The marks of 200 students are as given below :

Marks	No. of students
20-29	7
30-39	11
40-49	20
50-59	46
60-69	57
70-79	37
80-89	15
90-99	7

Draw a cumulative frequency curve and find :

(i) median marks

(ii) if 80% of the students passed, find the passing marks.

The above distribution is discontinuous converting into continuous distribution, we get :

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\dfrac{30 - 29}{2} = \dfrac{1}{2}$

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Cumulative frequency distribution table is as follows :

(i) Here, n = 200, which is even.

By formula,

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

Steps of construction :

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

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

3. Plot the point (19.5, 0) as ogive starts on x-axis representing lower limit of first class.

4. Plot the points (29.5, 7), (39.5, 18), (49.5, 38), (59.5, 84), (69.5, 141), (79.5, 178), (89.5, 193) and (99.5, 200).

5. Join the points by a free-hand curve.

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

From graph, C = 63

Hence, median = 63.

(ii) Given,

80% of students pass the exam or 20% students fail the exam.

$\dfrac{20}{100} \times 200 = 40$.

40 students fail the exam.

Draw a line parallel to x-axis from point D (no. of students) = 40, touching the graph at point E. From point E draw a line parallel to y-axis touching x-axis at point F.

From graph, F = 51

∴ 40 students score less than 51 marks.

Hence, passing marks = 51.