Two Sections with Different Means and SDs
Two Class XI sections in a higher secondary school have 40 students each and took the same 100-mark statistics test. Section A's mean is 62 with standard deviation 8; Section B's mean is 50 with standard deviation 5. The head of department says Section B is more consistent because its SD is smaller, so the remedial-teaching fund should go to Section A. Another teacher says Section A is clearly the better-performing section and needs no help. The principal must decide where to spend a single remedial fund and also wants a fair comparison for the school magazine. Separately, revaluation adds 5 marks to every student in Section A. A parent then argues that because Section B's SD is only 5, Section B must be the better section overall. The school also has to decide whether to compare the sections using absolute dispersion or relative dispersion, since their means are not equal.
1. Compute the coefficient of variation for each section and interpret the comparison. Does Section B's smaller SD make it 'better'?
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CV_A = (8/62) × 100 = 12.90%; CV_B = (5/50) × 100 = 10.00%. Section B is more consistent relative to its own mean, but it also has a much lower average. SD alone says B's marks are less spread in absolute marks; CV says B's spread is smaller per unit of mean. Neither shows that B is 'better' in achievement. For performance, A's mean 62 > B's 50. For consistency, B is relatively more consistent. A fair magazine statement should mention both.
2. After 5 marks are added to every Section A mark, what happens to Section A's mean, SD and CV? Does the consistency ranking change?
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Adding 5 to every Section A mark shifts the mean to 62 + 5 = 67. Each deviation x_i − mean is unchanged, so SD remains 8. New CV_A = (8/67) × 100 = 11.94%. Section B's CV remains 10.00%, so B is still relatively more consistent, even though A's average rose. The absolute SD of A is still 8, so B's absolute SD is still smaller. Adding a constant changes location, not dispersion.
3. A parent argues that Section B is better because its SD is only 5. Give a reasoned response using absolute and relative dispersion.
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SD is an absolute measure and depends on the units and the mean; it should not be used alone to compare groups with different means. CV = (SD/mean) × 100 puts dispersion on a relative scale. Here B has smaller SD and smaller CV, so B is less variable both absolutely and relatively, but A has the higher average. If the decision concerns consistency, B is preferable; if it concerns overall marks, A is preferable. The remedial fund should be targeted by the school's stated criterion, not by SD alone.