RD Sharma Solutions Class 12 Mean and Variance of a Random Variable Exercise 32.1

RD Sharma Solutions for Class 12 Maths Exercise 32.2 Chapter 32 Mean and Variance of a Random Variable are available here. This exercise of RD Sharma Solutions for Class 12 Maths Chapter 32 contains topics related to the Mean and Standard Deviation. Our subject experts have formulated these solutions in a comprehensive and detailed form. These solutions are among the top materials when it comes to providing a question bank to practise.

RD Sharma Solution for Class 12 Maths Chapter 32 Exercise 1

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EXERCISE 32.1

Question. 1(i)

Solution:

From the question distribution given,

x : 3 2 1 0 -1

P(x): 0.3 0.2 0.4 0.1 0.05

Then,

Sum of probabilities = P(x = 3) + P(x = 2) + P(x = 1) + P(x = 0) + P(x = -1)

= 0.3 + 0.2 + 0.4 + 0.1 + 0.05

= 1.05

So, 1.05 ≠ 1

Therefore, the given distribution is not a probability distribution.

Question. 1(ii)

Solution:

From the question distribution given,

x : 0 1 2

P(x): 0.6 0.4 0.2

Then,

Sum of probabilities = P(x = 0) + P(x = 1) + P(x = 2)

= 0.6 + 0.4 + 0.2

= 1.2

So, 1.2 ≠ 1

Therefore, the given distribution is not a probability distribution.

Question. 1(iii)

Solution:

From the question distribution given,

x : 0 1 2 3 4

P(x): 0.1 0.5 0.2 0.1 0.1

Then,

Sum of probabilities = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)

= 0.1 + 0.5 + 0.2 + 0.1 + 0.1

= 1

So, 1 = 1

Therefore, the given distribution is a probability distribution.

Question. 1(iv)

Solution:

From the question distribution given,

x : 0 1 2 3

P(x): 0.3 0.2 0.4 0.1

Then,

Sum of probabilities = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)

= 0.3 + 0.2 + 0.4 + 0.1

= 1

So, 1 = 1

Therefore, the given distribution is a probability distribution.

Question. 2

Solution:

From the question distribution given,

x : -2 -1 0 1 2 3

P(x) : 0.1 k 0.2 2k 0.3 k

Then, the Sum of the probabilities

0.1 + k + 0.2 + 2k + 0.3 + k = 1

4k + 0.6 = 1

4k = 1 – 0.6

4k = 0.4

K = 0.4/4

K = 4/40

K = 1/10

K = 0.1

Therefore, the value of K is 0.1.

Question. 3

Solution:

From the question distribution given,

x : 0 1 2 3 4 5 6 7 8

P(x) : a 3a 5a 7a 9a 11a 13a 15a 17a

(i)

Then, the Sum of the probabilities

P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) + P(x = 8)= 1

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

81a = 1

a = 1/81

Therefore, the value of a is 1/81.

(ii) P(x < 3) = P(0) + P(1) + P(2)

= a + 3a + 5a

= 9a

Now substitute the value of a, and we get

= 9(1/81)

So, P(x < 3) = 1/9

P(x ≥ 3) = 1 – P(x < 3)

= 1 – 1/9

= (9 – 1)/9

= 8/9

Then, P(0 < x < 5) = P(1) + P(2) + P(3) + P(4)

= 3a + 5a + 7a + 9a

= 24a

Substituting the value of a, we get,

= 24(1/81)

= 8/27

Therefore, P(0 < x < 5) = 8/27

Question. 4

Solution:

From the question distribution given,

x : 0 1 2

P(x) : 3c³ 4c – 10c² 5c – 1

(i) First, we have to find the value of c,

Sum of probabilities = P(x = 0) + P(x = 1) + P(x = 2 = 1

3c³ + 4c – 10c² + 5c – 1 = 1

3c³ – 10c² + 9c – 1 – 1 = 0

3c³ – 10c² + 9c – 2 = 0

Above terms can be written as,

3c³ – 3c² – 7c² + 7c + 2c – 2 = 0

Take out common in above terms, and we get,

3c² (c – 1) – 7c(c – 1) + 2(c – 1) = 0

(c – 1) (3c² – 7c + 2) = 0

(c – 1) (3c² – 6c – c + 2) = 0

(c – 1) (3c (c – 2) – 1(c – 2)) = 0

(c – 1) (3c – 1) (c – 2) = 0

C – 1 = 0, 3c – 1 = 0, c – 2 = 0

C = 1, C = 1/3, C = 2

So, it is clear that c = 1/3 is possible because if c = 1, or c = 2, then P(2) will become negative.

(ii) Now, P(x < 2) = P(0) + P(1)

= 3c³ + 4c – 10c²

= 3(1/3)³ + 4(1/3) – 10(1/3)²

= 3/27 + 4/3 – 10/9

= 1/9 + 4/3 – 10/9

= 3/9

Therefore, the value of P(x < 2) is 3/9.

(iii) P(1< c ≤ 2) = P(2)

= 5c – 1

Substitute the value of c we get,

= 5(1/3) – 1

= 2/3

Therefore, the value P(1< c ≤ 2) is 2/3

Question. 5

Solution:

From the question, it is given that,

2P(x₁) = 3P (x₂) = P(x₃) = 5P(x₄)

So, let us assume P(x₃) = a

Then, 2P(x₁) = P(x₃)

P(x₁) = a/2

3P(x₂) = P(x₃)

P(x₂) = a/3

5P(x₄) = P(x₃)

P(x₄) = a/5

So, P(x₁) + P(x₂) + P(x₃) + P(x₄) = 1

a/2 + a/3 + a/1 + a/5 = 1

LCM of 2, 3, 1 and 5 is 30.

(15a + 10a + 30a + 6a)/30 = 1

61a = 30

a = 30/61

Therefore,

x : x₁ x₂ x₃ x₄

P(x) : 15/61 10/61 30/61 6/61