Table 2.2: Scores of candidates. Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution. Use your curve to estimate the: (i) Median mark (ii) Upper and lower quartiles (iii) Semi-inter quartile range.

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Asked on August 19, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-08-19T16:36:08.607Z

Step 1: Prepare the cumulative frequency table. First, determine the class boundaries for the given marks. Since the classes are 21-30, 31-40, etc., the class boundaries are found by taking the midpoint between the upper limit of one class and the lower limit of the next. For example, between 30 and 31, the boundary is 30.5. | Marks (%) | Class Boundaries | No. of students (Frequency, f) | Cumulative Frequency (CF) | | :-------- | :--------------- | :------------------------------- | :------------------------ | | 21-30 | 20.5 - 30.5 | 9 | 9 | | 31-40 | 30.5 - 40.5 | 20 | 9 + 20 = 29 | | 41-50 | 40.5 - 50.5 | 31 | 29 + 31 = 60 | | 51-60 | 50.5 - 60.5 | 42 | 60 + 42 = 102 | | 61-70 | 60.5 - 70.5 | 50 | 102 + 50 = 152 | | 71-80 | 70.5 - 80.5 | 32 | 152 + 32 = 184 | | 81-90 | 80.5 - 90.5 | 16 | 184 + 16 = 200 | The total number of students, N = 200. Step 2: Draw the cumulative frequency curve (ogive). To draw the cumulative frequency curve, plot the cumulative frequency against the upper class boundaries. • Plot the points: (30.5, 9), (40.5, 29), (50.5, 60), (60.5, 102), (70.5, 152), (80.5, 184), (90.5, 200). • Start the curve from the lower boundary of the first class with a cumulative frequency of 0, i.e., (20.5, 0). • Join these points with a smooth curve. The x-axis represents "Marks (%)" and the y-axis represents "Cumulative Frequency". Step 3: Estimate the median mark (Q_2). The median mark is the value corresponding to the (N)/(2)-th position. Median position = (200)/(2) = 100 On your cumulative frequency curve, locate 100 on the y-axis (cumulative frequency). Draw a horizontal line from this point to intersect the curve. From the intersection point, draw a vertical line down to the x-axis (marks). Read the value on the x-axis. Using interpolation (which approximates reading from a smooth curve): Q_2 = L + ((N)/(2) - CF_bf_m) × c Where L = 50.5, CF_b = 60, f_m = 42, c = 10. Q_2 = 50.5 + ((100 - 60)/(42)) × 10 = 50.5 + ((40)/(42)) × 10 ≈ 50.5 + 9.52 = 60.02 The estimated median mark is 60.0 marks. Step 4: Estimate the lower quartile (Q_1). The lower quartile is the value corresponding to the (N)/(4)-th position. Lower quartile position = (200)/(4) = 50 On your cumulative frequency curve, locate 50 on the y-axis. Draw a horizontal line to the curve, then a vertical line down to the x-axis. Read the value. Using interpolation: Q_1 = L + ((N)/(4) - CF_bf_q) × c Where L = 40.5, CF_b = 29, f_q = 31, c = 10. Q_1 = 40.5 + ((50 - 29)/(31)) × 10 = 40.5 + ((21)/(31)) × 10 ≈ 40.5 + 6.77 = 47.27 The estimated lower quartile is 47.3 marks. Step 5: Estimate the upper quartile (Q_3). The upper quartile is the value corresponding to the (3N)/(4)-th position. Upper quartile position = (3 × 200)/(4) = 150 On your cumulative frequency curve, locate 150 on the y-axis. Draw a horizontal line to the curve, then a vertical line down to the x-axis. Read the value. Using interpolation: Q_3 = L + ((3N)/(4) - CF_bf_q) × c Where L = 60.5, CF_b = 102, f_q = 50, c = 10. Q_3 = 60.5 + ((150 - 102)/(50)) × 10 = 60.5 + ((48)/(50)) × 10 ≈ 60.5 + 9.6 = 70.1 The estimated upper quartile is 70.1 marks. Step 6: Calculate the semi-interquartile range (SIQR). The semi-interquartile range is half the difference between the upper and lower quartiles. SIQR = (Q_3 - Q_1)/(2) SIQR = (70.1 - 47.3)/(2) = (22.8)/(2) = 11.4 The semi-interquartile range is 11.4 marks. Send me the next one 📸

Accepted Answer · 2026-08-19T16:36:08.607Z

Step 1: Prepare the cumulative frequency table. First, determine the class boundaries for the given marks. Since the classes are 21-30, 31-40, etc., the class boundaries are found by taking the midpoint between the upper limit of one class and the lower limit of the next. For example, between 30 and 31, the boundary is 30.5. | Marks (%) | Class Boundaries | No. of students (Frequency, f) | Cumulative Frequency (CF) | | :-------- | :--------------- | :------------------------------- | :------------------------ | | 21-30 | 20.5 - 30.5 | 9 | 9 | | 31-40 | 30.5 - 40.5 | 20 | 9 + 20 = 29 | | 41-50 | 40.5 - 50.5 | 31 | 29 + 31 = 60 | | 51-60 | 50.5 - 60.5 | 42 | 60 + 42 = 102 | | 61-70 | 60.5 - 70.5 | 50 | 102 + 50 = 152 | | 71-80 | 70.5 - 80.5 | 32 | 152 + 32 = 184 | | 81-90 | 80.5 - 90.5 | 16 | 184 + 16 = 200 | The total number of students, N = 200. Step 2: Draw the cumulative frequency curve (ogive). To draw the cumulative frequency curve, plot the cumulative frequency against the upper class boundaries. • Plot the points: (30.5, 9), (40.5, 29), (50.5, 60), (60.5, 102), (70.5, 152), (80.5, 184), (90.5, 200). • Start the curve from the lower boundary of the first class with a cumulative frequency of 0, i.e., (20.5, 0). • Join these points with a smooth curve. The x-axis represents "Marks (%)" and the y-axis represents "Cumulative Frequency". Step 3: Estimate the median mark (Q_2). The median mark is the value corresponding to the (N)/(2)-th position. Median position = (200)/(2) = 100 On your cumulative frequency curve, locate 100 on the y-axis (cumulative frequency). Draw a horizontal line from this point to intersect the curve. From the intersection point, draw a vertical line down to the x-axis (marks). Read the value on the x-axis. Using interpolation (which approximates reading from a smooth curve): Q_2 = L + ((N)/(2) - CF_bf_m) × c Where L = 50.5, CF_b = 60, f_m = 42, c = 10. Q_2 = 50.5 + ((100 - 60)/(42)) × 10 = 50.5 + ((40)/(42)) × 10 ≈ 50.5 + 9.52 = 60.02 The estimated median mark is 60.0 marks. Step 4: Estimate the lower quartile (Q_1). The lower quartile is the value corresponding to the (N)/(4)-th position. Lower quartile position = (200)/(4) = 50 On your cumulative frequency curve, locate 50 on the y-axis. Draw a horizontal line to the curve, then a vertical line down to the x-axis. Read the value. Using interpolation: Q_1 = L + ((N)/(4) - CF_bf_q) × c Where L = 40.5, CF_b = 29, f_q = 31, c = 10. Q_1 = 40.5 + ((50 - 29)/(31)) × 10 = 40.5 + ((21)/(31)) × 10 ≈ 40.5 + 6.77 = 47.27 The estimated lower quartile is 47.3 marks. Step 5: Estimate the upper quartile (Q_3). The upper quartile is the value corresponding to the (3N)/(4)-th position. Upper quartile position = (3 × 200)/(4) = 150 On your cumulative frequency curve, locate 150 on the y-axis. Draw a horizontal line to the curve, then a vertical line down to the x-axis. Read the value. Using interpolation: Q_3 = L + ((3N)/(4) - CF_bf_q) × c Where L = 60.5, CF_b = 102, f_q = 50, c = 10. Q_3 = 60.5 + ((150 - 102)/(50)) × 10 = 60.5 + ((48)/(50)) × 10 ≈ 60.5 + 9.6 = 70.1 The estimated upper quartile is 70.1 marks. Step 6: Calculate the semi-interquartile range (SIQR). The semi-interquartile range is half the difference between the upper and lower quartiles. SIQR = (Q_3 - Q_1)/(2) SIQR = (70.1 - 47.3)/(2) = (22.8)/(2) = 11.4 The semi-interquartile range is 11.4 marks. Send me the next one 📸

Table 2.2: Scores of candidates. Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution. Use your curve to estimate the: (i) Median mark (ii) Upper and lower quartiles (iii) Semi-inter quartile range.

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Table 2.2: Scores of candidates. Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution. Use your curve to estimate the: (i) Median mark (ii) Upper and lower quartiles (iii) Semi-inter quartile range.

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