RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.3

RD Sharma Class 10 Solutions Chapter 7 Ex 7.3 PDF Free Download

This exercise exposes 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 on par 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 Chapter 7 Statistics Exercise 7.3 Download PDF

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

Access RD Sharma Solutions for Class 10 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 (fi)

Expenditure (in rupees) (xi)

Frequency (fi)

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 (xi)

d= xi – 275

ui = (xi – 275)/50

Frequency fi

fiui

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 fiui = -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

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 (fi)

xi

fixi

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 fiui = 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.

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 xi

d= xi – 150

ui = (x– 150)/20

Frequency fi

fiui

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 fiui = -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

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 (fi)

xi

di = xi – 75.5

ui = (x– 755)/h

fiui

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 fiui = 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.

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 xi

d= x– 15

u= (x– 15)/6

fi

fiui

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 fiui = 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

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 xi

d= x– 100

u= (x– 100)/20

fi

fiui

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 fiui = 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

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 xi

d= x– 20

u= (x– 20)/8

fi

fiui

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 fiui = 7

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

= 20.4

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 xi

d= x– 15

u= (x– 15)/6

fi

fiui

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 fiui = 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

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 xi

d= x– 25

u= (x– 25)/10

fi

fiui

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 fiui = 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

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 xi

d= x– 20

u= (x– 20)/8

fi

fiui

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 fiui = 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

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 xi

d= x– 20

u= (x– 20)/8

fi

fiui

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 fiui = -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

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 xi

d= x–60

u= (x– 60)/20

fi

fiui

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 fiui = 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

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 xi

d= x– 50

u= (x– 50)/10

fi

fiui

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 fiui = -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

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 xi

d= x– 42

u= (x– 42)/5

fi

fiui

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 fiui = -79

From the table it’s seen that,

A = 42 and h = 5

Mean = A + h x (Σfi ui/N)

= 42 + 5 x (-79/70)

= 42 – 79/14

= 42 – 5.643

= 36.357

Also, access other exercise solutions of RD Sharma Class 10 Maths Chapter 7 Statistics

Exercise 7.1 Solutions

Exercise 7.2 Solutions

Exercise 7.4 Solutions

Exercise 7.5 Solutions

Exercise 7.6 Solutions

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