EXERCISE 14.4

Graphical Representation of Cumulative Frequency Distribution:

The graph makes the data easy to understand. So to make the graph of the cumulative frequency distribution, we need to find the cumulative frequency of the given table, then we can plot the points on the graph.

The cumulative frequency distribution can be of two types –

  1. Less than Ogive

To draw the graph of less than ogive we take the lower limits of the class interval and mark the respective less than frequency. Then join the dots by a smooth curve.

  1. More than Ogive

To draw the graph of more than ogive we take the upper limits of the class interval on the x-axis and mark the respective more than frequency. Then join the dots.

Example : Draw the cumulative frequency distribution curve for the following table.

Marks of Students
0-10
10-20
20-30
30–40
40–50
50–60
Number of students
7
10
14
20
6
3

 

Solution: To draw the less than and more than ogive, we need to find the less than cumulative frequency and more than cumulative frequency.

Marks
Number of students
less than cumulative frequency
 
More than cumulative frequency
 
0 -10
7
Less than 10
7
More than 0
60
10 -20
10
Less than 20
17
More than 10
53
20 -30
14
Less than 30
31
More than 20
43
30 -40
20
Less than 40
51
More than 30
29
40 -50
6
Less than 50
57
More than 40
9
50 -60
3
Less than 60
60
More than 50
3
 
 
 
 
More than 60
0

 

Now we plot all the points on the graph and we get two curves.

Remark:

  • The class interval should be continuous to make the ogive curve.
  • The x-coordinate at the intersection of the less than and more than ogive is the median of the given data.

EXERCISE 14.4

  1. The following distribution gives the daily income of 50 workers if a factory. Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
Daily Income (in Rupees)
100-120
120–140
140–160
160–180
180–200
Number of workers
12
14
8
6
10

 

Solution: Convert the given distribution table to a less than type cumulative frequency distribution, and we get

Daily Income
Frequency
Cumulative Frequency
Less than 120
12
12
Less than 140
14
26
Less than 160
8
34
Less than 180
6
40
Less than 200
10
50

 

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve.

2.During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in Kg)
Number of Students
Less than 38
0
Less than 40
3
Less than 42
5
Less than 44
9
Less than 46
14
Less than 48
28
Less than 50
32
Less than 52
35

 

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution: From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale. Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them to get a smooth curve. The curve obtained is known as less than type ogive.

Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the mode by making a table.

Class Interval
Number of students (Frequency)
Cumulative Frequency
Less than 38
0
0
Less than 40
3 -0 = 3
3
Less than 42
5 -3 = 2
8
Less tan 44
9 -5 = 4
9
Less than 46
14 – 9= 5
14
Less than 48
28 – 14 =14
28
Less than 50
32 – 28 =4
32
Less than 52
35 – 22 =3
35

The class 46 – 48 has the maximum frequency, therefore, this is modal class
Where, l = 46,   h = 2,   f1 = 14 ,  f0 = 5,   f2 = 4
The mode formula is given as:
Mode = l + (f1  − f0)/(2f1 − f0 − f2 ) ×h
= 46 + 0.95 = 46.95
Thus, mode is verified.

3. The following tables gives production yield per hectare of wheat of 100 farms of a village.

Production Yeild
50-55
55-60
60-65
65–70
70–75
75–80
Number of Farms
2
8
12
24
38
16

 

Change the distribution to a more than type distribution and draw its ogive.

Solution: Converting the given distribution to a more than type distribution, we get

Production (in KG/hr)
Number of Farms
More than or Equal to 50
100
More than or Equal to 55
100 – 2 = 98
More than or Equal to 60
98 – 8 = 90
More than or Equal to 65
90 – 12 =78
More than or Equal to 70
78 – 24 =54
More than or Equal to 75
54 – 38 = 16

 

From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper. The graph obtained is known as more than type ogive curve.