Statistics is purely about data analysis. So, with the given data, students will have to find the mean by using different methods. In this exercise, students will have to find the mean of grouped data by using the direct method. The RD Sharma Solutions Class 10 prepared by subject experts at BYJU’S is a very useful resource for students to refer to and prepare well for their exams. Download RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.1, PDF provided below, for detailed solutions to problems.
RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.1
Access RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.1
1. Calculate the mean for the following distribution:
| x: | 5 | 6 | 7 | 8 | 9 |
| f: | 4 | 8 | 14 | 11 | 3 |
Solution:
| x | f | fx |
| 5 | 4 | 20 |
| 6 | 8 | 48 |
| 7 | 14 | 98 |
| 8 | 11 | 88 |
| 9 | 3 | 27 |
| N = 40 | Σ fx = 281 |
Mean = Σ fx/ N = 281/40
∴ Mean = 7.025
2. Find the mean of the following data:
| x: | 19 | 21 | 23 | 25 | 27 | 29 | 31 |
| f: | 13 | 15 | 16 | 18 | 16 | 15 | 13 |
Solution:
| x | f | fx |
| 19 | 13 | 247 |
| 21 | 15 | 315 |
| 23 | 16 | 368 |
| 25 | 18 | 450 |
| 27 | 16 | 432 |
| 29 | 15 | 435 |
| 31 | 13 | 403 |
| N = 106 | Σ fx = 2620 |
Mean = Σ fx/ N = 2620/106
∴ Mean = 25
3. If the mean of the following data is 20.6. Find the value of p.
| x: | 10 | 15 | p | 25 | 35 |
| f: | 3 | 10 | 25 | 7 | 5 |
Solution:
| x | f | fx |
| 10 | 3 | 30 |
| 15 | 10 | 150 |
| p | 25 | 25p |
| 25 | 7 | 175 |
| 35 | 5 | 175 |
| N = 50 | Σ fx = 530 + 25p |
We know that,
Mean = Σ fx/ N = (2620 + 25p)/ 50
Given,
Mean = 20.6
⇒ 20.6 = (530 + 25p)/ 50
(20.6 x 50) – 530 = 25p
p = 500/ 25
∴ p = 20
4. If the mean of the following data is 15, find p.
| x: | 5 | 10 | 15 | 20 | 25 |
| f: | 6 | p | 6 | 10 | 5 |
Solution:
| x | f | fx |
| 5 | 6 | 30 |
| 10 | p | 10p |
| 15 | 6 | 90 |
| 20 | 10 | 200 |
| 25 | 5 | 125 |
| N = p + 27 | Σ fx = 445 + 10p |
We know that,
Mean = Σ fx/ N = (445 + 10p)/ (p + 27)
Given,
Mean = 15
⇒ 15 = (445 + 10p)/ (p + 27)
15p + 405 = 445 + 10p
5p = 40
∴ p = 8
5. Find the value of p for the following distribution whose mean is 16.6
| x: | 8 | 12 | 15 | p | 20 | 25 | 30 |
| f: | 12 | 16 | 20 | 24 | 16 | 8 | 4 |
Solution:
| x | f | fx |
| 8 | 12 | 96 |
| 12 | 16 | 192 |
| 15 | 20 | 300 |
| P | 24 | 24p |
| 20 | 16 | 320 |
| 25 | 8 | 200 |
| 30 | 4 | 120 |
| N = 100 | Σ fx = 1228 + 24p |
We know that,
Mean = Σ fx/ N = (1228 + 24p)/ 100
Given,
Mean = 16.6
⇒ 16.6 = (1228 + 24p)/ 100
1660 = 1228 + 24p
24p = 432
∴ p = 18
6. Find the missing value of p for the following distribution whose mean is 12.58
| x: | 5 | 8 | 10 | 12 | p | 20 | 25 |
| f: | 2 | 5 | 8 | 22 | 7 | 4 | 2 |
Solution:
| x | f | fx |
| 5 | 2 | 10 |
| 8 | 5 | 40 |
| 10 | 8 | 80 |
| 12 | 22 | 264 |
| P | 7 | 7p |
| 20 | 4 | 80 |
| 25 | 2 | 50 |
| N = 50 | Σ fx = 524 + 7p |
We know that,
Mean = Σ fx/ N = (524 + 7p)/ 50
Given,
Mean = 12.58
⇒ 12.58 = (524 + 7p)/ 50
629 = 524 + 7p
7p = 629 – 524 = 105
∴ p = 15
7. Find the missing frequency (p) for the following distribution whose mean is 7.68
| x: | 3 | 5 | 7 | 9 | 11 | 13 |
| f: | 6 | 8 | 15 | p | 8 | 4 |
Solution:
| x | f | fx |
| 3 | 6 | 18 |
| 5 | 8 | 40 |
| 7 | 15 | 105 |
| 9 | p | 9p |
| 11 | 8 | 88 |
| 13 | 4 | 52 |
| N = 41 + p | Σ fx = 303 + 9p |
We know that,
Mean = Σ fx/ N = (303 + 9p)/ (41 + p)
Given,
Mean = 7.68
⇒ 7.68 = (303 + 9p)/ (41 + p)
7.68(41 + p) = 303 + 9p
7.68p + 314.88 = 303 + 9p
1.32p = 11.88
∴ p = 11.88/1.32 = 9
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