RD Sharma Solutions for Class 12 Maths Exercise 32.1 Chapter 32 Mean and Variance of a Random Variable are available here. Our subject experts have formulated the exercises to assist students with their exam preparation in attaining good marks in the Maths exam. Students who wish to score good marks in Maths should practise RD Sharma Solutions for Class 12 Maths Chapter 32.
RD Sharma Solution for Class 12 Maths Chapter 32 Exercise 2
Access Answers for Rd Sharma Solution Class 12 Maths Chapter 32 Exercise 2
EXERCISE 32.2
Question. 1(i)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| 2 | 0.2 | 0.4 | 0.8 |
| 3 | 0.5 | 1.5 | 4.5 |
| 4 | 0.3 | 1.2 | 4.8 |
| ∑xipi = 3.1 | ∑xi2pi = 10.1 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 10.1 – (3.1)2
= 0.49
Therefore, Standard deviation = √Variance
= √0.49
= 0.7
Question. 1(ii)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| 1 | 0.4 | 0.4 | 0.4 |
| 3 | 0.1 | 0.3 | 0.9 |
| 4 | 0.2 | 0.8 | 3.2 |
| 5 | 0.3 | 1.5 | 7.5 |
| ∑xipi = 3 | ∑xi2pi = 12 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 12 – (3)2
= 3
Therefore, Standard deviation = √Variance
= √3
= 1.732
Question. 1(iii)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| -5 | ¼ | -5/4 | 25/4 |
| -4 | 1/8 | -½ | 2 |
| 1 | ½ | ½ | ½ |
| 2 | 1/8 | ¼ | ½ |
| ∑xipi = -1 | ∑xi2pi = 37/4 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 37/4 – (-1)2
= 33/4
Therefore, Standard deviation = √Variance
= √(33/4)
= 2.9
Question. 1(iv)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| -1 | 0.3 | -0.3 | 0.3 |
| 0 | 0.1 | 0 | 0 |
| 1 | 0.1 | 0.1 | 0.1 |
| 2 | 0.3 | 0.6 | 1.2 |
| 3 | 0.2 | 0.6 | 1.8 |
| ∑xipi = 1 | ∑xi2pi = 3.4 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 3.4 – (1)2
= 2.4
Therefore, Standard deviation = √Variance
= √(2.4)
= 1.5
Question. 1(v)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| 1 | 0.4 | 0.4 | 0.4 |
| 2 | 0.3 | 0.6 | 1.2 |
| 3 | 0.2 | 0.6 | 1.8 |
| 4 | 0.1 | 0.4 | 1.6 |
| ∑xipi = 2 | ∑xi2pi = 5 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 5 – (2)2
= 1
Therefore, Standard deviation = √Variance
= √1
= 1
Question. 1(vi)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| 0 | 0.2 | 0 | 0 |
| 1 | 0.5 | 0.5 | 0.5 |
| 3 | 0.2 | 0.6 | 1.8 |
| 5 | 0.1 | 0.4 | 2.5 |
| ∑xipi = 1.6 | ∑xi2pi = 4.8 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 4.8 – (1.6)2
= 4.8 – 2.56
= 2.24
Therefore, Standard deviation = √Variance
= √2.24
= 1.497
Question. 1(vii)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| -2 | 0.1 | -0.2 | 0.4 |
| -1 | 0.2 | -0.2 | 0.2 |
| 0 | 0.4 | 0 | 0 |
| 1 | 0.2 | 0.2 | 0.2 |
| 2 | 0.1 | 0.2 | 0.4 |
| ∑xipi = 0 | ∑xi2pi = 1.2 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 1.2 – (0)2
= 1.2 – 0
= 1.2
Therefore, Standard deviation = √Variance
= √1.2
= 1.095
Question. 1(viii)
Solution:
We know that, Mean of any probability distribution = ∑xipi
| xi | pi | xipi | xi2pi |
| -3 | 0.05 | -0.15 | 0.45 |
| -1 | 0.45 | -0.45 | 0.45 |
| 0 | 0.20 | 0 | 0 |
| 1 | 0.25 | 0.2 5 | 0.25 |
| 3 | 0.05 | 0.15 | 0.45 |
| ∑xipi = -0.2 | ∑xi2pi = 1.6 |
Then, Variance = ∑pixi2 – (∑xipi)2
= 1.6 – (-0.2)2
= 1.6 – 0.04
= 1.56
Therefore, Standard deviation = √Variance
= √1.56
= 1.249
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