Marks obtained by 200 students in an examination are given below :

Marks	No. of students
0 - 10	5
10 - 20	11
20 - 30	10
30 - 40	20
40 - 50	28
50 - 60	37
60 - 70	40
70 - 80	29
80 - 90	14
90 - 100	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 ?

(i) Cumulative frequency distribution table :

Here, n = 200, which is even.

By formula,

Median = $\dfrac{n}{2}$ th term

= $\dfrac{200}{2}$ = 100th term.

Steps of construction :

1. Take 1 cm along x-axis = 10 marks.

2. Take 1 cm along y-axis = 20 students.

3. Plot the point (0, 0) as ogive starts from x-axis representing lower limit of first class.

4. Plot the points (10, 5), (20, 16), (30, 26), (40, 46), (50, 74), (60, 111), (70, 151), (80, 180), (90, 194) and (100, 200).

5. Join the points by a free hand curve.

6. Draw a line parallel to x-axis from point L (no. of students) = 100, touching the graph at point M. From point M draw a line parallel to y-axis touching x-axis at point N.

From graph, N = 57

Hence, median = 57.

(ii) Minimum marks required to pass = 40.

Draw a line parallel to y-axis from point O (marks) = 40, touching the graph at point Q. From point Q draw a line parallel to x-axis touching y-axis at point P.

From graph,

P = 46.

∴ 46 students scored less than 40 marks.

Hence, 46 students failed the examination.

(iii) 85 or more marks is considered grade one.

Draw a line parallel to y-axis from point Q (marks) = 85, touching the graph at point R. From point R draw a line parallel to x-axis touching y-axis at point S.

From graph,

S = 188.

∴ 188 students scored less than 85 marks.

∴ 12 (200 - 188) students scored more than 85 marks.

Hence, 12 students secured grade one.