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Entrance scores of three institutes
Short answer
Institute A: mean 865.56, SD 93.42, CV 10.79%. B: mean 919.67, SD 139.79, CV 15.20%. C: mean 856.89, SD 135.34, CV 15.79%. Institute B is best (highest mean). Institute A is most consistent (lowest CV).
The entrance score of three engineering institutes are as follows:
| Institute | Entrance scores |
|---|---|
| A | 740 800 830 840 860 890 830 930 1070 |
| B | 655 775 825 978 989 1025 950 980 1100 |
| C | 850 825 749 870 565 978 925 950 1000 |
| a) Calculate mean and standard deviation (and coefficient of variation) for institute A, B and C. | |
| b) Which institute is good? | |
| c) Which institute is consistent? |
Use mean x̄ = Σx/n, sample SD s = √[(Σx² − n·x̄²)/(n − 1)] and CV = (s/x̄) × 100. Each institute has n = 9 scores. A higher mean means better performance. A lower CV means more consistent scores.
a) Mean, SD and CV
| Institute | Σx | Σx² | Mean | Sxx | s | CV |
|---|---|---|---|---|---|---|
| A | 7790 | 6,812,500 | 865.56 | 69,822.2 | 93.42 | 10.79 % |
| B | 8277 | 7,768,405 | 919.67 | 156,324.0 | 139.79 | 15.20 % |
| C | 7712 | 6,754,860 | 856.89 | 146,532.9 | 135.34 | 15.79 % |
Working for A:
x̄ = 7790 / 9 = 865.56
Sxx = 6,812,500 − 9 × (865.56)² = 69,822.2
s² = 69,822.2 / 8 = 8727.8
s = √8727.8 = 93.42
CV = 93.42 / 865.56 × 100 = 10.79 %
Institutes B and C are worked the same way (s² = 19,540.5 and 18,316.6).
b) Which institute is good?
Institute B is the best because it has the highest mean entrance score (919.67).
c) Which institute is consistent?
Institute A is the most consistent because it has the smallest CV (10.79 %), against 15.20 % for B and 15.79 % for C.
If the population SD (divisor n) is used instead, the values are 88.08, 131.79 and 127.60 and the conclusions stay the same.
✓ Answer checked by Claude Sonnet 5.5
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