Class 10 RD Sharma Solutions Chapter 7 Statistics Exercise 7.3 | Set 1 (original) (raw)

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 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 (Σfi ui/N)

= 275 + 50 (-35/200)

= 275 – 8.75

= 266.25

**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

**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 (xi) = (upper class limit + lower class limit)/2

Now, let’s compute xi and fixi 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 fiui/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 xi and fi is very small.

**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

**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 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 (Σfi ui/N)

= 150 + 20 x (-12/50)

= 150 – 24/5

= 150 = 4.8

= 145.20

**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

**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 (xi) = (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 di, ui, fiui 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 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 table, it’s seen that

N = 30 and h = 3

So, the mean = A + h x (Σfi ui/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.

**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

**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 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 (Σfi ui/N)

= 15 + 6 x (3/40)

= 15 + 0.45

= 15.45

**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

**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 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 (Σfi ui/N)

= 100 + 20 x (61/100)

= 100 + 12.2

= 112.2

**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

**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 (Σfi ui/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 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 (Σfi ui/N)

= 15 + 6 x (5/40)

= 15 + 0.75

= 15.75

**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

**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 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 (Σfi ui/N)

= 25 + 10 x (8/60)

= 25 + 4/3

= 79/3 = 26.333

**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

**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 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 (Σfi ui/N)

= 20 + 8 x (5/40)

= 20 + 1

= 21

**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

**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 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 (Σfi ui/N)

= 20 + 6 x (-9/20)

= 20 – 72/20

= 20 – 3.6

= 16.4

**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

**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 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 (Σfi ui/N)

= 60 + 20 x (14/50)

= 60 + 28/5

= 60 + 5.6

= 65.6

**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

**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 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 (Σfi ui/N)

= 50 + 10 x (-2/40)

= 50 – 0.5

= 49.5

**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

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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.