PART II, QUESTION 1: Given the frequency distribution of marks, (a) draw the cumulative frequency curve (ogive), (b) find the median, and (c) determine the semi-interquartile range.

Welcome back 🙏👇🍊🍉Hustler😥😓🤗 — been a while! Let's pick up where you left off. Here is the solution for the first question in PART II, covering parts (a), (b), and (c). First, we need to calculate the cumulative frequencies and identify the upper class boundaries from the given frequency distribution table. | Marks (%) | Frequency | Upper Class Boundary | Cumulative Frequency | | :-------- | :-------- | :------------------- | :------------------- | | 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 | The total number of students is N=80. a) To draw the cumulative frequency curve (ogive): Step 1: Plot the cumulative frequencies against the upper class boundaries. The points to plot are: (9.5, 1), (19.5, 4), (29.5, 10), (39.5, 21), (49.5, 37), (59.5, 57), (69.5, 68), (79.5, 75), (89.5, 78), (99.5, 80). Step 2: Join these points with a smooth curve. The curve should start from (0,0) or the lower boundary of the first class with a cumulative frequency of 0. b) Using the data (as a graph is not available, we use linear interpolation): b)(i) The median mark: The median is the 50^th percentile. The position of the median is (N)/(2) = (80)/(2) = 40^th student. The 40^th student falls in the 50-59 class (cumulative frequency goes from 37 to 57). The formula for the median (or k^th percentile