| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | | | | | |
| | | | | | |
Class Mark
Frequency
12.5
12
17.5
17
22.5
22
27. 5
27
32 .5
30
37 . 5
31
Q. Draw an ogive to represent the following data:
Sol:
Class marks
No. of students
Class interval
12.5
17.5
22.5
27.5
32.5
37.5
12
17
22
27
30
31
First find the class width by subtracting two consecutive class mark
27.5 – 22.5 = 5
5 ÷ 2 = 2.5
12.5
– 2.5
= 10
+ 2.5
= 15
12.5
10 – 15
17.5
– 2.5
= 15
+ 2.5
= 20
17.5
15 – 20
22.5
– 2.5
= 20
+ 2.5
= 25
22.5
20 – 25
27.5
– 2.5
= 25
+ 2.5
= 30
27.5
25 – 30
32.5
– 2.5
= 30
+ 2.5
= 35
32.5
30 – 35
37.5
– 2.5
= 35
+ 2.5
= 40
37.5
35 – 40
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
Class interval
Frequency
fi
Cumulative Frequency
12
17
22
27
0
0
10
20
30
40
50
60
Scale :
Class interval
Cumulative frequency
X
2 cm = 5 units on x-axis
1 cm = 10 units on y-axis
70
80
Upper
Boundary
ui
90
(15, 12)
Let us find co-ordinates of points to be plotted
x co-ordinate is upper boundary of class
y co-ordinate is cumulative frequency less than type
Since frequency of 1st class is not zero
So, introduce a class before 1st class with frequency zero
+
+
+
+
Mark c.f. less than on Y-axis
What is highest c.f.?
139
Since we have 20 cms on
Y-axis, we will consider simplest scale 1 cm = 10 units
Now plot the points
10
15
20
25
30
31
35
40
(10, 0)
(20 , 29)
(25, 51)
(30, 78)
(35, 109)
(40, 139)
30
+
Mark upper class boundaries on X-axis
100
110
120
130
140
Join these points with smooth curve
+
{
1 cm
1 cm
=
10 boxes
10 boxes
=
10 units
1 box = 1 unit
30
40
30
31
32
5 – 10
10 – 15
15 – 20
20 – 25
25 – 30
30 – 35
35 – 40
From class mark subtract 2.5 to get lower limit and add 2.5 to get upper limit
15
20
25
30
10
40
35
12
29
51
78
0
139
109
10
20
12
50
60
51
20
30
29
70
80
78
100
110
109
130
140
139
Class mark & frequency is given
Class mark is mid point of classes
Let us find class interval using given class marks
Divide class width by 2
Additional Example
HOMEWORK