A Smudged Frequency in the Class X Result Analysis
The mathematics teacher of a government high school in Patiala is preparing the Class X board-result analysis for the annual report. She has her students' marks grouped into class intervals, but a cup of tea has smudged one entry in the register. The readable entries are: 0–10, 4 students; 10–20, 6 students; 20–30, unknown; 30–40, 8 students; 40–50, 2 students. From the mark sheet she can still read that the mean of the whole batch is 24.5 marks. The headmaster wants the average, the median and the mode for his report, and he also wants to write two sentences: 'more than half the class scored above 25 marks' and 'at least 90 per cent of students passed', the pass mark being 20. The teacher is uneasy about both sentences. She wants to know exactly what the grouped table can prove before the report goes to the printer, because a wrong claim in a board-result analysis would embarrass the school.
1. Find the smudged frequency and hence the total number of students.
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Let the unknown frequency of the class 20–30 be x, and use class marks (mid-values) 5, 15, 25, 35, 45.
Σf = 4 + 6 + x + 8 + 2 = 20 + x Σfx = 4(5) + 6(15) + x(25) + 8(35) + 2(45) = 20 + 90 + 25x + 280 + 90 = 480 + 25x
Mean = Σfx / Σf = (480 + 25x)/(20 + x) = 24.5
Cross-multiply: 480 + 25x = 24.5(20 + x) = 490 + 24.5x 25x − 24.5x = 490 − 480 0.5x = 10 → x = 20
So the 20–30 class has 20 students and the batch has 20 + 20 = 40 students.
Check: Σfx = 480 + 25(20) = 980; 980 ÷ 40 = 24.5 ✓ - the mean agrees, so the value is consistent.
2. Compute the median and the mode from the completed table.
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Cumulative frequencies: 4, 10, 30, 38, 40. N = 40, so N/2 = 20.
Median: the 20th student falls in the class 20–30 (cf jumps from 10 to 30). Median = l + [(N/2 − cf)/f] × h = 20 + [(20 − 10)/20] × 10 = 20 + (10/20) × 10 = 20 + 5 = 25 marks
Mode: the highest frequency is 20, in the class 20–30, so the modal class is 20–30 with l = 20, f₁ = 20, f₀ = 6 (preceding class), f₂ = 8 (succeeding class), h = 10. Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h = 20 + [(20 − 6)/(40 − 6 − 8)] × 10 = 20 + (14/26) × 10 = 20 + 5.38 = 25.38 marks (≈ 25.4)
Report figures: mean 24.5, median 25, mode ≈ 25.4 - the three are close together, which is a fair summary of the batch.
3. Evaluate the headmaster's two sentences: 'more than half the class scored above 25 marks' and 'at least 90 per cent passed'. What should the teacher actually write?
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Sentence 1 - 'more than half scored above 25': not supported. The median is exactly 25, which by definition means the middle student scored 25, i.e. roughly half the class lies at or below 25. Grouped data cannot even tell us how many scored above 25, because the 20 students in the 20–30 class are not individually visible. The honest wording is: 'The median mark is 25, so about half the class scored 25 or less.'
Sentence 2 - 'at least 90% passed': false. Students scoring below 20 are the 4 in 0–10 and the 6 in 10–20 = 10 students; they cannot have reached the pass mark of 20. So at most 40 − 10 = 30 students passed: 30/40 × 100 = 75%, not 90%. (On the usual convention that a boundary value 20 belongs to the higher class, exactly 30 passed, i.e. 75%.) The pass claim overstates the result by 15 percentage points.
What she should print: mean 24.5, median 25, mode ≈ 25.4, pass percentage 75%. She may add that 20 of 40 students lie in the 20–30 band, which is where most of the teaching effort should go next year.