Normal Distribution Cut-off for a Scholarship
Scores on a selection test are approximately normally distributed with mean 62 and standard deviation 11. A scholarship is awarded to the top 8% of candidates. A second, smaller cohort sits a harder version with mean 50 and standard deviation 14, and the same top 8% rule applies. A coordinator wants a single raw score that would qualify a candidate in the first cohort, and then asks whether a score of 78 in the second cohort would qualify, and whether comparing raw scores across cohorts is fair.
1. Find the raw score marking the top 8% in the first cohort (mean 62, SD 11).
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Top 8% means the 92nd percentile. Find z with P(Z < z) = 0.92. From standard normal tables, z ≈ 1.405 (since P(Z<1.40)=0.9192 and P(Z<1.41)=0.9207). Raw score x = μ + zσ = 62 + 1.405(11) = 62 + 15.46 ≈ 77.5. So a candidate needs about 77.5, i.e. 78 on a whole-number scale, to be in the top 8%. The judgement step is converting a tail probability to a z-value and then to the raw scale, not just quoting a z-score.
2. Determine the top-8% cut-off in the second cohort and decide whether 78 qualifies there. Comment on fairness of comparing raw scores.
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Second cohort: x = 50 + 1.405(14) = 50 + 19.67 ≈ 69.7, so cut-off ≈ 70. A score of 78 in the second cohort: z = (78 − 50)/14 = 28/14 = 2.0, which corresponds to P(Z<2.0)=0.9772, so top 2.28% - comfortably inside the top 8%. So 78 qualifies in the second cohort. Fairness: raw scores are not comparable because the distributions differ in mean and spread. A raw 78 is top 2.3% in cohort 2 but only around the 92nd percentile in cohort 1. Comparing z-scores or percentiles is fairer than comparing raw marks; the coordinator should standardise before ranking across cohorts.