Complete the cumulative frequency table and draw the cumulative frequency curve (ogive).

The cumulative frequency is the running total of the frequencies. The upper class boundary is used for plotting the cumulative frequency curve.

Marks (%)	Frequency (f)	Upper Class Boundary	Cumulative Frequency (cf)
0-9	1	9.5	1
10-19	3	19.5	1 + 3 = 4
20-29	6	29.5	4 + 6 = 10
30-39	11	39.5	10 + 11 = 21
40-49	16	49.5	21 + 16 = 37
50-59	20	59.5	37 + 20 = 57
60-69	11	69.5	57 + 11 = 68
70-79	7	79.5	68 + 7 = 75
80-89	3	89.5	75 + 3 = 78
90-99	2	99.5	78 + 2 = 80

To draw the cumulative frequency curve (ogive):

• Plot the upper class boundaries on the x-axis (Marks %).
• Plot the corresponding cumulative frequencies on the y-axis (Number of Candidates).
• Start the curve at the lower boundary of the first class with a cumulative frequency of 0 (i.e., plot (0, 0) or (minus 0.5, 0) if using true class boundaries).
• Connect the plotted points with a smooth curve.

(i) The median mark.
The total number of candidates is $N = 80$. The median corresponds to the $\frac{N}{2}$th value, which is the $\frac{80}{2} = 40$th candidate.
On your graph, locate 40 on the cumulative frequency (y-axis). Draw a horizontal line from 40 to intersect the curve. From the intersection point, draw a vertical line down to the marks (x-axis). Read the value on the x-axis.
Based on the table, the 40th value falls within the 50-59 class.
The median mark is approximately 51.5% (this is an estimate from a typical ogive for this data).

(ii) The 60th percentile mark.
The 60th percentile corresponds to $0.60 \times N = 0.60 \times 80 = 48$th candidate.
On your graph, locate 48 on the cumulative frequency (y-axis). Draw a horizontal line from 48 to intersect the curve. From the intersection point, draw a vertical line down to the marks (x-axis). Read the value on the x-axis.
Based on the table, the 48th value falls within the 50-59 class.
The 60th percentile mark is approximately 55.5% (this is an estimate from a typical ogive for this data).

To find the number of students who passed (scored 50% or more) using the graph:
Locate 50% on the marks (x-axis). Draw a vertical line from 50% up to intersect the curve. From the intersection point, draw a horizontal line to the cumulative frequency (y-axis). Read this value, which represents the number of students who scored less than 50%. Let's call this $CF_{<50}$.
From the table, we know 37 students scored less than 49.5%. Using interpolation or a graph, $CF_{<50}$ would be slightly above 37. Let's assume $CF_{<50} \approx 38$.
Number of students who passed = Total students - $CF_{<50}$
Number of students who passed $\approx 80 - 38 = 42$.
Percentage of students who passed = $\frac{\text{Number of students who passed}}{\text{Total students}} \times 100\%$.
Percentage passed $\approx \frac{42}{80} \times 100\% = 52.5\%$.

Alternatively, by summing frequencies for marks $\ge 50$:
Number of students who scored 50% or more = $20 + 11 + 7 + 3 + 2 = 43$.
Percentage of students who passed:

$$\frac{43}{80} \times 100\% = 53.75\%$$

Since the question asks to "Use your drawn graph", the first method (estimating $CF_{<50}$ from the graph) is the intended approach. However, without the graph, the direct calculation from the table provides a precise value for students in the given class intervals.

Using the direct calculation from the table for students scoring 50% or more:
Number of students who passed = $20 + 11 + 7 + 3 + 2 = 43$.
Percentage of students who passed:

$$\frac{43}{80} \times 100\% = \frac{4300}{80}\% = 53.75\%$$

The percentage of students who passed is \boxed{\text{53.75%}}.