Construct a cumulative frequency table and a cumulative frequency polygon from the given frequency distribution of marks.

Here are the solutions to the problems based on the provided frequency distribution table. a) Make a cumulative frequency table and use it to draw a cumulative frequency polygon. Step 1: Create the cumulative frequency table. To do this, we add the frequency of each class to the sum of the frequencies of all preceding classes. | Marks % | Frequency (f) | Upper Class Boundary | Cumulative Frequency (cf) | | :-------- | :------------ | :------------------- | :------------------------ | | 10-20 | 18 | 20 | 18 | | 20-30 | 34 | 30 | 18 + 34 = 52 | | 30-40 | 58 | 40 | 52 + 58 = 110 | | 40-50 | 42 | 50 | 110 + 42 = 152 | | 50-60 | 24 | 60 | 152 + 24 = 176 | | 60-70 | 10 | 70 | 176 + 10 = 186 | | 70-80 | 6 | 80 | 186 + 6 = 192 | | 80-90 | 8 | 90 | 192 + 8 = 200 | Step 2: Identify points for the cumulative frequency polygon. To draw a cumulative frequency polygon, you would plot the cumulative frequency against the upper class boundary. The points to plot are: (10, 0), (20, 18), (30, 52), (40, 110), (50, 152), (60, 176), (70, 186), (80, 192), (90, 200). As an AI, I cannot draw the graph for you. b) Estimate the median from the graph and by calculation. Step 1: State inability to draw graph. As an AI, I cannot draw the graph to estimate the median visually. Step 2: Calculate the median. The total number of students is N = 200. The median position is (N)/(2) = (200)/(2) = 100. From the cumulative frequency table, the 100th student falls into the 30-40 class (since cf at 30 is 52 and cf at 40 is 110). The median class is 30-40. Using the formula for the median of grouped data: Median = L + ( (N)/(2) - CFf ) × w Where: L = 30 (lower boundary of the median class) N = 200 (total frequency) CF = 52 (cumulative frequency of the class before the median class) f = 58 (frequency of the median class) w = 10 (width of the median class, 40-30) Substitute the values into the formula: