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Class 10 SELINA Solutions Maths Chapter 23 - Graphical Representation (Histograms and Ogives)

Graphical Representation (Histograms and Ogives) Exercise Ex. 23

Solution 2

(i)

Class Interval

Frequency

0-10

12

10-20

20

20-30

26

30-40

18

40-50

10

50-60

06

Steps of construction:

(a) Taking suitable scales, mark class intervals on x-axis and frequency on y-axis.

(b)Construct rectangles with class intervals as bases and corresponding frequencies as heights.

(ii)

Class Interval

Frequency

10-16

15

16-22

23

22-28

30

28-34

20

34-40

16

Steps of construction:

(a) Taking suitable scales, mark class intervals on x-axis and frequency on y-axis.

(b) Construct rectangles with class intervals as bases and corresponding frequencies as heights.


(iii)

Since the difference between each class marks is 8, therefore, subtract 8/2 = 4 from each class mark to get the lower limit of the corresponding class interval and add 5 to each class mark to get the upper limit.

Class Marks

Class Intervals

Frequency

16

12-20

08

24

20-28

12

32

28-36

15

40

36-44

18

48

44-52

25

56

52-60

19

64

60-68

10

Steps of construction:

(a) Convert the class marks into class intervals.

(b) Taking suitable scales, mark class intervals on x-axis and frequency on y-axis.

(c) Construct rectangles with class intervals as bases and corresponding frequencies as heights.

Solution 3

(i)

The cumulative frequency table for the given distribution is as follows:

 

Class Interval

Frequency

Cumulative frequency

10-15

10

10

15-20

15

25

20-25

17

42

25-30

12

54

30-35

10

64

35-40

8

72

Taking upper limits along X-axis and corresponding cumulative frequencies along Y-axis, mark the points (15,10), (20,25), (25,42), (30,54), (35, 64) and (40, 72).

Join the points marked by a free hand curve.

Solution 4

(i)

Marks Obtained

No. of students (c.f.)

less than 10

8

less than 20

25

less than 30

38

less than 40

50

less than 50

67

 

Steps Of construction:

(a) Plot the points (10,8), (20, 25), (30, 38), (40, 50) and (50, 67) on the graph.

(b) Join them with free hand to obtain an ogive.

(ii)

Age in years

(less than)

Cumulative

Frequency

10

0

20

17

30

32

40

37

50

53

60

58

70

65

 

 

Steps Of construction:

(a) Plot the points (10, 0), (20, 17), (30, 32), (40, 37), (50, 53), (60, 58) and (70, 65) on the graph.

(b) Join them with free hand to obtain an ogive.

Solution 5

(a)

Class Interval

Frequency

c.f.

8-12

9

9

12-16

16

25

16-20

22

47

20-24

18

65

24-28

12

77

28-32

4

81

(b) Now plot the points (12, 9), (16, 25), (20, 47), (24, 65), (28, 77), (32, 81) and join them to obtain an ogive.

Solution 6

(i)

 

Wages

No. Of workers

c.f.

6500-7000

10

10

7000-7500

18

28

7500-8000

22

50

8000-8500

25

75

8500-9000

17

92

9000-9500

10

102

9500-10000

8

110

Total = 110

(ii)

Now plot the points (7000,10), (7500,28), (8000,50), (8500,75), (9000,92), (9500,102) and (10000,110) and join them to obtain an ogive.

Solution 7

 (i)

Height (in cm)

No. Of workers

c.f.

150-155

6

6

155-160

12

18

160-165

18

36

165-170

20

56

170-175

13

69

175-180

8

77

180-185

6

83

(ii)

We plot the points (155, 6), (160, 18), (165, 36), (170, 56), (175, 69), (180, 77) and (185, 83) on the graph and join them in free hand to obtain an ogive.

Solution 8

(i)

Marks

(less than)

Cumulative frequency

Frequency

0-10

7

7

10-20

28

28-7=21

20-30

54

54-28=26

30-40

71

71-54=17

40-50

84

84-71=13

50-60

105

105-84=21

60-70

147

147-105=42

70-80

180

180-147=33

80-90

196

196-180=16

90-100

200

200-196=4

Total

 

200

 

(ii)

Marks

(more than)

Cumulative frequency

Frequency

0-10

100

13

10-20

87

22

20-30

65

10

30-40

55

13

40-50

42

6

50-60

36

5

60-70

31

10

70-80

21

3

80-90

18

11

90-100

7

7

Total

 

100

 

 

Solution 1(a)

Correct option: (i) a = 16, b = 52 and c = 8

24 + a = 40 a = 16

b = 40 + 12 = 52

b + c = 60

52 + c = 60 c = 8

Solution 1(b)

Correct option: (ii) 120

Frequency of class 30-50 = 50

Frequency of class 50-70 = 30

Frequency of class 70-90 = 40

Then, cumulative frequency of the class 70-90 = 50 + 30 + 40 = 120

Solution 1(c)

Correct option: (i)

The adjustment factor is half the difference between the upper limit of the class and the lower limit of the next class.

Note: The adjustment factor is used to convert inclusive classes into exclusive form.

Solution 1(d)

Correct option: (iii) lower limit of 1st class

The cumulative curve for a frequency distribution starts from the lower limit of 1st class.

Solution 1(e)

Correct option: (iv) upper limit of last class

The cumulative curve for a frequency distribution terminates at the upper limit of last class.

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