Draw a cumulative frequency curve for the following data :

Marks obtained	No. of students
0 - 10	8
10 - 20	10
20 - 30	22
30 - 40	40
40 - 50	20

Hence, determine :

(i) the median

(ii) the pass marks if 85% of the students pass.

(iii) the marks which 45% of the students exceed.

1. The cumulative frequency table for the given continuous distribution is :

2. Take 2 cm along x-axis = 10 marks

3. Take 1 cm along y-axis = 10 (students)

4. Plot the points (10, 8), (20, 18), (30, 40), (40, 80) and (50, 100) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 0 - 10.

5. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 100.

To find the median :

Let A be the point on y-axis representing frequency = $\dfrac{n}{2} = \dfrac{100}{2}$ = 50.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 32.5 marks.

(ii) Total no. of students = 100.

85% of students pass i.e. 85 students pass.

Remaining no. of students = 15.

Let B be the point on y-axis representing frequency 15.

Through B draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 18 marks.

Hence, passing marks = 18 if 85% students pass.

(iii) Total no. of students = 100.

45% of students i.e. 45 students.

Remaining no. of students = 55.

Let C be the point on y-axis representing frequency 55.

Through C draw a horizontal line to meet the ogive at R. Through R, draw a vertical line to meet the x-axis at O. The abscissa of the point O represents 34 marks.

Hence, 45% of students exceed 34 marks.