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

# 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.

## Mean and Variance of a Random Variable Excercise: 31 MCQ

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

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

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

$0.77$

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

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

$\frac{1}{10}$
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

$\frac{3}{4}$
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

$E\left ( X \right )=\frac{122}{32} or 3.82$
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

$-1.8$
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

$10$
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

$E\left ( X^{2} \right )-\left ( E\left ( X \right ) \right )^{2}$
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

$E\left ( X \right )=4$
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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