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Two main statistical methods are used in data analysis: descriptive statistics, which summarizes data using indexes such as mean and median and another is inferential statistics, which draw conclusions from data using statistical tests such as student's t-test.
What is the basic concept of statistics?
In general, statistics is a study of data: describing properties of the data, which is called descriptive statistics, and drawing conclusions about a population of interest from information extracted from a sample, which is called inferential statistics.
What does ∑ mean in statistics?
summation
What is the N symbol in statistics?
N = population size. ND = normal distribution, whose graph is a bell-shaped curve; also “normally distributed”. p = probability value. The specific meaning depends on context.
RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 PDF
RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 Statistics has been provided here. Students can refer to these questions before their examinations for better preparation.
Neha Tanna15 Nov, 2024
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RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3:
Chapter 7 of RD Sharma's Class 10 Maths book, which focuses on "Statistics," introduces methods to systematically organize and interpret data. Exercise 7.3 specifically covers the computation of cumulative frequency distributions, cumulative frequency curves, and their applications in analyzing grouped data.
Students learn how to convert data into cumulative frequencies, construct graphs, and utilize these visuals to estimate median values. This exercise strengthens skills in understanding data distribution patterns and prepares students for practical applications of statistics. It emphasizes visualizing data trends, a fundamental aspect of real-world data analysis and interpretation.
Chapter 7 of RD Sharma’s Class 10 Maths book covers Statistics, which is crucial for developing data interpretation skills. Exercise 7.3, in particular, focuses on calculating measures of central tendency, such as mean, median, and mode, which help summarize data into a single representative value.
These concepts are foundational for higher-level statistics and practical data analysis in various fields, including economics, science, and social studies. Understanding and practicing this exercise equips students with the skills to organize, interpret, and make informed conclusions from data, an essential competency in our data-driven world.
RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 PDF
Below, we have provided a PDF for RD Sharma Solutions for Class 10 Maths Chapter 7, Exercise 7.3, focusing on Statistics. This exercise is essential for students aiming to understand data interpretation, mean, median, and mode calculations, and graphical representation techniques. These solutions are comprehensive, making it easier to grasp concepts and solve problems accurately. Perfect for revising before exams, this PDF guide will help you tackle all statistics-related questions confidently. Download the PDF to access step-by-step solutions for every problem in Exercise 7.3.
Below is the RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 Statistics -
1. The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.
Expenditure (in rupees) (x)
Frequency (f
i
)
Expenditure (in rupees) (x
i
)
Frequency (f
i
)
100 – 150
24
300 – 350
30
150 – 200
40
350 – 400
22
200 – 250
33
400 – 450
16
250 – 300
28
450 – 500
7
Find the average expenditure (in rupees) per household.
Solution:
Let the assumed mean (A) = 275
Class interval
Mid value (x
i
)
d
i
= x
i
– 275
u
i
= (x
i
– 275)/50
Frequency f
i
f
i
u
i
100 – 150
125
-150
-3
24
-72
150 – 200
175
-100
-2
40
-80
200 – 250
225
-50
-1
33
-33
250 – 300
275
0
0
28
0
300 – 350
325
50
1
30
30
350 – 400
375
100
2
22
44
400 – 450
425
150
3
16
48
450 – 500
475
200
4
7
28
N = 200
Σ f
i
u
i
= -35
It’s seen that A = 275 and h = 50
So,
Mean = A + h x (Σf
i
u
i
/N)
= 275 + 50 (-35/200)
= 275 – 8.75
= 266.25
2. A survey was conducted by a group of students as a part of their environmental awareness program, in which they collected the following data regarding the number of plants in 200 houses in a locality. Find the mean number of plants per house.
Number of plants:
0 – 2
2 – 4
4 – 6
6 – 8
8 – 10
10 – 12
12 – 14
Number of house:
1
2
1
5
6
2
3
Which method did you use for finding the mean, and why?
Solution:
From the given data,
To find the class interval, we know that,
Class marks (x
i
) = (upper class limit + lower class limit)/2
Now, let’s compute x
i
and f
i
x
i
by the following
Number of plants
Number of house (f
i
)
x
i
f
i
x
i
0 – 2
1
1
1
2 – 4
2
3
6
4 – 6
1
5
5
6 – 8
5
7
35
8 – 10
6
9
54
10 – 12
2
11
22
12 – 14
3
13
39
Total
N = 20
Σ f
i
u
i
= 162
Here,
Mean = Σ f
i
u
i
/ N
= 162/ 20
= 8.1
Thus, the mean number of plants in a house is 8.1
We have used the direct method as the values of class mark x
i
and f
i
are very small.
3. Consider the following distribution of daily wages of workers of a factory
Daily wages (in ₹)
100 – 120
120 – 140
140 – 160
160 – 180
180 – 200
Number of workers:
12
14
8
6
10
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Let the assume mean (A) = 150
Class interval
Mid value x
i
d
i
= x
i
– 150
u
i
= (x
i
– 150)/20
Frequency f
i
f
i
u
i
100 – 120
110
-40
-2
12
-24
120 – 140
130
-20
-1
14
-14
140 – 160
150
0
0
8
0
160 – 180
170
20
1
6
6
180 – 200
190
40
2
10
20
N= 50
Σ f
i
u
i
= -12
It’s seen that,
A = 150 and h = 20
So,
Mean = A + h x (Σf
i
u
i
/N)
= 150 + 20 x (-12/50)
= 150 – 24/5
= 150 = 4.8
= 145.20
4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Number of heart beats per minute:
65 – 68
68 – 71
71 – 74
74 – 77
77 – 80
80 – 83
83 – 86
Number of women:
2
4
3
8
7
4
2
Solution:
Using the relation (x
i
) = (upper class limit + lower class limit)/ 2
And, class size of this data = 3
Let the assumed mean (A) = 75.5
So, let’s calculate d
i
, u
i
, f
i
u
i
as follows:
Number of heart beats per minute
Number of women (f
i
)
x
i
d
i
= x
i
– 75.5
u
i
= (x
i
– 755)/h
f
i
u
i
65 – 68
2
66.5
-9
-3
-6
68 – 71
4
69.5
-6
-2
-8
71 – 74
3
72.5
-3
-1
-3
74 – 77
8
75.5
0
0
0
77 – 80
7
78.5
3
1
7
80 – 83
4
81.5
6
2
8
83 – 86
2
84.5
9
3
6
N = 30
Σ f
i
u
i
= 4
From the table, it’s seen that
N = 30 and h = 3
So, the mean = A + h x (Σf
i
u
i
/N)
= 75.5 + 3 x (4/30
= 75.5 + 2/5
= 75.9
Therefore, the mean heartbeats per minute for those women are 75.9 beats per minute.
Find the mean of each of the following frequency distributions: (5 – 14)
5.
Class interval:
0 – 6
6 – 12
12 – 18
18 – 24
24 – 30
Frequency:
6
8
10
9
7
Solution:
Let’s consider the assumed mean (A) = 15
Class interval
Mid – value x
i
d
i
= x
i
– 15
u
i
= (x
i
– 15)/6
f
i
f
i
u
i
0 – 6
3
-12
-2
6
-12
6 – 12
9
-6
-1
8
-8
12 – 18
15
0
0
10
0
18 – 24
21
6
1
9
9
24 – 30
27
12
2
7
14
N = 40
Σ f
i
u
i
= 3
From the table, it’s seen that,
A = 15 and h = 6
Mean = A + h x (Σf
i
u
i
/N)
= 15 + 6 x (3/40)
= 15 + 0.45
= 15.45
6.
Class interval:
50 – 70
70 – 90
90 – 110
110 – 130
130 – 150
150 – 170
Frequency:
18
12
13
27
8
22
Solution:
Let’s consider the assumed mean (A) = 100
Class interval
Mid – value x
i
d
i
= x
i
– 100
u
i
= (x
i
– 100)/20
f
i
f
i
u
i
50 – 70
60
-40
-2
18
-36
70 – 90
80
-20
-1
12
-12
90 – 110
100
0
0
13
0
110 – 130
120
20
1
27
27
130 – 150
140
40
2
8
16
150 – 170
160
60
3
22
66
N= 100
Σ f
i
u
i
= 61
From the table, it’s seen that,
A = 100 and h = 20
Mean = A + h x (Σf
i
u
i
/N)
= 100 + 20 x (61/100)
= 100 + 12.2
= 112.2
7.
Class interval:
0 – 8
8 – 16
16 – 24
24 – 32
32 – 40
Frequency:
6
7
10
8
9
Solution:
Let’s consider the assumed mean (A) = 20
Class interval
Mid – value x
i
d
i
= x
i
– 20
u
i
= (x
i
– 20)/8
f
i
f
i
u
i
0 – 8
4
-16
-2
6
-12
8 – 16
12
-8
-1
7
-7
16 – 24
20
0
0
10
0
24 – 32
28
8
1
8
8
32 – 40
36
16
2
9
18
N = 40
Σ f
i
u
i
= 7
From the table, it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σf
i
u
i
/N)
= 20 + 8 x (7/40)
= 20 + 1.4
= 21.4
8.
Class interval:
0 – 6
6 – 12
12 – 18
18 – 24
24 – 30
Frequency:
7
5
10
12
6
Solution:
Let’s consider the assumed mean (A) = 15
Class interval
Mid – value x
i
d
i
= x
i
– 15
u
i
= (x
i
– 15)/6
f
i
f
i
u
i
0 – 6
3
-12
-2
7
-14
6 – 12
9
-6
-1
5
-5
12 – 18
15
0
0
10
0
18 – 24
21
6
1
12
12
24 – 30
27
12
2
6
12
N = 40
Σ f
i
u
i
= 5
From the table, it’s seen that,
A = 15 and h = 6
Mean = A + h x (Σf
i
u
i
/N)
= 15 + 6 x (5/40)
= 15 + 0.75
= 15.75
9.
Class interval:
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
Frequency:
9
12
15
10
14
Solution:
Let’s consider the assumed mean (A) = 25
Class interval
Mid – value x
i
d
i
= x
i
– 25
u
i
= (x
i
– 25)/10
f
i
f
i
u
i
0 – 10
5
-20
-2
9
-18
10 – 20
15
-10
-1
12
-12
20 – 30
25
0
0
15
0
30 – 40
35
10
1
10
10
40 – 50
45
20
2
14
28
N = 60
Σ f
i
u
i
= 8
From the table, it’s seen that,
A = 25 and h = 10
Mean = A + h x (Σf
i
u
i
/N)
= 25 + 10 x (8/60)
= 25 + 4/3
= 79/3 = 26.333
10.
Class interval:
0 – 8
8 – 16
16 – 24
24 – 32
32 – 40
Frequency:
5
9
10
8
8
Solution:
Let’s consider the assumed mean (A) = 20
Class interval
Mid – value x
i
d
i
= x
i
– 20
u
i
= (x
i
– 20)/8
f
i
f
i
u
i
0 – 8
4
-16
-2
5
-10
8 – 16
12
-4
-1
9
-9
16 – 24
20
0
0
10
0
24 – 32
28
4
1
8
8
32 – 40
36
16
2
8
16
N = 40
Σ f
i
u
i
= 5
From the table, it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σf
i
u
i
/N)
= 20 + 8 x (5/40)
= 20 + 1
= 21
11.
Class interval:
0 – 8
8 – 16
16 – 24
24 – 32
32 – 40
Frequency:
5
6
4
3
2
Solution:
Let’s consider the assumed mean (A) = 20
Class interval
Mid – value x
i
d
i
= x
i
– 20
u
i
= (x
i
– 20)/8
f
i
f
i
u
i
0 – 8
4
-16
-2
5
-12
8 – 16
12
-8
-1
6
-8
16 – 24
20
0
0
4
0
24 – 32
28
8
1
3
9
32 – 40
36
16
2
2
14
N = 20
Σ f
i
u
i
= -9
From the table, it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σf
i
u
i
/N)
= 20 + 6 x (-9/20)
= 20 – 72/20
= 20 – 3.6
= 16.4
12.
Class interval:
10 – 30
30 – 50
50 – 70
70 – 90
90 – 110
110 – 130
Frequency:
5
8
12
20
3
2
Solution:
Let’s consider the assumed mean (A) = 60
Class interval
Mid – value x
i
d
i
= x
i
–60
u
i
= (x
i
– 60)/20
f
i
f
i
u
i
10 – 30
20
-40
-2
5
-10
30 – 50
40
-20
-1
8
-8
50 – 70
60
0
0
12
0
70 – 90
80
20
1
20
20
90 – 110
100
40
2
3
6
110 – 130
120
60
3
2
6
N = 50
Σ f
i
u
i
= 14
From the table, it’s seen that,
A = 60 and h = 20
Mean = A + h x (Σf
i
u
i
/N)
= 60 + 20 x (14/50)
= 60 + 28/5
= 60 + 5.6
= 65.6
13.
Class interval:
25 – 35
35 – 45
45 – 55
55 – 65
65 – 75
Frequency:
6
10
8
12
4
Solution:
Let’s consider the assumed mean (A) = 50
Class interval
Mid – value x
i
d
i
= x
i
– 50
u
i
= (x
i
– 50)/10
f
i
f
i
u
i
25 – 35
30
-20
-2
6
-12
35 – 45
40
-10
-1
10
-10
45 – 55
50
0
0
8
0
55 – 65
60
10
1
12
12
65 – 75
70
20
2
4
8
N = 40
Σ f
i
u
i
= -2
From the table, it’s seen that,
A = 50 and h = 10
Mean = A + h x (Σf
i
u
i
/N)
= 50 + 10 x (-2/40)
= 50 – 0.5
= 49.5
14.
Class interval:
25 – 29
30 – 34
35 – 39
40 – 44
45 – 49
50 – 54
55 – 59
Frequency:
14
22
16
6
5
3
4
Solution:
Let’s consider the assumed mean (A) = 42
Class interval
Mid – value x
i
d
i
= x
i
– 42
u
i
= (x
i
– 42)/5
f
i
f
i
u
i
25 – 29
27
-15
-3
14
-42
30 – 34
32
-10
-2
22
-44
35 – 39
37
-5
-1
16
-16
40 – 44
42
0
0
6
0
45 – 49
47
5
1
5
5
50 – 54
52
10
2
3
6
55 – 59
57
15
3
4
12
N = 70
Σ f
i
u
i
= -79
From the table, it’s seen that,
A = 42 and h = 5
Mean = A + h x (Σf
i
u
i
/N)
= 42 + 5 x (-79/70)
= 42 – 79/14
= 42 – 5.643
= 36.357
Benefits of Solving RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3
Solving RD Sharma Solutions for Class 10 Maths Chapter 7 Exercise 7.3 on Statistics offers several benefits for students preparing for their exams. Here’s how it helps:
1.
Strengthens Conceptual Understanding
Exercise 7.3 focuses on the application of statistical measures, such as mean, median, and mode. Working through these solutions deepens the understanding of these fundamental statistical concepts and helps students remember the formulas and methods.
2.
Enhances Problem-Solving Skills
The exercise includes various types of questions, allowing students to practice and strengthen their problem-solving abilities. By attempting diverse problems, students learn to apply the correct methods in a structured way.
3.
Builds Analytical Skills
Solving statistics problems involves analyzing data and choosing appropriate techniques. This analytical approach to solving questions can benefit students not just in mathematics but in other subjects involving data interpretation as well.
4.
Improves Accuracy and Speed
Regular practice with these solutions improves calculation accuracy and speeds up the time taken to solve problems, which is crucial in a timed exam setting. Consistency in solving RD Sharma exercises helps students become faster and more precise.
5.
Prepares for Competitive Exams
This chapter is fundamental in many entrance exams and competitive tests. Understanding the basic statistical concepts here helps students prepare for exams beyond school, as they form the foundation for more advanced topics in statistics.
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