RD Sharma Class 12 Exercise 31.2 Mean and Variance of a Random variable Solutions Maths

RD Sharma Class 12 Exercise 31.2 Mean and Variance of a Random variable Solutions Maths

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 04:38 PM IST

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

Mean and Variance of a Random Variable Excercise: 32 2

Mean and Variance of a random Variable Exercise 31.2 Question 1(i)

Answer:3.1,0.7
Hint: Mean E(X)=\sum_{i=1}^{n} x_{i} \cdot p_{i}, E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} \cdot p_{i}(x)
Given:
\begin{array}{|c|c|c|c|} \hline x_{i} & 2 & 3 & 4 \\ \hline p_{i} & 0.2 & 0.5 & 0.3 \\ \hline \end{array}
Solution:
Mean:=E(X)=\sum_{i=1}^{n} x_{i} p_{i}
\begin{aligned} &=2 \times 0.2+3 \times 0.5+4 \times 0.3 \\ &=0.4+1.5+1.2 \\ &=3.1 \end{aligned}
\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} \cdot p_{i} \\ &=2^{2} \times 0.2+3^{2} \times 0.5+4^{2} \times 0.3 \\ &=0.8+4.5+4.8 \\ &=10.1 \\ \end{aligned}
\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=10.1-3.1^{2} \\ &=10.1-9.61 \\ &=0.49 \end{aligned}
Standard deviation =\sqrt{var\left ( X \right )}=\sqrt{0.49}=0.7

Mean and Variance of a random Variable Exercise 31.2 Question 1(iii)

Answer:
-1,2.9
Hint: Mean E(X)=\sum_{i=1}^{n} x_{i} \cdot p_{i}, E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} \cdot p_{i}
Standard Deviation =\sqrt{var\left ( X \right )}
Given:
\begin{array}{|c|c|c|c|c|} \hline x_{i} & -5 & -4 & 1 & 2 \\ \hline p_{i} & \frac{1}{4} & \frac{1}{8} & \frac{1}{2} & \frac{1}{8} \\ \hline \end{array}
Solution:
\begin{aligned} &=E(X)=\sum_{i=1}^{n} x_{i} p_{i} \\ &=-5 \times \frac{1}{4}+(-4) \times \frac{1}{8}+1 \times \frac{1}{2}+2 \times \frac{1}{8} \\ \end{aligned}
\begin{aligned} &=-\frac{5}{4}-\frac{1}{2}+\frac{1}{2}+\frac{1}{4} \\ &=\frac{-5+1}{4}=-1 \\ \end{aligned}
\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} p_{i} \\ &=(-5)^{2} \times \frac{1}{4}+(-4)^{2} \times \frac{1}{8}+1^{2} \times \frac{1}{2}+2^{2} \times \frac{1}{8} \\ &=\frac{25}{4}+2+\frac{1}{2}+\frac{1}{2} \\ &=\frac{25+8+2+2}{4} \end{aligned}
\begin{aligned} &=\frac{37}{4} \\ \end{aligned}
\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\\\ &=\frac{37}{4}-1=\frac{33}{4} \\ \end{aligned}
\begin{aligned} &\text { Standard deviation }=\sqrt{\operatorname{Var}(X)}=\sqrt{\frac{33}{4}}=2.87=2.9 \end{aligned}

Mean and Variance of a random Variable Exercise 31.2 Question 1(iv)

Answer:
1,1.5
Hint:
Mean=E\left ( X \right )and Standard Deviation =\sqrt{var\left ( X \right )}
Given:
$$ \begin{array}{|c|c|c|c|c|c|} \hline x_{i} & -1 & 0 & 1 & 2 & 3 \\ \hline p_{i} & 0.3 & 0.1 & 0.1 & 0.3 & 0.2 \\ \hline \end{array}
Solution:
Mean=E(X)=\sum_{i=1}^{n} x_{i} p_{i}
=-1 \times 0.3+0 \times 0.1+1 \times 0.1+2 \times 0.3+3 \times 0.2
=-0.3+0.1+0.6+0.6
=1
E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} p_{i}
=(-1)^{2} \times 0.3+0^{2} \times 0.1+1^{2} \times 0.1+2^{2} \times 0.3+3^{2} \times 0.2
=0.3+0.1+1.2+1.8
=3.4
\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}
\begin{aligned} &=3.4-(1)^{2} \\ &=2.4 \end{aligned}
Standard deviation =\sqrt{\operatorname{Var}(X)}=\sqrt{2.4}=1.5

Mean and Variance of a random Variable Exercise 31.2 Question 1(v)

Answer:
2,1
Hint:
Mean=E\left ( X \right )=\sum_{i-1}^{n}p_{i}x_{i}and Standard Deviation =\sqrt{var\left ( X \right )}
Given:
\begin{array}{|c|c|c|c|c|} \hline x_{i} & 1 & 2 & 3 & 4 \\ \hline p_{i} & 0.4 & 0.3 & 0.2 & 0.1 \\ \hline \end{array}
Solution:
Mean=E(X)=\sum_{i=1}^{n} x_{i} p_{i}
=1 \times 0.4+2 \times 0.3+3 \times 0.2+4 \times 0.1\\
=0.4+0.6+0.6+0.4\\
=2
E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} p_{i}\\
=1^{2} \times 0.4+2^{2} \times 0.3+3^{2} \times 0.2+4^{2} \times 0.1\\
=0.4+1.2+1.8+1.6\\
=5
\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=5-4=1 \end{aligned}
Standard deviation =\sqrt{\operatorname{Var}(X)}=\sqrt{1}=1

Mean and Variance of a random Variable Exercise 31.2 Question 1(vi)

Answer:
1.6,1.49
Hint:
Mean=E(X)=\sum_{i=1}^{n} p_{i} x_{i}and Standard Deviation =\sqrt{\operatorname{Var}(X)}
Given:
\begin{array}{|c|c|c|c|c|} \hline x_{i} & 0 & 1 & 3 & 5 \\ \hline p_{i} & 0.2 & 0.5 & 0.2 & 0.1 \\ \hline \end{array}
Solution:
Mean =E(X)=\sum_{i=1}^{n} p_{i} x_{i}
=0 \times 0.2+1 \times 0.5+3 \times 0.2+5 \times 0.1
=0+0.5+0.6+0.5
=1.6
E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} p_{i}
=0^{2} \times 0.2+1^{2} \times 0.5+3^{2} \times 0.2+5^{2} \times(0.1)
=0+0.5+1.8+2.5
=4.8
\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=4.8-2.6 \\ &=2.2 \end{aligned}
Standard deviation =\sqrt{\operatorname{Var}(X)}=\sqrt{2.2}=1.49

Mean and Variance of a random Variable Exercise 31.2 Question 1(vii)

Answer:
0,1.095
Hint:
Mean=E\left ( X \right )=\sum_{i-1}^{n}p_{i}x_{i}and Standard Deviation =\sqrt{var\left ( X \right )}
Given:
\begin{array}{|c|c|c|c|c|c|} \hline x_{i} & -2 & -1 & 0 & 1 & 2 \\ \hline p_{i} & 0.1 & 0.2 & 0.4 & 0.2 & 0.1 \\ \hline \end{array}
Solution:
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} x_{i} p_{i} \\ &=(-2) \times 0.1+(-1) \times 0.2+0 \times 0.4+1 \times 0.2+2 \times 0.1 \\ &=-0.2-0.2+0+0.2+0.2 \\ &=0 \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} p_{i} \end{aligned}
\begin{aligned} &=(-2)^{2} \times 0.1+(-1)^{2} \times 0.2+0^{2} \times 0.4+1^{2} \times 0.2+2^{2} \times 0.1 \\ &=0.4+0.2+0+0.2+0.4 \\ &=1.2 \\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=1.2-0 \\ &=1.2 \\ &\text { Standard deviation }=\sqrt{\operatorname{Var}(X)}=\sqrt{1.2}=1.095 \end{aligned}

Mean and Variance of a random Variable Exercise 31.2 Question 1(viii)

Answer:
-0.2.1.249
Hint:
MeanE\left ( X \right )=\sum_{i-1}^{n}p_{i}x_{i}and Standard Deviation =\sqrt{var\left ( X \right )}
Given:
\begin{array}{|c|c|c|c|c|c|} \hline x_{i} & -3 & -1 & 0 & 1 & 3 \\ \hline p_{i} & 0.05 & 0.45 & 0.20 & 0.25 & 0.05 \\ \hline \end{array}
Solution:
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} x_{i} p_{i} \\ &=(-3) \times 0.05+(-1) \times 0.45+0 \times 0.20+1 \times 0.25+3 \times 0.05 \\ &=-0.15-0.45+0+0.25+0.15 \\ &=-0.2 \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} p_{i} \\ &=(-3)^{2} \times 0.05+(-1)^{2} \times 0.45+0^{2} \times 0.20+1^{2} \times 0.25+3^{2} \times 0.05 \\ &=0.45+0.45+0+0.25+0.45 \\ &=1.6 \\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \end{aligned}
=1.6-(-0.2)
=1.6-(0.04)
=1.56
Standard deviation \sqrt{var}=\sqrt{1.56}=1.249

Mean and Variance of a random Variable Exercise 31.2 Question 1(ix)

Answer:
\frac{35}{18},\frac{\sqrt{665}}{18}
Hint:
Mean=E\left ( X \right )=\sum_{i-1}^{n}p_{i}x_{i}and Standard Deviation =\sqrt{var\left ( X \right )}
Given:
\begin{array}{|c|c|c|c|c|c|c|} \hline x_{i} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline p_{i} & \frac{1}{6} & \frac{5}{18} & \frac{2}{9} & \frac{1}{6} & \frac{1}{9} & \frac{1}{18} \\ \hline \end{array}
Solution:
Mean E\left ( X \right )=\sum_{i-1}^{n}p_{i}x_{i}
\begin{aligned} &=0 \times \frac{1}{6}+1 \times \frac{5}{18}+2 \times \frac{2}{9}+3 \times \frac{1}{6}+4 \times \frac{1}{9}+5 \times \frac{1}{18} \\ &=0+\frac{5}{18}+\frac{4}{9}+\frac{1}{2}+\frac{4}{9}+\frac{5}{18} \\ &=\frac{5+8+8+5+9}{18} \\ &=\frac{35}{18} \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} p_{i} \\ &=0^{2} \times \frac{1}{6}+1^{2} \times \frac{5}{18}+2^{2} \times \frac{2}{9}+3^{2} \times \frac{1}{6}+4^{2} \times \frac{1}{9}+5^{2} \times \frac{1}{18} \\ &=0+\frac{5}{18}+\frac{8}{9}+\frac{3}{2}+\frac{16}{9}+\frac{25}{18} \end{aligned}
\begin{aligned} &=\frac{5+16+27+32+25}{18} \\ &=\frac{105}{18} \\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=\frac{105}{18}-\left(\frac{35}{18}\right)^{2} \\ &=\frac{105}{18}-\frac{1225}{324} \\ &=\frac{1890-1225}{324}=\frac{665}{324} \\ &\text { Standard deviation }=\sqrt{\operatorname{Var}(X)}=\sqrt{\frac{665}{324}}=\frac{\sqrt{665}}{18} \end{aligned}

Mean and Variance of a random Exercise 31.2 Question 3

Answer:
ap+bq,\left ( a^{2}-b^{2} \right )pq,\mid a-b\mid pq
Hint:
E\left ( X \right )=ap+bq
V\left ( X \right )=\left ( a-b \right )^{2}pq
Standard deviation=\mid a-b\mid \sqrt{pq}
Given:
\begin{aligned} &\begin{array}{|c|c|c|} \hline x_{i} & a & b \\ \hline p_{i} & p & q \\ \hline \end{array}\\ &\text { Where } p+q=1 \end{aligned}
Solution:
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} x_{i} p_{i} \\ \end{aligned}
\begin{aligned} &=a \times p+b \times q \\ \end{aligned}
\begin{aligned} &=a p+b q \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} p_{i} \\ \end{aligned}
\begin{aligned} &=a^{2} \times p+b^{2} \times q \\ &=a^{2} p+b^{2} q \\ \end{aligned}
\begin{aligned} &V(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=\left(a^{2} p+b^{2} q\right)-(a p+b q)^{2} \\ &=a^{2} p+b^{2} q-a^{2} p^{2}-b^{2} q^{2}-2 a b p q \\ &=a^{2} p(1-p)+b^{2} q(1-q)-2 a b p q \\ &=a^{2} p q+b^{2} p q-2 a b p q \\ &=(a-b)^{2} p q \\ &\text { Standard deviation }=\sqrt{V(X)}=\sqrt{(a-b)^{2} p q}=|a-b| \sqrt{p q} \end{aligned}

Mean and Variance of a random Exercise 31.2 Question 4

Answer:
3,\frac{3}{4}
Hint:
\begin{aligned} &E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \end{aligned}
Given:
Three coins are tossed
Solution:
S=\left \{ HHH,HHT,HTH,THH,TTH,THT,HTT,TTT \right \}
Set ‘X’ be the random variable for tail
\begin{array}{|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\ \hline \end{array}
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8} \\ &=\frac{3}{8}+\frac{6}{8}+\frac{3}{8} \\ &=\frac{12}{8}=\frac{3}{2} \end{aligned}
\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=0^{2} \times \frac{1}{8}+1^{2} \times \frac{3}{8}+2^{2} \times \frac{3}{8}+3^{2} \times \frac{1}{8} \\ &=0+\frac{3}{8}+\frac{12}{8}+\frac{9}{8} \\ &=\frac{24}{8}=3 \end{aligned}
\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=3-\left(\frac{3}{2}\right)^{2} \\ &=3-\frac{9}{4} \\ &=\frac{3}{4} \end{aligned}

Mean and Variance of a random Exercise 31.2 Question 5

Answer:
Mean=\frac{34}{221} , variance =\frac{400}{2873}, standard deviation=0.37
Hint:
E\left ( X \right )=\sum_{i-1}^{n}X_{i}P_{i}
Standard deviation=\sqrt{var\left ( X \right )}
Given:
Number of kings in 52 cards
Solution:
Let number of kings be X
P (X=0) = P (Number of kings)
=\frac{48C_{2}}{52C_{2}}=\frac{48\times 47}{52\times 51}=\frac{188}{221}
\begin{aligned} &P(X=1)=P(1 \mathrm{King}) \\\\ &=\frac{4 C_{1} 48 C_{1}}{52 C_{2}}=\frac{4 \times 48 \times 2}{52}=\frac{32}{221} \\\\ &P(X=2)=P(2 \mathrm{King}) \\\\ &=\frac{4 C_{2}}{52 C_{2}}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221} \end{aligned}
\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & \frac{188}{221} & \frac{32}{221} & \frac{1}{221} \\ \hline \end{array}
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0 \times \frac{188}{221}+1 \times \frac{32}{221}+2 \times \frac{1}{221} \\ &=\frac{32}{221}+\frac{2}{221} \\ &=\frac{34}{221} \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=0^{2} \times \frac{188}{221}+1^{2} \times \frac{32}{221}+2^{2} \times \frac{1}{221} \\ &=\frac{36}{221} \\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \end{aligned}
\begin{aligned} &=\frac{36}{221}-\left(\frac{34}{221}\right)^{2} \\ &=\frac{400}{2873} \\ &\text { Standard deviation }=\sqrt{\operatorname{var}(X)}=\sqrt{\frac{400}{2873}}=0.37 \end{aligned}

Mean and Variance of a random Exercise 31.2 Question 6

Answer:
3,\frac{3}{4},\frac{\sqrt{3}}{2}
Hint:
E(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { And } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}
Given:
Three coins are tossed
Solution:
S=\left \{ HHH,HHT,HTH,THH,TTH,THT,HTT,TTT \right \}
Let ‘X’ be the random variable for tail
\begin{array}{|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\ \hline \end{array}
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8} \\ &=\frac{3}{8}+\frac{6}{8}+\frac{3}{8} \\ &=\frac{12}{8}=\frac{3}{2} \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \end{aligned}
\begin{aligned} &=0^{2} \times \frac{1}{8}+1^{2} \times \frac{3}{8}+2^{2} \times \frac{3}{8}+3^{2} \times \frac{1}{8} \\ &=0+\frac{3}{8}+\frac{12}{8}+\frac{9}{8} \\ &=\frac{24}{8}=3 \\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=3-\left(\frac{3}{2}\right)^{2}=3-\frac{9}{4} \\ &=\frac{3}{4} \\ \end{aligned}
\begin{aligned} &\text { Standard deviation }=\sqrt{\operatorname{Var}(X)}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2} \end{aligned}

Mean and Variance of a random Exercise 31.2 Question 8

Answer:
2.53,1.96
Hint:
E(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { and } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}
Given:
Let X be the random variable which denotes the minimum of two numbers which appears
Solution:
\begin{aligned} &S=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\\ \end{aligned}
\begin{aligned} &\begin{array}{|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(X) & \frac{11}{36} & \frac{9}{36} & \frac{7}{36} & \frac{5}{36} & \frac{3}{36} & \frac{1}{36} \\ \hline \end{array} \end{aligned}
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=1 \times \frac{11}{36}+2 \times \frac{9}{36}+3 \times \frac{7}{36}+4 \times \frac{5}{36}+5 \times \frac{3}{36}+6 \times \frac{1}{36} \\ &=\frac{91}{36}=2.53 \\ \end{aligned}
\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=1^{2} \times \frac{11}{36}+2^{2} \times \frac{9}{36}+3^{2} \times \frac{7}{36}+4^{2} \times \frac{5}{36}+5^{2} \times \frac{3}{36}+6^{2} \times \frac{1}{36} \\ &=\frac{301}{36} \\ &=8.36 \end{aligned}
\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=8.36-2.53^{2} \\ &=1.96 \end{aligned}

Mean and Variance of a Random Variable Exercise 31.2 Question 9

Answer:
2,1
Hint:
E(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { and } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}
Given:
A fair coin tossed 4 times
Solution:
\begin{aligned} &\begin{array}{|c|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & \frac{1}{16} & \frac{1}{4} & \frac{3}{8} & \frac{1}{4} & \frac{1}{16} \\ \hline \end{array}\\ \end{aligned}
\begin{aligned} &\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0 \times \frac{1}{16}+1 \times \frac{1}{4}+2 \times \frac{3}{8}+3 \times \frac{1}{4}+4 \times \frac{1}{16} \\ &=2 \\ \end{aligned} \end{aligned}
\begin{aligned} &\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=\frac{1}{4}+\frac{3}{2}+\frac{9}{4}+1=5 \\ \end{aligned} \end{aligned}
\begin{aligned} &\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=5-2^{2} \\ &=5-4=1 \end{aligned} \end{aligned}

Mean and Variance of a Random Variable Exercise 31.2 Question 10

Answer:
7,11.67
Hint:
E(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { and } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}
Given:
A fair dice is tossed
Solution:
\begin{aligned} &\begin{aligned} &S=\{1,2,3,4,5,6\} \\ &X=\{2,4,6,8,10,12\} \end{aligned}\\ &\begin{array}{|c|c|c|c|c|c|c|} \hline X & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline P(X) & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\ \hline \end{array} \end{aligned}
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=\frac{1}{6}[2+4+6+8+10+12] \\ &=\frac{42}{6}=7 \\ \end{aligned}
\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=\frac{1}{6}\left[2^{2}+4^{2}+6^{2}+8^{2}+10^{2}+12^{2}\right] \\ &=\frac{1}{6}[4+16+36+64+100+144] \\ &=60.67 \\ \end{aligned}
\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\\\ &=60.67-49 \\\\ &=11.67 \end{aligned}

Mean and Variance of a Random Variable Exercise 31.2 Question 11

Answer:
2,1
Hint:
E(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { and } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}
Given:
A fair dice is tossed
Solution:
\begin{aligned} &S=\{1,2,3,4,5,6\}\\ &\begin{array}{|c|c|c|} \hline X & 1 & 3 \\ \hline P(X) & \frac{1}{2} & \frac{1}{2} \\ \hline \end{array} \end{aligned}
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=1 \times \frac{1}{2}+3 \times \frac{1}{2} \\ &=\frac{1}{2}+\frac{3}{2}=2 \\ \end{aligned}
\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=1^{2} \times \frac{1}{2}+3^{2} \times \frac{1}{2} \\ &=5 \\ \end{aligned}
\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=5-2^{2} \\ &=5-4 \\ &=1 \end{aligned}

Mean and Variance of a Random Variable Exercise 31.2 Question 12

Answer:
2,1,1
Hint:
E(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { and } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}
Given:
A fair coin tossed four times
Solution:
X denotes the occurrence of head
Sample space \left \{ \text { HHHH,HHHT,HHTH,HHTT,HTHH,HTHT,HTTH,HTTT,THHH,THHT,THTH,THTT, TTHH, TTHT, TTTH, TTTT \} } \right \}
\begin{aligned} &\begin{array}{|c|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & \frac{1}{16} & \frac{1}{4} & \frac{3}{8} & \frac{1}{4} & \frac{1}{16} \\ \hline \end{array}\\ \end{aligned}
\begin{aligned} &\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0 \times \frac{1}{16}+1 \times \frac{1}{4}+2 \times \frac{3}{8}+3 \times \frac{1}{4}+4 \times \frac{1}{16} \\ &=\frac{1}{4}+\frac{3}{4}+\frac{3}{4}+\frac{1}{4} \\ &=2 \end{aligned} \end{aligned}
\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=0 \times \frac{1}{16}+1^{2} \times \frac{1}{4}+2^{2} \times \frac{3}{8}+3^{2} \times \frac{1}{4}+4^{2} \times \frac{1}{16} \\ &=0+\frac{1}{4}+\frac{3}{2}+\frac{9}{4}+1 \\ &=5 \\ \end{aligned}
\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=5-2^{2} \\ &=5-4 \\ &=1 \end{aligned}
Standard deviation =\sqrt{\operatorname{Var}(X)}=\sqrt{1}=1

Mean and Variance of a Random Variable Exercise 31.2 Question 13

Answer:
Mean = 2.3
Variance = 5
Hint:
E(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { and } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}
Given:
There are 5 cards numbered 1, 1, 2, 2 and 3
Solution:
There are 5 cards numbered 1, 1, 2, 2 and 3
Let, x = sum of two numbers on card =2, 3, 4, 5
Y = Maximum of two numbers = 1, 2, 3
Thus the probability distribution of x is given by
\begin{array}{|c|c|c|c|c|} \hline x & 2 & 3 & 4 & 5 \\ \hline P(x) & \frac{1}{10} & \frac{4}{10} & \frac{3}{10} & \frac{2}{10} \\ \hline \end{array}
Total number of cards = 5 (2 cards with “1” + 2 cards with “2” + 1 card with “3”
Total number of possible choice (in drawing the two cards) = number of ways in which threw two cards can be drawn from the total 5
\begin{aligned} &=5 C_{2} \\ &=\frac{5 !}{2 ! 3 !}=10 \end{aligned}
Probability that the two car drawn would be having the number 1 and 1 on them is P (11)
P(1,1)=\frac{2 C_{2} \times 2 C_{0} \times 1 C_{0}}{5 C_{2}}=\frac{1 \times 1 \times 1}{10}=\frac{1}{10}
Probability that two cards drawn would be having the number 1 and 2 on them is P (11)
\begin{aligned} &P(1,1)=\frac{2 C_{2} \times 2 C_{0} \times 1 C_{0}}{5 C_{2}} \\ &=\frac{1 \times 1 \times 1}{10}=\frac{1}{10} \end{aligned}
"1" and "2" on them
P(1,2)=\frac{2 C_{1} \times 2 C_{1} \times 1 C_{0}}{5 C_{2}}
\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline & 1^{\prime} s & 2^{\prime} s & 3^{\prime} s & \text { Total } \\ \hline \text { Available } & 2 & 2 & 1 & 5 \\ \hline \text { To choose } & 1 & 1 & 0 & 2 \\ \hline \text { Choices } & 2 C_{1} & 2 C_{1} & 1 C_{0} & 5 C_{2} \\ \hline \end{array}\\ &\begin{aligned} &=\frac{\frac{2}{1} \times \frac{2}{1} \times 1}{10} \\ &=\frac{2 \times 2 \times 1}{10}=\frac{4}{10}=\frac{2}{5} \end{aligned} \end{aligned}
"1" and "3" on them
P(1,3)=\frac{2 C_{1} \times 2 C_{0} \times 1 C_{1}}{5 C_{2}}
\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline & 1^{\prime} s & 2^{\prime} s & 3^{\prime} s & \text { Total } \\ \hline \text { Available } & 2 & 2 & 1 & 5 \\ \hline \text { To choose } & 1 & 0 & 1 & 2 \\ \hline \text { Choices } & 2 C_{1} & 2 C_{0} & 1 C_{1} & 5 C_{2} \\ \hline \end{array}\\ &=\frac{\frac{2}{1} \times 1 \times \frac{1}{1}}{10}=\frac{2 \times 1 \times 1}{10}=\frac{2}{10}=\frac{1}{5} \end{aligned}
"2" and "2" on them
P(2,2)=\frac{2 C_{1} \times 2 C_{2} \times 1 C_{1}}{5 C_{2}}
\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline & 1^{\prime} s & 2^{\prime} s & 3^{\prime} s & \text { Total } \\ \hline \text { Available } & 2 & 2 & 1 & 5 \\ \hline \text { To choose } & 0 & 2 & 0 & 2 \\ \hline \text { Choices } & 2 C_{0} & 2 C_{2} & 1 C_{0} & 5 C_{2} \\ \hline \end{array}\\ &=\frac{1 \times 1 \times 1}{10}=\frac{1}{10} \end{aligned}
"2" and "3" on them
P(2,3)=\frac{2 C_{0} \times 2 C_{1} \times 1 C_{1}}{5 C_{2}}\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline & 1^{\prime} s & 2^{\prime} s & 3^{\prime} s & \text { Total } \\ \hline \text { Available } & 2 & 2 & 1 & 5 \\ \hline \text { To choose } & 0 & 1 & 1 & 2 \\ \hline \text { Choices } & 2 C_{0} & 2 C_{1} & 1 C_{1} & 5 C_{2} \\ \hline \end{array}\\ &=\frac{1 \times \frac{2}{1} \times 1}{10}=\frac{1 \times 2 \times 1}{10}=\frac{2}{10}=\frac{1}{5} \end{aligned}

Probability for the sum of the members on the cards drawn to be

\begin{aligned} &2 \Rightarrow P(x=2)=P(1,1)=\frac{1}{10} \\\\ &3 \Rightarrow P(x=3)=P(1,2)=\frac{4}{10} \\\\ &4 \Rightarrow P(x=4)=P(1,3 \text { or } 2,2) \\\\ &\text { i. } e, P(1,3 \text { or } 2,2)=P(1,3)+P(2,2) \\\\ \end{aligned}

\begin{aligned} &=\frac{2}{10}+\frac{1}{10} \\\\ &=\frac{2+1}{10}=\frac{3}{10} \\\\ &5 \Rightarrow P(x=2)=P(2,3)=\frac{2}{10} \end{aligned}

The probability distribution of x would be

\begin{array}{|c|c|c|c|c|} \hline x & 2 & 3 & 4 & 5 \\ \hline P(X=x) & \frac{1}{10} & \frac{4}{10} & \frac{3}{10} & \frac{2}{10} \\ \hline \end{array}

Calculations for mean and standard deviation

\begin{array}{|c|c|c|c|c|} \hline x & P(X=x) & P x=x \cdot P(X=x) & x^{2} & P x^{2}=x^{2} \cdot P(X=x) \\ \hline 2 & \frac{1}{10} & \frac{2}{10} & 4 & \frac{4}{10} \\ \hline 3 & \frac{4}{10} & \frac{12}{10} & 9 & \frac{36}{10} \\ \hline 4 & \frac{3}{10} & \frac{12}{10} & 16 & \frac{48}{10} \\ \hline 5 & \frac{2}{10} & \frac{10}{10} & 25 & \frac{50}{10} \\ \hline \text { Total } & 1 & \frac{36}{10}=3.6 & & \frac{138}{10}=13.6 \\ \hline \end{array}

The expected value of the sum

Expectation of x

\begin{aligned} &E(x)=\sum P x=3.6 \\ \end{aligned}

Variance of the sum of the numbers on the cards

\begin{aligned} &\operatorname{Var}(x)=E\left(x^{2}\right)-(E(x))^{2} \\ &=\sum P x-\left(\sum P x\right)^{2} \\ &=13.8-3.6^{2} \\ &=13.8-12.96 \\ &=0.84 \end{aligned}

Standard deviation of the sum of the numbers on the cards

\begin{aligned} &\text { S.D }(x)=+\sqrt{\operatorname{Var}(x)} \\ &=+\sqrt{0.84} \\ &=+0.917 \end{aligned}

Computation of mean and variance

\begin{array}{|c|c|c|c|} \hline x_{i} & p_{i} & p_{i} x_{i} & p_{i} x_{i}^{2} \\ \hline 2 & \frac{1}{10} & \frac{2}{10} & \frac{4}{10} \\ \hline 3 & \frac{4}{10} & \frac{12}{10} & \frac{36}{10} \\ \hline 4 & \frac{3}{10} & \frac{12}{10} & \frac{48}{10} \\ \hline 5 & \frac{2}{10} & 1 & \frac{50}{10} \\ \hline \end{array}

\begin{aligned} &\sum p_{i} x_{i}=\frac{36}{10}=3.6 \\ &\sum p_{i} x_{i}^{2}=\frac{138}{10}=13.8 \\ &\text { Mean }=\sum p_{i} x_{i}=3.6 \\ &\text { Variance }=\sum p_{i} x_{i}^{2}-(\text { Mean })^{2}=1 \end{aligned}

Thus the probability distribution of Y is given by

\begin{array}{|c|c|c|c|} \hline Y & 1 & 2 & 3 \\ \hline P(Y) & 0.1 & 0.5 & 0.4 \\ \hline \end{array}

Computation of Mean and Variance

\begin{array}{|c|c|c|c|} \hline y_{i} & p_{i} & p_{i} y_{i} & p_{i} y_{i}{ }^{2} \\ \hline 1 & 0.1 & 0.1 & 0.1 \\ \hline 2 & 0.5 & 1 & 2 \\ \hline 3 & 0.4 & 1.2 & 3.6 \\ \hline \end{array}

\begin{aligned} &\sum p_{i} y_{i}=2.3 \\\\ &\sum p_{i} y_{i}^{2}=5.7 \\\\ &\text { Mean }=\sum p_{i} y_{i}=2.3 \\\\ &\text { Variance }=\sum p_{i} y_{i}^{2}-(\text { mean })^{2}=5.7-5.29=0.41 \end{aligned}

Mean and Variance of a Random Variable Exercise 31.2 Question 14

Answer:
Variance=\frac{1}{2}
Hint:
Var\left ( X \right )=E\left ( X^{2} \right )-\left ( E\left ( X \right ) \right )^{2}
Given:
A die is tossed twice a “success” of getting odd number of toss
Solution:
\begin{aligned} &S=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,1) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\\ \end{aligned}
\begin{aligned} &\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & \frac{9}{36} & \frac{18}{36} & \frac{9}{36} \\ \hline \end{array}\\ \end{aligned}
\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \end{aligned}
\begin{aligned} &=\frac{18}{36}+1-\left(\frac{18}{36}+\frac{18}{36}\right)^{2} \\ &=1.5-1=0.5 \\ &=\frac{1}{2} \end{aligned}

Mean and Variance of a Random Variable Exercise 31.2 Question 16

Answer:
-\frac{30}{13}
Hint:
E\left ( X \right )=\sum_{i-1}^{n}X_{i}P_{i}
Given:
Person loss 10 rupees and get 90 rupees.
Solution:
P (Wheels/Getting same slot) =\frac{1}{13}=p
\begin{aligned} &p^{\prime}=q=1-p=1-\frac{1}{13}=\frac{12}{13}\\ &\begin{array}{|c|c|c|} \hline x_{i} & 90 & -10 \\ \hline p\left(x_{i}\right) & \frac{1}{13} & \frac{12}{13} \\ \hline \end{array} \end{aligned}
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} x_{i} p_{i} \\ &=90 \times \frac{1}{13}-10 \times \frac{12}{13} \\ &=\frac{90}{13}-\frac{120}{13} \\ &=-\frac{30}{13} \end{aligned}

Mean and Variance of a Random Variable Exercise 31.2 Question 17

Answer:
1.5
Hint:
MeanE\left ( X \right )=\sum_{i-1}^{n}X_{i}P_{i}
Given:
Three cards are drawn at randomly from a shuffle pack of 52 cards
Solution:
In 52 cards
26 = Red
26 = Black
Number of outcomes
3 Red 0 Black
2 Red 1 Black
1 Red 2 Black
0 Red 3 Black
\begin{aligned} &P(X=0)+P(X=1)+P(X=2)+P(X=3) \\\\ &=\frac{26 C_{0} \times 26 C_{3}}{52 C_{3}}+\frac{26 C_{1} \times 26 C_{2}}{52 C_{3}}+\frac{26 C_{2} \times 26 C_{1}}{52 C_{3}}+\frac{26 C_{3} \times 26 C_{0}}{52 C_{3}} \\\\ &=1.5 \\\\ &\text { Mean } E(X)=\sum_{i=1}^{n} x_{i} p_{i}=\frac{3}{2}=1.5 \end{aligned}

Mean and Variance of a Random Variable Exercise 31.2 Question 18

Answer:
\frac{4}{7},0.34
Hint:
MeanE(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { and } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}
Given:
An urn contains 5 red 2 black balls. Two balls are randomly drawn, without replacement.
Solution:
\begin{aligned} &\begin{aligned} &P(X=0)=\frac{5 C_{2}}{7 C_{2}}=\frac{10}{21} \\ &P(X=1)=\frac{2 C_{1} \times 5 C_{1}}{7 C_{2}}=\frac{10}{21} \\ &P(X=2)=\frac{2 C_{2}}{7 C_{2}}=\frac{1}{21} \end{aligned}\\ &\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & \frac{10}{21} & \frac{10}{21} & \frac{1}{21} \\ \hline \end{array} \end{aligned}
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0+\frac{10}{21}+\frac{2}{21} \\ &=\frac{12}{21}=\frac{4}{7} \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=\frac{10}{21}+\frac{4}{21} \\ &=\frac{14}{21}=\frac{2}{3} \\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=\frac{2}{3}-\frac{16}{49} \\ &=\frac{50}{147}=0.34 \end{aligned}

Mean and Variance of a Random Variable Exercise 31.2 Question 19

Answer:
\frac{14}{9}
Hint:
Mean E(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { and } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}
Given:
Positive integer 2, 3,4,5,6 and 7
Solution:
\begin{aligned} &P(X=3)=\frac{1}{15} \\ &P(X=4)=\frac{2}{15} \\ &P(X=5)=\frac{3}{15} \\ &P(X=6)=\frac{4}{15} \\ &P(X=7)=\frac{5}{15} \end{aligned}
\begin{aligned} &\begin{array}{|c|c|c|c|c|c|} \hline X & 3 & 4 & 5 & 6 & 7 \\ \hline P(X) & \frac{1}{15} & \frac{2}{15} & \frac{3}{15} & \frac{4}{15} & \frac{5}{15} \\ \hline \end{array}\\ \end{aligned}
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i}\\ \end{aligned}
\begin{aligned} &\begin{aligned} &=\frac{85}{15}=\frac{17}{3} \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=3^{2} \times \frac{1}{15}+4^{2} \times \frac{2}{15}+5^{2} \times \frac{3}{15}+6^{2} \times \frac{4}{15}+7^{2} \times \frac{5}{15} \\ &=\frac{101}{3} \end{aligned}\\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \end{aligned}
=\frac{101}{3}-\left ( \frac{17}{3} \right )^{2}
=\frac{14}{9}

Mean and Variance of a Random Variable Exercise 31.2 Question 20

Answer:
\frac{19}{9}
Hint:
Mean E\left ( X \right )=\sum_{i-1}^{n}X_{i}P_{i}
Given:
A die has thrown
Solution:
\begin{aligned} &P(X=5)=\frac{9}{27} \\ &P(X=4)=\frac{6}{27} \\ &P(X=3)=\frac{4}{27} \\ &P(X=-3)=\frac{8}{27} \\ &E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ \end{aligned}
\begin{aligned} &=5 \times \frac{8}{27}+4 \times \frac{6}{27}+3 \times \frac{4}{27}+(-3) \times \frac{8}{27} \\ &=\frac{45+24+12-24}{27} \\ &=\frac{57}{27}=\frac{19}{9} \end{aligned}

Mean and Variance of a Random Variable Exercise 31.2 Question 21

Answer:
\frac{2}{3}
Hint:
Mean E\left ( X \right )=\sum_{i-1}^{n}X_{i}P_{i}
Given:
A pair of dice are thrown
Solution:
\begin{aligned} &S=\{(1,1),(2,2),(3,3,),(4,4),(5,5),(6,6)\} \\\\ &p=\frac{6}{36}=\frac{1}{6}, q=\frac{5}{6} \\ \end{aligned}
\begin{aligned} &P(X=0)=4 C_{0}\left(\frac{1}{6}\right)^{0} \times\left(\frac{5}{6}\right)^{4}=\frac{625}{1296} \\\\ &P(X=1)=4 C_{1}\left(\frac{1}{6}\right) \times\left(\frac{5}{6}\right)^{3}=\frac{125}{324} \\\\ &P(X=2)=4 C_{2}\left(\frac{1}{6}\right)^{2} \times\left(\frac{5}{6}\right)^{2}=\frac{25}{216} \\ \end{aligned}
\begin{aligned} &P(X=3)=4 C_{3}\left(\frac{1}{6}\right)^{3} \times\left(\frac{5}{6}\right)=\frac{5}{324} \\\\ &P(X=4)=4 C_{4}\left(\frac{1}{6}\right)^{4} \times\left(\frac{5}{6}\right)^{0}=\frac{1}{1296} \end{aligned}
\begin{array}{|c|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & \frac{625}{1296} & \frac{125}{324} & \frac{25}{216} & \frac{5}{324} & \frac{1}{1296} \\ \hline X \cdot P(X) & 0 & \frac{125}{324} & \frac{25}{108} & \frac{5}{108} & \frac{1}{324} \\ \hline \end{array}
\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0+\frac{125}{324}+\frac{25}{108}+\frac{5}{108}+\frac{1}{324} \\ &=\frac{216}{324} \\ &=\frac{2}{3} \end{aligned}

Mean and Variance of a Random Variable Exercise 31.2 Question 22

Answer:
(i) c=\frac{1}{5} (ii) Mean=3.2 (iii) Variance=2.9
Hint:
P\left ( X \right )=1
Given:
\begin{array}{|c|c|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(X) & 4 c^{2} & 3 c^{2} & 2 c^{2} & c^{2} & c & 2 c \\ \hline \end{array}
Solution:
(i) We know that
\begin{aligned} &4 c^{2}+3 c^{2}+2 c^{2}+c^{2}+c+2 c=1 \\ &10 c^{2}+3 c-1=0 \\ &10 c^{2}+5 c-2 c-1=0 \\ &5 c(2 c+1)-1(2 c+1)=0 \\ &(2 c+1)(5 c-1)=0 \\ &c=\frac{1}{5} \because c \geq 0 \end{aligned}
(ii) Mean E\left ( X \right )=\sum_{i-1}^{n}X_{i}P_{i}
\begin{aligned} &=3 c^{2}+4 c^{2}+3 c^{2}+4 c+10 c \\ &=10 c^{2}+14 c \\ &=10 \times \frac{1}{25}+14 \times \frac{1}{5} \\ &=\frac{10}{25}+\frac{14}{5} \\ &=\frac{16}{5}=3.2 \end{aligned}
\begin{aligned} &\text { (iii) } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=14-(3.32)^{2} \\ &=2.9 \end{aligned}

The RD Sharma class 12 solution of Mean and Variance of a Random Variable exercise 31.2 comprises of two-level questions which varies from easy to moderate to tough, which consists of a total of 33 questions covering the essential concepts of the chapter mentioned below-

  • Probability distribution

  • Mean of a discrete random variable

  • Standard deviation

  • Mean of any probability distribution

Well, the RD Sharma class 12th exercise 31.2 is recommended by experts for several reasons mentioned below:-

  • The RD Sharma class 12 solutions chapter 31 exercise 31.2 can be downloaded from the Careers360 website for free of cost.

  • The RD Sharma class 12th exercise 31.2 consists of questions that have the chances that they might be asked in the board exams as observed previously.

  • The RD Sharma class 12 chapter 31 exercise 31.2 can also be used by teachers for preparing lectures and also question papers for school exams.

  • The RD Sharma class 12th exercise 31.2 also contains solved questions which cannot be found in the NCERT textbooks.

  • The RD Sharma class 12th exercise 31.2 provides helpful tips to solve questions in easy methods which cannot be taught in any school.

  • The RD Sharma class 12 chapter 31 exercise 31.2 is also helpful in solving homeworks as it contains solved questions and tips.

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