The Scholarship Score and the Weighted Average
A school awards a scholarship to the candidate with the highest weighted score. Three candidates apply. Candidate A has subject scores 90, 85, 70 and 60; candidate B has 80, 80, 80 and 80; candidate C has 95, 60, 60, 95. The weights are 40% for the first subject, 30% for the second, 20% for the third and 10% for the fourth. A rule states that no candidate may receive the scholarship if any single subject score is below 65. The committee also applies a tie-break: the higher score in the most heavily weighted subject wins. One committee member argues that B should win because B is 'the most consistent'. Another argues that C should win because C has the highest single scores.
1. Compute each candidate's weighted score and determine who is eligible under the minimum-score rule.
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Weighted score = 0.4·s1 + 0.3·s2 + 0.2·s3 + 0.1·s4. Candidate A: 0.4·90 = 36; 0.3·85 = 25.5; 0.2·70 = 14; 0.1·60 = 6; total 81.5. Candidate B: 0.4·80 = 32; 0.3·80 = 24; 0.2·80 = 16; 0.1·80 = 8; total 80. Candidate C: 0.4·95 = 38; 0.3·60 = 18; 0.2·60 = 12; 0.1·95 = 9.5; total 77.5. Eligibility: A has a 60 in the fourth subject, below 65, so A is ineligible despite the highest score. C has 60 in the second and third subjects, so C is ineligible. B has no score below 65 and is eligible. Therefore B wins - not because of consistency as a vague virtue, but because the minimum-score rule eliminates the higher-scoring candidates.
2. The committee considers waiving the minimum-score rule for A. What would the weighted scores then be, and what principle is at stake?
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If the rule is waived, A's 81.5 beats B's 80 and C's 77.5, so A would win by 1.5 points. The principle at stake is whether a stated rule binds the committee or is advisory. Waiving it after seeing the results is retroactive rule-changing, which undermines fairness: candidates may have made choices based on the published rule. A defensible process either applies the rule as written or, if the rule is judged wrong, changes it for future cycles with notice. The judgement tested is that procedural fairness can outweigh a marginally higher score.
3. Suppose A's fourth score were 65 instead of 60. Recompute and state the outcome, including any tie-break.
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A becomes eligible. New A total: 36 + 25.5 + 14 + 0.1·65 = 36 + 25.5 + 14 + 6.5 = 82. B remains 80. C remains ineligible. A wins with 82 against 80, a margin of 2 points. No tie-break is needed. The tie-break (higher score in the most heavily weighted subject) would only matter if totals were equal; here A also leads on the first subject (90 vs 80), so A would win on the tie-break as well. This shows the result is robust to the tie-break rule.