Comparing Two Class Test Results
Two Grade 8 classes took the same 20-point mathematics quiz. Class A (28 students) had a mean of 14.2 and a median of 15. Class B (22 students) had a mean of 13.6 and a median of 12. A teacher notices Class A's mean is below its median, while Class B's mean is above its median. The combined group of 50 students is being reported to the school board, and a parent claims 'Class A clearly did better because its mean is higher.'
1. Compute the combined mean of all 50 students and explain why it is not simply the average of 14.2 and 13.6.
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Class A total = 28 × 14.2 = 397.6. Class B total = 22 × 13.6 = 299.2. Combined total = 696.8 over 50 students → mean = 13.936. The simple average (14.2 + 13.6)/2 = 13.9 is close but wrong because the classes have different sizes; the combined mean must be weighted by enrollment.
2. Interpret the relationship between mean and median in each class. What does it suggest about the shape of each distribution?
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Class A: mean (14.2) < median (15) suggests a left-skewed distribution - a few low scores pull the mean down. Class B: mean (13.6) > median (12) suggests right-skewness - a few high scores pull the mean up. So Class A likely has a cluster of strong scores with some low outliers, while Class B has many modest scores with a few high performers.
3. Evaluate the parent's claim that Class A 'clearly did better.' Is the evidence sufficient?
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The claim is only partly supported. Class A's mean is higher (14.2 vs 13.6) and its median is higher (15 vs 12), so on both measures Class A tends to score higher. However, 'clearly did better' overstates it: the difference in means is 0.6 points on a 20-point quiz, and we lack spread (range, standard deviation) and distribution data. The medians differ more (3 points), suggesting Class A's typical student did better, but a few high scores in Class B could narrow the gap. A defensible statement: Class A performed somewhat better on average and at the median, but the margin is modest and would benefit from spread data.