Two Smudged Frequencies and a Parent's Complaint
The principal of a government senior secondary school in Haryana is preparing a report on the weekly self-study hours of the 100 students of Class X. The survey sheet gives five class intervals with the following counts: 0–10 h, 10 students; 10–20 h, 20 students; 20–30 h, f₁ (smudged); 30–40 h, 30 students; 40–50 h, f₂ (smudged). Only the sum of the two smudged entries survives. It is certain that exactly 100 students were surveyed and that the median of the distribution is 28 hours. A parent has already written claiming that 'the average student studies about 27 hours a week, so a child studying 29 hours is comfortably above average and needs no support'. The principal must also choose one summary figure for the school magazine, knowing that the data are bunched towards the higher hours.
1. Find f₁ and f₂ from the total number of students and the given median. Verify that the median class is consistent with your answer.
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Total: 10 + 20 + f₁ + 30 + f₂ = 100 ⇒ f₁ + f₂ = 40.
Cumulative frequency up to the end of 10–20 is 10 + 20 = 30. Since the median is 28, it falls in 20–30, so 20–30 is the median class, with l = 20, cf = 30, h = 10 and n/2 = 50.
Median = l + [(n/2 − cf)/f] × h ⇒ 28 = 20 + [(50 − 30)/f₁] × 10 ⇒ 8 = 200/f₁ ⇒ f₁ = 25.
Then f₂ = 40 − 25 = 15.
Consistency check: cumulative frequency up to 20–30 is 30 + 25 = 55 ≥ 50, and before the class it is 30 < 50, so 20–30 genuinely is the median class. The answer is self-consistent.
2. Compute the mean and the mode, then assess the parent's claim that 29 hours is 'comfortably above average'.
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Midpoints: 5, 15, 25, 35, 45.
Σfx = 10(5) + 20(15) + 25(25) + 30(35) + 15(45) = 50 + 300 + 625 + 1050 + 675 = 2700, so mean = 2700/100 = 27 hours.
Modal class is 30–40 (frequency 30). Mode = 30 + [(30 − 25)/(2(30) − 25 − 15)] × 10 = 30 + (5/20) × 10 = 32.5 hours.
Assessing the claim: the mean of 27 is dragged down by the 10 students below 10 hours and the 20 between 10 and 20 hours. The median is 28 and the mode is 32.5, so a student at 29 hours is only 1 hour above the median and below the most crowded band. The phrase 'comfortably above average' is not supported, and the conclusion 'needs no support' is unsafe: the correct comparison is against the median, not the mean.
3. Test the data against the empirical relation 3 × median = mode + 2 × mean, comment on the direction of skew, and advise which single figure the magazine should print.
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3 × median = 3 × 28 = 84. Mode + 2 × mean = 32.5 + 2(27) = 32.5 + 54 = 86.5.
Difference = 2.5, about 3% of 84. The empirical relation is an approximation that holds exactly only for a moderately skewed distribution, so a small gap is expected and is not evidence of an arithmetic error.
Ordering: mean (27) < median (28) < mode (32.5). This ordering shows a negatively skewed distribution with a long tail towards the low hours - a small group of very low study times pulling the mean down.
Recommendation: print the median, 28 hours. It is not affected by the low-hour tail, it matches what a typical student actually does, and it is the figure the parent's comparison should have used. The mean understates typical study time; the mode describes only the most crowded band.