marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
No. of students | 5 | 11 | 10 | 20 | 28 | 37 | 40 | 29 | 14 | 6 |
Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis.
Using the graph, determine
(i) The median marks.
(ii) The number of students who failed if minimum marks required to pass is 40.
(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.
Solution:
We write the given data in cumulative frequency table.
Marks | No of students f | Cumulative frequency c.f |
0-10 | 5 | 5 |
10-20 | 11 | 16 |
20-30 | 10 | 26 |
30-40 | 20 | 46 |
40-50 | 28 | 74 |
50-60 | 37 | 111 |
60-70 | 40 | 151 |
70-80 | 29 | 180 |
80-90 | 14 | 194 |
90-100 | 6 | 200 |
To represent the data in the table graphically, we mark the upper limits of the class intervals on
the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis),
Plot the points (10, 5), (20, 16), (30, 26), (40, 46), (50, 74), (60, 111) , (70, 151) (80, 180) (90, 194) and (100, 200) on the graph.
Join the points with the free hand. We get an ogive as shown:
(i)Here n= 200
Median = (n/2)th term
= 200/2
= 100th term
= 57 [from graph]
(ii)number of students failed if minimum marks required to pass is 40 = 44 [from graph]
(iii)Number of students who got grade 1 = number of students who scored 85 and more
= 200-188
= 12[From graph]