RD Sharma Class 12 Exercise 31.2 Mean and Variance of a Random variable Solutions Maths-Download PDF Free Online

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 (X^{2}) = \sum_{i = 1}^{n} x_{i}^{2} \cdot p_{i} (x)

Given:

\begin{array}{cccc} x_{i} & 2 & 3 & 4 \\ p_{i} & 0.2 & 0.5 & 0.3 \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 (X^{2}) = \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} Var (X) = E (X^{2}) - (E (X))^{2} \\ = 10.1 - {3.1}^{2} \\ = 10.1 - 9.61 \\ = 0.49 \end{aligned}

Standard deviation

= \sqrt{v a r (X)} = \sqrt{0.49} = 0.7

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

Answer:

3, 1.7

Hint:
Mean

E (X) = \sum_{i = 1}^{n} x_{i} \cdot p_{i}, E (X^{2}) = \sum_{i = 1}^{n} x_{i}^{2} \cdot p_{i}

Given:

\begin{array}{ccccc} x_{i} & 1 & 3 & 4 & 5 \\ p_{i} & 0.4 & 0.1 & 0.2 & 0.3 \end{array}

Solution:
Mean

\begin{aligned} = E (X) = \sum_{i = 1}^{n} p_{i} x_{i} \end{aligned}

\begin{aligned} = 0.4 \times 1 + 0.1 \times 3 + 0.2 \times 4 + 0.3 \times 5 \\ = 0.4 + 0.3 + 0.8 + 1.5 \\ = 3 \end{aligned}

\begin{aligned} E (X^{2}) = \sum_{i = 1}^{n} x_{i}^{2} p_{i} \end{aligned}

\begin{aligned} = 1^{2} \times 0.4 + 3^{2} \times 0.1 + 4^{2} \times 0.2 + 5^{2} \times 0.3 \\ = 0.4 + 0.9 + 3.2 + 7.5 \\ = 12 \end{aligned}

\begin{aligned} Var (X) = E (X^{2}) - (E (X))^{2} \\ = 12 - 9 \\ = 3 \\ Standard deviation = \sqrt{Var (X)} = \sqrt{3} = 1.7 \end{aligned}

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 (X^{2}) = \sum_{i = 1}^{n} x_{i}^{2} \cdot p_{i}

Standard Deviation

= \sqrt{v a r (X)}

Given:

\begin{array}{ccccc} x_{i} & - 5 & - 4 & 1 & 2 \\ p_{i} & \frac{1}{4} & \frac{1}{8} & \frac{1}{2} & \frac{1}{8} \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 (X^{2}) = \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} Var (X) = E (X^{2}) - (E (X))^{2} \\ = \frac{37}{4} - 1 = \frac{33}{4} \end{aligned}

\begin{aligned} Standard deviation = \sqrt{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 (X)

and Standard Deviation

= \sqrt{v a r (X)}

Given:
$$

\begin{array}{cccccc} x_{i} & - 1 & 0 & 1 & 2 & 3 \\ p_{i} & 0.3 & 0.1 & 0.1 & 0.3 & 0.2 \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 (X^{2}) = \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

Var (X) = E (X^{2}) - (E (X))^{2}

\begin{aligned} = 3.4 - (1)^{2} \\ = 2.4 \end{aligned}

S t a n d a r d d e v i a t i o n = \sqrt{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 (X) = \sum_{i - 1}^{n} p_{i} x_{i}

and Standard Deviation

= \sqrt{v a r (X)}

Given:

\begin{array}{ccccc} x_{i} & 1 & 2 & 3 & 4 \\ p_{i} & 0.4 & 0.3 & 0.2 & 0.1 \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 (X^{2}) = \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} Var (X) = E (X^{2}) - (E (X))^{2} \\ = 5 - 4 = 1 \end{aligned}

S t a n d a r d d e v i a t i o n = \sqrt{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{Var (X)}

Given:

\begin{array}{ccccc} x_{i} & 0 & 1 & 3 & 5 \\ p_{i} & 0.2 & 0.5 & 0.2 & 0.1 \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 (X^{2}) = \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} Var (X) = E (X^{2}) - (E (X))^{2} \\ = 4.8 - 2.6 \\ = 2.2 \end{aligned}

S t a n d a r d d e v i a t i o n = \sqrt{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 (X) = \sum_{i - 1}^{n} p_{i} x_{i}

and Standard Deviation

= \sqrt{v a r (X)}

Given:

\begin{array}{cccccc} x_{i} & - 2 & - 1 & 0 & 1 & 2 \\ p_{i} & 0.1 & 0.2 & 0.4 & 0.2 & 0.1 \end{array}

Solution:

\begin{aligned} 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 (X^{2}) = \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 \\ Var (X) = E (X^{2}) - (E (X))^{2} \\ = 1.2 - 0 \\ = 1.2 \\ Standard deviation = \sqrt{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:
Mean

E (X) = \sum_{i - 1}^{n} p_{i} x_{i}

and Standard Deviation

= \sqrt{v a r (X)}

Given:

\begin{array}{cccccc} x_{i} & - 3 & - 1 & 0 & 1 & 3 \\ p_{i} & 0.05 & 0.45 & 0.20 & 0.25 & 0.05 \end{array}

Solution:
$\begin{aligned} Mean E (X) = \sum_{i = 1}^{n} x_{i} p_{i} \\ = (- 3) \times 0.05 + (- 1) \times 0.45 + 0 \times 0.20 + 1 \times 0.25 + 3 \times 0.05 \\ = - 0.15 - 0.45 + 0 + 0.25 + 0.15 \\ = - 0.2 \\ E (X^{2}) = \sum_{i = 1}^{n} x_{i}^{2} p_{i} \\ = (- 3)^{2} \times 0.05 + (- 1)^{2} \times 0.45 + 0^{2} \times 0.20 + 1^{2} \times 0.25 + 3^{2} \times 0.05 \\ = 0.45 + 0.45 + 0 + 0.25 + 0.45 \\ = 1.6 \\ Var (X) = E (X^{2}) - (E (X))^{2} \end{aligned}$
$= 1.6 - (- 0.2)$
$= 1.6 - (0.04)$
$= 1.56$
Standard deviation

\sqrt{v a r} = \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 (X) = \sum_{i - 1}^{n} p_{i} x_{i}

and Standard Deviation

= \sqrt{v a r (X)}

Given:

\begin{array}{ccccccc} x_{i} & 0 & 1 & 2 & 3 & 4 & 5 \\ p_{i} & \frac{1}{6} & \frac{5}{18} & \frac{2}{9} & \frac{1}{6} & \frac{1}{9} & \frac{1}{18} \end{array}

Solution:
Mean

E (X) = \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 (X^{2}) = \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} \\ Var (X) = E (X^{2}) - (E (X))^{2} \\ = \frac{105}{18} - {(\frac{35}{18})}^{2} \\ = \frac{105}{18} - \frac{1225}{324} \\ = \frac{1890 - 1225}{324} = \frac{665}{324} \\ Standard deviation = \sqrt{Var (X)} = \sqrt{\frac{665}{324}} = \frac{\sqrt{665}}{18} \end{aligned}

Mean and Variance of a random Variable Exercise 31.2 Question 2

Answer:
$\frac{1}{3} . \frac{23}{18}$
Hint:
$\sum_{i - 1}^{n} p_{i} = 1$
Given:
$\begin{array}{ccccc} X & 0.5 & 1 & 1.5 & 2 \\ P (X) & k & k^{2} & 2 k^{2} & k \end{array}$
Solution:
(i)
$\begin{aligned} \sum_{i = 1}^{n} P (X_{i}) = 1 \\ k + k^{2} + 2 k^{2} + k = 1 \\ 3 k^{2} + 2 k - 1 = 0 \\ 3 k^{2} + 3 k - k - 1 = 0 \\ (3 k - 1) (k + 1) = 0 \\ k = \frac{1}{3}, - 1 \\ k \geq 0, k \neq - 1 \\ k = \frac{1}{3} \end{aligned}$
$\begin{aligned} (ii) E (X) = \sum_{i = 1}^{n} X_{i} P_{i} \\ = 0.5 \times k + 1 \times k^{2} + 1.5 \times 2 k^{2} + 2 k \\ = 4 k^{2} + 2.5 k \\ k = \frac{1}{3} \\ = 4 \times {(\frac{1}{3})}^{2} + 2.5 \times \frac{1}{3} = \frac{23}{18} \end{aligned}$

Mean and Variance of a random Exercise 31.2 Question 3

Answer:

a p + b q, (a^{2} - b^{2}) p q, ∣ a - b ∣ p q

Hint:

E (X) = a p + b q

V (X) = {(a - b)}^{2} p q

Standard deviation

=∣ a - b ∣ \sqrt{p q}

Given:

\begin{aligned} \begin{array}{ccc} x_{i} & a & b \\ p_{i} & p & q \end{array} \\ Where p + q = 1 \end{aligned}

Solution:

\begin{aligned} 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 (X^{2}) = \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 (X^{2}) - (E (X))^{2} \\ = (a^{2} p + b^{2} q) - (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 \\ 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} \\ Var (X) = E (X^{2}) - (E (X))^{2} \end{aligned}

Given:
Three coins are tossed
Solution:
$S = {H H H, H H T, H T H, T H H, T T H, T H T, H T T, T T T}$
Set ‘X’ be the random variable for tail

\begin{array}{ccccc} X & 0 & 1 & 2 & 3 \\ P (X) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \end{array}

\begin{aligned} 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 (X^{2}) = \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} Var (X) = E (X^{2}) - (E (X))^{2} \\ = 3 - {(\frac{3}{2})}^{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 (X) = \sum_{i - 1}^{n} X_{i} P_{i}

Standard deviation

= \sqrt{v a r (X)}

Given:
Number of kings in 52 cards
Solution:
Let number of kings be X
P (X=0) = P (Number of kings)

= \frac{48 C_{2}}{52 C_{2}} = \frac{48 \times 47}{52 \times 51} = \frac{188}{221}

\begin{aligned} P (X = 1) = P (1 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 King) \\ = \frac{4 C_{2}}{52 C_{2}} = \frac{4 \times 3}{52 \times 51} = \frac{1}{221} \end{aligned}

\begin{array}{cccc} X & 0 & 1 & 2 \\ P (X) & \frac{188}{221} & \frac{32}{221} & \frac{1}{221} \end{array}

\begin{aligned} 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 (X^{2}) = \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} \\ Var (X) = E (X^{2}) - (E (X))^{2} \end{aligned}

\begin{aligned} = \frac{36}{221} - {(\frac{34}{221})}^{2} \\ = \frac{400}{2873} \\ Standard deviation = \sqrt{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} And Var (X) = E (X^{2}) - (E (X))^{2}

Given:
Three coins are tossed
Solution:
$S = {H H H, H H T, H T H, T H H, T T H, T H T, H T T, T T T}$
Let ‘X’ be the random variable for tail

\begin{array}{ccccc} X & 0 & 1 & 2 & 3 \\ P (X) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \end{array}

\begin{aligned} 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 (X^{2}) = \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 \\ Var (X) = E (X^{2}) - (E (X))^{2} \\ = 3 - {(\frac{3}{2})}^{2} = 3 - \frac{9}{4} \\ = \frac{3}{4} \end{aligned}

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

\begin{aligned} Var (X) = E (X^{2}) - (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:
$E (X) = \sum_{i = 1}^{n} X_{i} P_{i} and Var (X) = E (X^{2}) - (E (X))^{2}$
Given:
A fair coin tossed 4 times
Solution:
$\begin{aligned} \begin{array}{cccccc} X & 0 & 1 & 2 & 3 & 4 \\ P (X) & \frac{1}{16} & \frac{1}{4} & \frac{3}{8} & \frac{1}{4} & \frac{1}{16} \end{array} \end{aligned}$
$\begin{aligned} \begin{aligned} Mean E (X) = \sum_{i = 1}^{n} X_{i} P_{i} \\ = 0 \times \frac{1}{16} + 1 \times \frac{1}{4} + 2 \times \frac{3}{8} + 3 \times \frac{1}{4} + 4 \times \frac{1}{16} \\ = 2 \end{aligned} \end{aligned}$
$\begin{aligned} \begin{aligned} E (X^{2}) = \sum_{i = 1}^{n} X_{i}^{2} P_{i} \\ = \frac{1}{4} + \frac{3}{2} + \frac{9}{4} + 1 = 5 \end{aligned} \end{aligned}$
$\begin{aligned} \begin{aligned} Var (X) = E (X^{2}) - (E (X))^{2} \\ = 5 - 2^{2} \\ = 5 - 4 = 1 \end{aligned} \end{aligned}$

Mean and Variance of a Random Variable Exercise 31.2 Question 10

Answer:

7, 11.67

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

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} and Var (X) = E (X^{2}) - (E (X))^{2}

Given:
A fair dice is tossed
Solution:
$\begin{aligned} S = {1, 2, 3, 4, 5, 6} \\ \begin{array}{ccc} X & 1 & 3 \\ P (X) & \frac{1}{2} & \frac{1}{2} \end{array} \end{aligned}$
$\begin{aligned} Mean E (X) = \sum_{i = 1}^{n} X_{i} P_{i} \\ = 1 \times \frac{1}{2} + 3 \times \frac{1}{2} \\ = \frac{1}{2} + \frac{3}{2} = 2 \end{aligned}$
$\begin{aligned} E (X^{2}) = \sum_{i = 1}^{n} X_{i}^{2} P_{i} \\ = 1^{2} \times \frac{1}{2} + 3^{2} \times \frac{1}{2} \\ = 5 \end{aligned}$
$\begin{aligned} Var (X) = E (X^{2}) - (E (X))^{2} \\ = 5 - 2^{2} \\ = 5 - 4 \\ = 1 \end{aligned}$

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} and Var (X) = E (X^{2}) - (E (X))^{2}

Given:
A fair coin tossed four times
Solution:
X denotes the occurrence of head
Sample space

{HHHH,HHHT,HHTH,HHTT,HTHH,HTHT,HTTH,HTTT,THHH,THHT,THTH,THTT, TTHH, TTHT, TTTH, TTTT }}

\begin{aligned} \begin{array}{cccccc} X & 0 & 1 & 2 & 3 & 4 \\ P (X) & \frac{1}{16} & \frac{1}{4} & \frac{3}{8} & \frac{1}{4} & \frac{1}{16} \end{array} \end{aligned}

\begin{aligned} \begin{aligned} 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 (X^{2}) = \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} Var (X) = E (X^{2}) - (E (X))^{2} \\ = 5 - 2^{2} \\ = 5 - 4 \\ = 1 \end{aligned}

S t a n d a r d d e v i a t i o n = \sqrt{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} and Var (X) = E (X^{2}) - (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}{ccccc} x & 2 & 3 & 4 & 5 \\ P (x) & \frac{1}{10} & \frac{4}{10} & \frac{3}{10} & \frac{2}{10} \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}{ccccc} 1^{'} s & 2^{'} s & 3^{'} s & Total \\ Available & 2 & 2 & 1 & 5 \\ To choose & 1 & 1 & 0 & 2 \\ Choices & 2 C_{1} & 2 C_{1} & 1 C_{0} & 5 C_{2} \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}{ccccc} 1^{'} s & 2^{'} s & 3^{'} s & Total \\ Available & 2 & 2 & 1 & 5 \\ To choose & 1 & 0 & 1 & 2 \\ Choices & 2 C_{1} & 2 C_{0} & 1 C_{1} & 5 C_{2} \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}{ccccc} 1^{'} s & 2^{'} s & 3^{'} s & Total \\ Available & 2 & 2 & 1 & 5 \\ To choose & 0 & 2 & 0 & 2 \\ Choices & 2 C_{0} & 2 C_{2} & 1 C_{0} & 5 C_{2} \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}{ccccc} 1^{'} s & 2^{'} s & 3^{'} s & Total \\ Available & 2 & 2 & 1 & 5 \\ To choose & 0 & 1 & 1 & 2 \\ Choices & 2 C_{0} & 2 C_{1} & 1 C_{1} & 5 C_{2} \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

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

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

The probability distribution of x would be

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

Calculations for mean and standard deviation

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

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

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

Standard deviation of the sum of the numbers on the cards

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

Computation of mean and variance

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

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

Thus the probability distribution of Y is given by

$\begin{array}{cccc} Y & 1 & 2 & 3 \\ P (Y) & 0.1 & 0.5 & 0.4 \end{array}$

Computation of Mean and Variance

$\begin{array}{cccc} y_{i} & p_{i} & p_{i} y_{i} & p_{i} y_{i}^{2} \\ 1 & 0.1 & 0.1 & 0.1 \\ 2 & 0.5 & 1 & 2 \\ 3 & 0.4 & 1.2 & 3.6 \end{array}$

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

Mean and Variance of a Random Variable Exercise 31.2 Question 14

Answer:
Variance

= \frac{1}{2}

Hint:

V a r (X) = E (X^{2}) - {(E (X))}^{2}

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

Mean and Variance of a Random Variable Exercise 31.2 Question 16

Answer:

- \frac{30}{13}

Hint:

E (X) = \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^{'} = q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13} \\ \begin{array}{ccc} x_{i} & 90 & - 10 \\ p (x_{i}) & \frac{1}{13} & \frac{12}{13} \end{array} \end{aligned}

\begin{aligned} 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:
Mean

E (X) = \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 \\ 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:
Mean

E (X) = \sum_{i = 1}^{n} X_{i} P_{i} and Var (X) = E (X^{2}) - (E (X))^{2}

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

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} and Var (X) = E (X^{2}) - (E (X))^{2}

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

Mean and Variance of a Random Variable Exercise 31.2 Question 20

Answer:

\frac{19}{9}

Hint:
Mean

E (X) = \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 (X) = \sum_{i - 1}^{n} X_{i} P_{i}

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

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:
$P (X) = 1$
Given:
$\begin{array}{ccccccc} X & 0 & 1 & 2 & 3 & 4 & 5 \\ P (X) & 4 c^{2} & 3 c^{2} & 2 c^{2} & c^{2} & c & 2 c \end{array}$
Solution:
(i) We know that
$\begin{aligned} 4 c^{2} + 3 c^{2} + 2 c^{2} + c^{2} + c + 2 c = 1 \\ 10 c^{2} + 3 c - 1 = 0 \\ 10 c^{2} + 5 c - 2 c - 1 = 0 \\ 5 c (2 c + 1) - 1 (2 c + 1) = 0 \\ (2 c + 1) (5 c - 1) = 0 \\ c = \frac{1}{5} ∵ c \geq 0 \end{aligned}$
(ii) Mean $E (X) = \sum_{i - 1}^{n} X_{i} P_{i}$
$\begin{aligned} = 3 c^{2} + 4 c^{2} + 3 c^{2} + 4 c + 10 c \\ = 10 c^{2} + 14 c \\ = 10 \times \frac{1}{25} + 14 \times \frac{1}{5} \\ = \frac{10}{25} + \frac{14}{5} \\ = \frac{16}{5} = 3.2 \end{aligned}$
$\begin{aligned} (iii) Var (X) = E (X^{2}) - (E (X))^{2} \\ = 14 - (3.32)^{2} \\ = 2.9 \end{aligned}$

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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