RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.3 Avail PDF For Free

This exercise helps students to find the arithmetic mean of a continuous frequency distribution. The RD Sharma Solutions Class 10 provides a stepwise solution to problems so that students attain strong conceptual knowledge. All the solutions are consistent with the latest CBSE pattern. Further, students can access the RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.3, PDF provided below.

RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.3

Download PDF

carouselExampleControls112

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.3 08

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.3 09

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.3 10

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.3 11

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.3 12

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.3 13

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.3 14

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.3 15

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.3 16

Previous Next

Access RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.3

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

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 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_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

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 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_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 the 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 heartbeats per minute for those women are 75.9 beats per minute.

Find the mean of each of the following frequency distributions: (5 – 14)

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

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

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_iu_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_iu_i= 7

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

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

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

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_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

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_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

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_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

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_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

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_iu_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_iu_i= -79