RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.5

RD Sharma Class 10 Solutions Chapter 7 Ex 7.5 PDF Free Download

In statistics, the mode is one important parameter. This exercise completely deals with finding mode/ modal value of the given continuous frequency distribution. Suitable methods are introduced for computing the modal value, and this can be clearly understood in the RD Sharma Solutions Class 10. If you want to get the grip over finding mode, the RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.5 PDF is available for free to download.

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.5 Download PDF

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Access RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.5

1. Find the mode of the following data: 

(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4 

(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4 

(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15 

Solution:

(i)

Value (x)

3

4

5

6

7

8

9

Frequency (f)

4

2

5

2

2

1

2

Thus, the mode = 5 since it occurs the maximum number of times.

(ii)

Value (x)

3

4

5

6

7

8

9

Frequency (f)

5

2

4

2

2

1

2

Thus, the mode = 3 since it occurs the maximum number of times.

(iii)

Value (x)

8

15

18

19

20

24

25

Frequency (f)

1

4

1

1

1

2

1

Thus, the mode = 15 since it occurs the maximum number of times.

2. The shirt size worn by a group of 200 persons, who bought the shirt from a store, are as follows:

Shirt size:

37

38

39

40

41

42

43

44

Number of persons:

15

25

39

41

36

17

15

12

Find the model shirt size worn by the group.

Solution:

Shirt size:

37

38

39

40

41

42

43

44

Number of persons:

15

25

39

41

36

17

15

12

From the data its observed that,

Model shirt size = 40 since it was the size which occurred for the maximum number of times.

3. Find the mode of the following distribution.

(i)

Class interval:

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

60 – 70

70 – 80

Frequency:

5

8

7

12

28

20

10

10

Solution:

Class interval:

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

60 – 70

70 – 80

Frequency:

5

8

7

12

28

20

10

10

It’s seen that the maximum frequency is 28.

So, the corresponding class i.e., 40 – 50 is the modal class.

And,

l = 40, h = 50 40 = 10, f = 28, f1 = 12, f= 20

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 1

= 40 + 160/ 24

= 40 + 6.67

= 46.67

(ii)

Class interval

10 – 15

15 – 20

20 – 25

25 – 30

30 – 35

35 – 40

Frequency

30

45

75

35

25

15

Solution:

Class interval

10 – 15

15 – 20

20 – 25

25 – 30

30 – 35

35 – 40

Frequency

30

45

75

35

25

15

It’s seen that the maximum frequency is 75.

So, the corresponding class i.e., 20 – 25 is the modal class.

And,

l = 20, h = 25 – 20 = 5, f = 75, f1 = 45, f= 35

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 2

= 20 + 150/70

= 20 + 2.14

= 22.14 

(iii)

Class interval

25 – 30

30 – 35

35 – 40

40 – 45

45 – 50

50 – 55

Frequency

25

34

50

42

38

14

Solution:

Class interval

25 – 30

30 – 35

35 – 40

40 – 45

45 – 50

50 – 55

Frequency

25

34

50

42

38

14

It’s seen that the maximum frequency is 50.

So, the corresponding class i.e., 35 – 40 is the modal class.

And,

l = 35, h = 40 – 35 = 5, f = 50, f1 = 34, f= 42

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 3

= 35 + 80/24

= 35 + 3.33

= 38.33 

4. Compare the modal ages of two groups of students appearing for an entrance test:

Age in years

16 – 18

18 – 20

20 – 22

22 – 24

24 – 26

Group A

50

78

46

28

23

Group B

54

89

40

25

17

Solution:

Age in years

16 – 18

18 – 20

20 – 22

22 – 24

24 – 26

Group A

50

78

46

28

23

Group B

54

89

40

25

17

For Group A:

It’s seen that the maximum frequency is 78.

So, the corresponding class 18 – 20 is the model class.

And,

l = 18, h = 20 – 18 = 2, f = 78, f1 = 50, f= 46 

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 4

= 18 + 56/60

= 18 + 0.93

= 18.93 years

For group B:

It’s seen that the maximum frequency is 89

So, the corresponding class 18 – 20 is the modal class.

And,

l = 18, h = 20 – 18 = 2, f = 89, f1 = 54, f= 40

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 5

Mode

= 18 + 70/84

= 18 + 0.83

= 18.83 years

Therefore, the modal age of the Group A is higher than that of Group B.

5. The marks in science of 80 students of class X are given below. Find the mode of the marks obtained by the students in science.

Marks

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

60 – 70

70 – 80

80 – 90

90 – 100

Frequency

3

5

16

12

13

20

5

4

1

1

Solution:

Marks

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

60 – 70

70 – 80

80 – 90

90 – 100

Frequency

3

5

16

12

13

20

5

4

1

1

It’s seen that the maximum frequency is 20.

So, the corresponding class 50 – 60 is the modal class.

And,

l = 50, h = 60 – 50 = 10, f = 20, f1 = 13, f= 5 

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 6

= 50 + 70/22

= 50 + 3.18

= 53.18

6. The following is the distribution of height of students of a certain class in a city:

Height (in cm):

160 – 162

163 – 165

166 – 168

169 – 171

172 – 174

No of students:

15

118

142

127

18

Find the average height of maximum number of students. 

Solution:

Heights(exclusive)

160 – 162

163 – 165

166 – 168

169 – 171

172 – 174

Heights (inclusive)

159.5 – 162.5

162.5 – 165.5

165.5 – 168.5

168.5 – 171.5

171.5 – 174.5

No of students

15

118

142

127

18

It’s seen that the maximum frequency is 142.

So, the corresponding class 165.5 – 168.5 is the modal class.

And,

l = 165.5, h = 168.5 – 165.5 = 3, f = 142, f1 = 118, f2 = 127 

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 7

= 165.5 + 72/39

= 165.5 + 1.85

= 167.35 cm 

7. The following table shows the ages of the patients admitted in a hospital during a year:

Ages (in years):

5 – 15

15 – 25

25 – 35

35 – 45

45 – 55

55 – 65

No of students:

6

11

21

23

14

5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. 

Solution:

To find the mean:

For the given data let the assumed mean (A) = 30

Age (in years)

Number of patients fi

Class marks xi

di = xi – 275

fidi

5 – 15

6

10

– 20

-120

15 – 25

11

20

– 10

-110

25 – 35

21

30

0

0

35 – 45

23

40

10

230

45 – 55

14

50

20

280

55 – 65

5

60

30

150

N = 80

Σfdi = 430

It’s observed from the table that Σfi = N = 80 and Σfdi = 430.

Using the formula for mean,

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 8

= 30 + 430/80

= 30 + 5.375

= 35.375

= 35.38

Thus, the mean of this data is 35.38. It can also be interpreted as that on an average the age of a patients admitted to hospital was 35.38 years.

It is also observed that maximum class frequency is 23 and it belongs to class interval 35 – 45

So, modal class is 35 – 45 with the Lower limit (l) of modal class = 35

And, Frequency (f) of modal class = 23

Class size (h) = 10

Frequency (f1) of class preceding the modal class = 21

Frequency (f2) of class succeeding the modal class = 14

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 9

Mode

Therefore, the mode is 36.8. This represents that maximum number of patients admitted in hospital were of 36.8 years. 

Hence, it’s seen that mode is greater than the mean.

8. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Lifetimes (in hours):

0 – 20

20 – 40

40 – 60

60 – 80

80 – 100

100 – 120

No. of components:

10

35

52

61

38

29

Determine the modal lifetimes of the components.

Solution:

From the data given as above its observed that maximum class frequency is 61 which belongs to class interval 60 – 80.

So, modal class limit (l) of modal class = 60

Frequency (f) of modal class = 61

Frequency (f1) of class preceding the modal class = 52

Frequency (f2) of class succeeding the modal class = 38

Class size (h) = 20

Using the formula for find mode, we have

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 10

Mode 

Thus, the modal lifetime of electrical components is 65.625 hours 

 

9. The following table gives the daily income of 50 workers of a factory:

Daily income

100 – 120

120 – 140

140 – 160

160 – 180

180 – 200

Number of workers

12

14

8

6

10

Find the mean, mode and median of the above data.

Solution:

Class interval

Mid value (x)

Frequency (f)

fx

Cumulative frequency

100 – 120

110

12

1320

12

120 – 140

130

14

1820

26

140 – 160

150

8

1200

34

160 – 180

170

6

1000

40

180 – 200

190

10

1900

50

N = 50

Σfx = 7260

We know that,

Mean = Σfx / N

= 7260/ 50

= 145.2

Then,

We have, N = 50

⇒ N/2 = 50/2 = 25

So, the cumulative frequency just greater than N/2 is 26, then the median class is 120 – 140

Such that l = 120, h = 140 – 120 = 20, f = 14, F = 12 

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 11

= 120 + 260/14

= 120 + 18.57

= 138.57

From the data, its observed that maximum frequency is 14, so the corresponding class 120 – 140 is the modal class

And,

l = 120, h = 140 – 120 = 20, f = 14, f= 12, f= 8 

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 12

= 120 + 5

= 125 

Therefore, mean = 145.2, median = 138.57 and mode = 125 

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

Exercise 7.1 Solutions

Exercise 7.2 Solutions

Exercise 7.3 Solutions

Exercise 7.4 Solutions

Exercise 7.6 Solutions

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