Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.3 | Set 1

Last Updated : 23 Jul, 2025

Chapter 7 of RD Sharma’s Class 10 textbook focuses on Statistics a fundamental area in mathematics that deals with data collection, analysis, interpretation, and presentation. Exercise 7.3 | Set 1 of this chapter covers the various problems designed to enhance students' understanding and application of statistical methods. This exercise aims to reinforce the concepts learned and provide the practice in solving the statistical problems.

Statistics

Statistics is the branch of mathematics that involves collecting, analyzing, interpreting, presenting, and organizing data. It provides tools and techniques to summarize and make sense of the large datasets facilitating informed decision-making. In essence, statistics helps in understanding the patterns and trends within data which is crucial in various fields such as economics, social sciences, and natural sciences.

Question 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_iu_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_iu_i= -35
It’s seen that A = 275 and h = 50
So,
Mean = A + h x (Σf_iu_i/N)
= 275 + 50 (-35/200)
= 275 – 8.75
= 266.25

Question 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_ix_i by the following
Number of plants Number of house (f_i) x_i f_ix_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_iu_i= 162
Here,
Mean = Σ f_iu_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 is very small.

Question 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 us 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_iu_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_iu_i= -12
It’s seen that,
A = 150 and h = 20
So,
Mean = A + h x (Σf_iu_i/N)
= 150 + 20 x (-12/50)
= 150 – 24/5
= 150 = 4.8
= 145.20

Question 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_iu_i as following:
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_iu_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_iu_i= 4
From table, it’s seen that
N = 30 and h = 3
So, the mean = A + h x (Σf_iu_i/N)
= 75.5 + 3 x (4/30
= 75.5 + 2/5
= 75.9
Therefore, the mean heart beats per minute for those women are 75.9 beats per minute.

Question 5. Find the mean of each of the following frequency distributions:

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_iu_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_iu_i= 3
From the table it’s seen that,
A = 15 and h = 6
Mean = A + h x (Σf_iu_i/N)
= 15 + 6 x (3/40)
= 15 + 0.45
= 15.45

Question 6. Find the mean of the following frequency distribution:

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_iu_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_iu_i= 61
From the table it’s seen that,
A = 100 and h = 20
Mean = A + h x (Σf_iu_i/N)
= 100 + 20 x (61/100)
= 100 + 12.2
= 112.2

Question 7. Find the mean of the following frequency distribution:

Class interval:	0 – 8	8 – 16	16 – 24	24 – 32	32 – 40
Frequency:	6	7	10	8	9

Solution:

From the table it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σf_iu_i/N)
= 20 + 8 x (7/40)
= 20 + 1.4
= 21.4

Question 8. Find the mean of the following frequency distribution:

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_iu_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_iu_i= 5
From the table it’s seen that,
A = 15 and h = 6
Mean = A + h x (Σf_iu_i/N)
= 15 + 6 x (5/40)
= 15 + 0.75
= 15.75

Question 9. Find the mean of the following frequency distribution:

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_iu_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_iu_i= 8
From the table it’s seen that,
A = 25 and h = 10
Mean = A + h x (Σf_iu_i/N)
= 25 + 10 x (8/60)
= 25 + 4/3
= 79/3 = 26.333

Question 10. Find the mean of the following frequency distribution:

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_iu_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_iu_i= 5
From the table it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σf_iu_i/N)
= 20 + 8 x (5/40)
= 20 + 1
= 21

Question 11. Find the mean of the following frequency distribution:

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_iu_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_iu_i= -9
From the table it’s seen that,
A = 20 and h = 8
Mean = A + h x (Σf_iu_i/N)
= 20 + 6 x (-9/20)
= 20 – 72/20
= 20 – 3.6
= 16.4

Question 12. Find the mean of the following frequency distribution:

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_iu_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_iu_i= 14
From the table it’s seen that,
A = 60 and h = 20
Mean = A + h x (Σf_iu_i/N)
= 60 + 20 x (14/50)
= 60 + 28/5
= 60 + 5.6
= 65.6

Question 13. Find the mean of the following frequency distribution:

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_iu_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_iu_i= -2
From the table it’s seen that,
A = 50 and h = 10
Mean = A + h x (Σf_iu_i/N)
= 50 + 10 x (-2/40)
= 50 – 0.5
= 49.5

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Conclusion

Exercise 7.3 | Set 1 of RD Sharma’s Chapter 7 on Statistics offers practical problems that challenge students to the apply statistical methods effectively. By working through these problems students will gain a deeper insight into the statistical concepts and improve their problem-solving skills. Mastery of these exercises will build a solid foundation for the tackling more complex statistical analyses in the future studies.

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