Prepare a cumulative frequency table for the given data.

a) To prepare a cumulative frequency table, we first determine the upper class boundaries and then sum the frequencies. Step 1: Determine the upper class boundaries. For a class like 0-9, the upper class boundary is 9.5. For 10-19, it's 19.5, and so on. Step 2: Calculate the cumulative frequency. The cumulative frequency for a class is the sum of its frequency and the frequencies of all preceding classes. Here is the cumulative frequency table: | Marks | Frequency (f) | Upper Class Boundary | Cumulative Frequency (cf) | | :------ | :------------ | :------------------- | :------------------------ | | 0-9 | 4 | 9.5 | 4 | | 10-19 | 6 | 19.5 | 10 | | 20-29 | 10 | 29.5 | 20 | | 30-39 | 10 | 39.5 | 30 | | 40-49 | 13 | 49.5 | 43 | | 50-59 | 17 | 59.5 | 60 | | 60-69 | 20 | 69.5 | 80 | | 70-79 | 15 | 79.5 | 95 | | 80-89 | 4 | 89.5 | 99 | | 90-99 | 1 | 99.5 | 100 | b) To draw a cumulative frequency curve (ogive): Step 1: Plot the cumulative frequency against the upper class boundaries. The points to plot are: (-0.5, 0), (9.5, 4), (19.5, 10), (29.5, 20), (39.5, 30), (49.5, 43), (59.5, 60), (69.5, 80), (79.5, 95), (89.5, 99), (99.5, 100). Step 2: Draw a smooth curve connecting these points. The x-axis represents the marks (upper class boundaries), and the y-axis represents the cumulative frequency. c) Since I cannot draw a curve, I will estimate the values using interpolation from the cumulative frequency table, which is equivalent to reading from a correctly drawn curve. The total frequency N = 100. • Median value: The median is the value at the (N)/(2) position. (N)/(2) = (100)/(2) = 50^th value. The 50^th value falls in the 50-59 class (cumulative frequency goes from 43 to 60). Using the formula for median: Median = L + ((N)/(2) - cf_bf_m) × c Where L = 49.5 (lower boundary of median class), cf_b = 43 (cumulative frequency before median class), f_m = 17 (frequency of median class), c = 10 (class width). Median = 49.5 + ((50 - 43)/(17)) × 10 Median = 49.5 + ((7)/(17)) × 10 Median = 49.5 + 4.1176... Median ≈ 53.62 • Interquartile range (IQR): IQR = Q_3 - Q_1. First Quartile (Q_1): The value at the (N)/(4) position. (N)/(4) = (100)/(4) = 25^th value. The 25^th value falls in the 30-39 class (cumulative frequency goes from 20 to 30). Q_1 = L + ((N)/(4) - cf_bf_q1) × c Where L = 29.5, cf_b = 20, f_q1 = 10, c = 10. Q_1 = 29.5 + ((25 - 20)/(10)) × 10 Q_1 = 29.5 + ((5)/(10)) × 10 Q_1 = 29.5 + 5 = 34.5 Third Quartile (Q_3): The value at the (3N)/(4) position. (3N)/(4) = (3 × 100)/(4) = 75^th value. The 75^th value falls in the 60-69 class (cumulative frequency goes from 60 to 80). Q_3 = L + ((3N)/(4) - cf_bf_q3) × c Where L = 59.5, cf_b = 60, f_q3 = 20, c = 10. Q_3 = 59.5 + ((75 - 60)/(20)) × 10 Q_3 = 59.5