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

Step 1: Complete the cumulative frequency table. 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 | Step 2: (a) Draw the cumulative frequency curve. 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. Step 3: (b) Use your drawn graph to estimate: (i) The median mark. The total number of candidates is N = 80. The median corresponds to the (N)/(2)th value, which is the (80)/(2) = 40th 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 × N = 0.60 × 80 = 48th 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). Step 4: (c) If the pass mark was 50%, what percentage of students passed? 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 ≈ 38. Number of students who passed = Total students - CF_<50 Number of students who passed ≈ 80 - 38 = 42. Percentage of students who passed = Number of students who passedTotal students × 100\%. Percentage passed ≈ (42)/(80) × 100\% = 52.5\%. Alternatively, by summing frequencies for marks 50: Number of students who scored 50% or more = 20 + 11 + 7 + 3 + 2 = 43. Percentage of students who passed: (43)/(80) × 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: (43)/(80) × 100\% = (4300)/(80)\% = 53.75\% The percentage of students who passed is 53.75%. That's 2 down. 3 left today — send the next one.