}$	$0$	$0$
$1$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{3}{8}$
$2$	$\frac{3}{8}$	$\frac{6}{8}$	$\frac{12}{8}$
$3$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{9}{8}$

$\begin{aligned} &\begin{array}{l} \operatorname{Var}=\sum X^{2} P(X)-\left(\sum X P(X)\right)^{2} \\ \end{array} \\ \end{aligned}$
$\begin{aligned} &\begin{array}{l} =3-(1.5)^{2} \\ =3-\frac{9}{4} \end{array} \\ \end{aligned}$
$=\frac{12-9}{4}=\frac{3}{4}$

Mean and Variance of a Random Variable exercise multiple choice questions question 6

Answer:
$E\left ( X \right )=\frac{122}{32} or 3.82$
Hint:
To solve this we add all variables
Given:

$X$	$2$	$3$	$4$	$5$
$P\left ( X \right )$	$\frac{5}{k}$	$\frac{7}{k}$	$\frac{9}{k}$	$\frac{11}{k}$

Solution:
$Aswe know, \sum_{i=0}^{n} P\left(X_{i}\right)=1$
i.e.
$\begin{aligned} &P(X=2)+P(X=3)+P(X=4)+P(X=5)=1 \\ &\frac{5}{k}+\frac{7}{k}+\frac{9}{k}+\frac{11}{k}=1 \\ &\frac{5+7+9+11}{k}=1 \\ &k=32 \\ \end{aligned}$ $E(X)=\sum X \cdot P(X)$

$X$	$P\left ( X \right )$	$X.P\left ( X \right )$
$2$	$\frac{5}{32}$	$\frac{10}{32}$
$3$	$\frac{7}{32}$	$\frac{21}{32}$
$4$	$\frac{9}{32}$	$\frac{36}{32}$
$5$	$\frac{11}{32}$	$\frac{55}{32}$

$E(X)=\sum X \cdot P(X)$

$\begin{aligned} &=\frac{10}{32}+\frac{21}{32}+\frac{36}{32}+\frac{55}{32} \\ &=\frac{122}{32} \end{aligned}$

Mean and Variance of a Random Variable exercise multiple choice questions question 7

Answer:
$-1.8$
Hint:
To solve this equation we use $\sum X.P\left ( X \right )$ formula
Given:
$\begin{array}{cccccc} X & -4 & -3 & -2 & -1 & 0 \\ P(X) & 0.1 & 0.2 & 0.3 & 0.2 & 0.2 \end{array}$
Solution:
$E\left ( X \right )=\sum X.P\left ( X \right )$

$X$	$P\left ( X \right )$	$X.P\left ( X \right )$
$-4$	$0.1$	$-0.4$
$-3$	$0.2$	$-0.6$
$-2$	$0.3$	$-0.6$
$-1$	$0.2$	$-0.2$
$0$	$0.2$	$0$

$E\left ( X \right )=\sum X.P\left ( X \right )$
$=-0.4-0.6-0.6-0.2$
$=-1.8$

Mean and Variance of a Random Variable exercise multiple choice questions question 8

Answer:
$10$
Hint:
To solve this equation we use $E\left(X^{2}\right)=\sum_{i=0}^{n} X_{i}^{2} P(X)$
Given:

$X$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$

Solution:
$E\left(X^{2}\right)=\sum_{i=0}^{n} X_{i}^{2} P(X)$
$E\left(X^{2}\right)=\frac{1}{10}+\frac{4}{5}+\frac{27}{10}+\frac{32}{5}$

$X$	$X^{2}$	$P\left ( X \right )$	$X^{2}P\left ( X \right )$
$1$	$1$	$\frac{1}{10}$	$\frac{1}{10}$
$2$	$4$	$\frac{1}{5}$	$\frac{4}{5}$
$3$	$9$	$\frac{3}{10}$	$\frac{27}{10}$
$4$	$16$	$\frac{2}{5}$	$\frac{32}{5}$

$=\frac{1+8+27+64}{10}$
$=\frac{100}{10}=10$

Mean and Variance of a Random Variable exercise multiple choice questions question 9

Answer:
$E\left ( X^{2} \right )-\left ( E\left ( X \right ) \right )^{2}$
Hint:
Fill the formula
Given:
Let X be a discrete random variable, then the variance of X is
Solution:
Var iance $\left ( X \right )=E\left ( X^{2} \right )-\left ( E\left ( X \right ) \right )^{2}$
i.e. Var $=\sum X^{2} P\left ( X\right )-\left ( \sum XP\left ( X \right ) \right )^{2}$

Mean and Variance of a Random Variable exercise multiple choice questions question 10

Answer:
$E\left ( X \right )=4$
Hint:
To solve this we use $\sum_{k=1}^{3}X.P\left ( X \right )$
Given:

$X$	$30$	$10$	$-10$
$P\left ( X \right )$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{1}{2}$

Solution:
$E\left ( X \right )=\sum_{k=1}^{3}X.P\left ( X \right )$
$=30\times \frac{1}{5}+10\times \frac{3}{10}+\left ( -10 \right )\frac{1}{2}$
$=6+3-5$
$=9-5$
$=4$

The RD Sharma class 12 solutions chapter 31 exercise MCQ consists of a total of 10 questions that covers the essential concepts of the chapter mentioned below-

Probability distribution
Random variables
Discrete random variable
Standard deviation
Expected Mean and Variance

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These solutions are trusted by thousands of students and teachers and are highly recommended by experts for scoring high marks in the board exams.
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Frequently Asked Questions (FAQs)

1. Where can I find the solution for RD Sharma class 12 maths chapter 31?

You can access the RD Sharma class 12 maths chapter 31 by downloading the E-book from the official website of Career360 and access all the other materials.

2. How much is the charge for the RD Sharma class 12 chapter 31 solution?

The solution is free of cost and can be downloaded online from the Career360 website.

3. Is this solution helpful for the preparation of exams?

Yes, the RD Sharma class 12 chapter 31 solution is designed to follow the syllabus of NCERT, which makes it worthy of preparation.

4. Is the RD Sharma class 12 chapter 31 of the latest version?

Yes, the solution is updated yearly following the syllabus of NCERT and is updated accordingly.

5. Do the solutions from Career360 cover all the questions?

Yes, the solutions from RD Sharma class 12 chapter 31 are designed by experts and cover all concepts. Moreover, it contains important questions that have the possible chances to be asked in the exams.