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Statistics is a branch of mathematics dealing with data collection, analysis, interpretation, and presentation. It provides methods for summarizing and drawing conclusions from the data which is essential in various fields such as economics, sociology, and science. Chapter 7 of RD Sharma's Class 10 textbook focuses on the fundamental statistical concepts and their applications. Exercise 7.5 | Set 1 contains problems designed to reinforce students' understanding of these concepts.
Statistics involves the collection, organization, and analysis of the data to make informed decisions. It encompasses various methods to summarize and interpret numerical information. Key concepts include mean, median, mode, variance, and standard deviation. Mastery of statistics enables individuals to understand data trends make predictions and support decision-making processes based on empirical evidence.
Solution:
(i)
Value (x) 3 4 5 6 7 8 9 Frequency (f) 4 2 5 2 2 1 2 Therefore, mode = 5 because 5 occurs the maximum number of times.
(ii)
Value (x) 3 4 5 6 7 8 9 Frequency (f) 5 2 4 2 2 1 2 Therefore, mode = 3 because 3 occurs the maximum number of times.
(iii)
Value (x) 8 15 18 19 20 24 25 Frequency (f) 1 4 1 1 1 2 1 Therefore, mode = 15 because 15 occurs the maximum number of times.
| Shirt size: | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 |
| Number of persons: | 15 | 25 | 39 | 41 | 36 | 17 | 15 | 12 |
Solution:
From the data present in the table we conclude that
Model shirt size = 40
Because shirt size 40 occurred for the maximum number of times.
| Class interval: | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | 70 - 80 |
| Frequency: | 5 | 8 | 7 | 12 | 28 | 20 | 10 | 10 |
Solution:
From the given table we conclude that
The maximum frequency = 28
So, the model class = 40 – 50
and,
l = 40, h = 50 40 = 10, f = 28, f1 = 12, f2 = 20
Using the formula of mode
Mode =
= 40 + 160/ 24
= 40 + 6.67
= 46.67
Hence, the mode = 46.67
| Class interval | 10 - 15 | 15 - 20 | 20 - 25 | 25 - 30 | 30 - 35 | 35 - 40 |
| Frequency | 30 | 45 | 75 | 35 | 25 | 15 |
Solution:
From the given table we conclude that
The maximum frequency = 75
So, the modal class = 20 – 25
And,
l = 20, h = 25 – 20 = 5, f = 75, f1 = 45, f2 = 35
Using the formula of mode
Mode =
= 20 + 150/70
= 20 + 2.14
= 22.14
Hence, the mode = 22.14
| Class interval | 25 - 30 | 30 - 35 | 35 - 40 | 40 - 45 | 45 - 50 | 50 - 55 |
| Frequency | 25 | 34 | 50 | 42 | 38 | 14 |
Solution:
From the given table we conclude that
The maximum frequency = 50
So, the modal class = 35 – 40
And,
l = 35, h = 40 – 35 = 5, f = 50, f1 = 34, f2 = 42
Using the formula of mode
Mode =
= 35 + 80/24
= 35 + 3.33
= 38.33
Hence, the mode = 38.33
| Age in years | 16 - 18 | 18 - 20 | 20 - 22 | 22 - 24 | 24 - 26 |
| Group A | 50 | 78 | 46 | 28 | 23 |
| Group B | 54 | 89 | 40 | 25 | 17 |
Solution:
For Group A:
From the given table we conclude that
The maximum frequency = 78.
So, the model class = 18 – 20
And,
l = 18, h = 20 – 18 = 2, f = 78, f1 = 50, f2 = 46
Using the formula of mode
Mode =
= 18 + 56/60
= 18 + 0.93
= 18.93 years
For Group B:
From the given table we conclude that
The maximum frequency = 89
The modal class = 18 – 20
And,
l = 18, h = 20 – 18 = 2, f = 89, f1 = 54, f2 = 40
Using the formula of mode
Mode =
= 18 + 70/84
= 18 + 0.83
= 18.83 years
After finding the mode of both A and B group we conclude that
the modal age of the Group A is greater than Group B.
| Marks | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | 70 - 80 | 80 - 90 | 90 - 100 |
| Frequency | 3 | 5 | 16 | 12 | 13 | 20 | 5 | 4 | 1 | 1 |
Solution:
From the given table we conclude that
The maximum frequency = 20
The modal class = 50 – 60
And,
l = 50, h = 60 – 50 = 10, f = 20, f1 = 13, f2 = 5
Using the formula of mode
Mode =
= 50 + 70/22
= 50 + 3.18
= 53.18
Hence, the mode = 53.18
| Height (in cm) | 160 - 162 | 163 - 165 | 166 - 168 | 169 - 171 | 172 - 174 |
| No. of students: | 15 | 118 | 142 | 127 | 18 |
Solution:
| Heights (exclusive) | 160 - 162 | 163 - 165 | 166 - 168 | 169 - 171 | 172 - 174 |
| Heights (inclusive) | 159.5 - 162.5 | 162.5 - 165.5 | 165.5 - 168.5 | 168.5 - 171.5 | 171.5 - 174.5 |
| No of students | 15 | 118 | 142 | 127 | 18 |
From the given table we conclude that
The maximum frequency = 142
The modal class = 165.5 – 168.5
And,
l = 165.5, h = 168.5 – 165.5 = 3, f = 142, f1 = 118, f2 = 127
Using the formula of mode
Mode =
= 165.5 + 72/39
= 165.5 + 1.85
= 167.35 cm
Hence, the average height of maximum number of students = 167.35 cm
| Ages (in years): | 5 - 15 | 15 - 25 | 25 - 35 | 35 - 45 | 45 - 55 | 55 - 65 |
| No of students: | 6 | 11 | 21 | 23 | 14 | 5 |
Solution:
For mean:
Let us considered mean (A) = 30
Age (in years) Number of patients fi Class marks xi di = xi - 275 fidi 5 - 15 6 10 -20 -120 15 - 25 11 20 -10 -110 25 - 35 21 30 0 0 35 - 45 23 40 10 230 45 - 55 14 50 20 280 55 - 65 5 60 30 150 N = 80 From the table we get
Σfi = N = 80 and Σfi di = 430.
Using the formula of mean
= 30 + 430/80
= 30 + 5.375
= 35.375
= 35.38
Therefore, the mean = 35.38. It represents the average age of the patients = 35.38 years.
For mode:
From the given table we conclude that
The maximum class frequency = 23
So, modal class = 35 – 45
and
l = 35, f = 23, h = 10, f1 = 21, f2 = 14
Using the formula of mode
Mode
= 35 + 1.81 = 36.8
Hence, the mode = 36.8. It represents the maximum number of patients admitted in hospital of age 36.8 years.
Therefore, mode is greater than mean
| Lifetimes (in hours): | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 - 120 |
| No. of components: | 10 | 35 | 52 | 61 | 38 | 29 |
Solution:
From the given table we conclude that
The maximum class frequency = 61
So, modal class = 60 – 80
and
l = 60, f = 61, h = 20, f1 = 52, f2 = 38
Using the formula of mode
Mode
= 60 + 5.625 = 65.625
Hence, the modal lifetime of electrical components = 65.625 hours
| Daily income | 100 - 120 | 120 - 140 | 140 - 160 | 160 - 180 | 180 - 200 |
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Solution:
| Class interval | Mid value (x) | Frequency (f) | fx | Cumulative Frequency |
| 100 - 120 | 110 | 12 | 1320 | 12 |
| 120 - 140 | 130 | 14 | 1820 | 26 |
| 140 - 160 | 150 | 8 | 1200 | 34 |
| 160 - 180 | 170 | 6 | 1000 | 40 |
| 180 - 200 | 190 | 10 | 1900 | 50 |
| N = 50 |
Finding Mean:
From the table we get
N = 50, fx = 7260
So using mean formula, we get
Mean = Σfx / N
= 7260/ 50
= 145.2
Hence, the mean = 145.2
Finding Median:
N/2 = 50/2 = 25
So, the cumulative frequency just greater than N/2 = 26,
The median class = 120 – 140
Such that l = 120, h = 140 – 120 = 20, f = 14, F = 12
By using the formula of median we get
Median =
= 120 + 260/14
= 120 + 18.57
= 138.57
Hence, the median = 138.57
Finding Mode:
From the table we get
The maximum frequency = 14,
So the modal class = 120 – 140
And,
l = 120, h = 140 – 120 = 20, f = 14, f1 = 12, f2 = 8
By using the formula of mode we get
Mode =
= 120 + 5
= 125
Hence, the mode = 125
Number of students per teacher | Number of states/U.T |
15 - 20 | 3 |
20 - 25 | 8 |
25 - 30 | 9 |
30 - 35 | 10 |
35 - 40 | 3 |
40 - 45 | 0 |
45 - 50 | 0 |
50 - 55 | 2 |
Solution:
From the given table we conclude that
The maximum class frequency = 10
So, modal class = 30 - 35
and
l = 30, h = 5, f = 10, f1 = 9, f2 = 3
By using the formula of mode we get
Mode = l + f - f1 2f - f1 - f2 × hl +
= 30 + 120 - 12 × 530 +
= 30 + 120 - 12 × 530 +
= 30 + 5/8
= 30.625
Hence, the mode = 30.6 and it represents that most of states/ U.T have a teacher-students ratio = 30.6
Now we are going to find class marks using the following formula
Class mark =
Let us considered mean(a) = 32.5, and now we are going to find di, ui, and fiui as following
Number of students per teacher Number of states/ U.T (fi) xi di = xi - 32.5 Ui fiui 15 - 20 3 17.5 -15 -3 -9 20 - 25 8 22.5 -10 -2 -16 25 - 30 9 27.5 -5 -1 -9 30 - 35 10 32.5 0 0 0 35 - 40 3 37.5 5 1 3 40 - 45 0 42.5 10 2 0 45 - 50 0 47.5 10 2 0 50 - 55 2 52.5 20 4 8 Total 35 -23 Using the mean formula, we get
= 32.5 - 23/7
= 32.5 - 3.28
= 29.2
Hence, the mean = 29.2 and it represents that on an average teacher-student ratio = 29.2.
Read More:
Understanding statistical concepts like mean, median and mode is crucial for the analyzing and interpreting data effectively. Exercise 7.5 | Set 1 from RD Sharma's Class 10 book provides the practice with these fundamental concepts reinforcing students' ability to the handle various data analysis problems.