Comparing Quiz Scores: Mean, Median, and Outlier
Two Algebra I classes take the same 100-point quiz. Class A's ten scores are 30, 88, 90, 92, 94, 96, 98, 99, 100, and 100. Class B's ten scores are 80, 82, 84, 86, 88, 90, 92, 94, 96, and 98. A student compares the means and says, 'Class B did better because its mean is 89, which is higher than Class A's mean of 88.7.' A second student looks at the medians and says Class A did better because its median is 95 compared with 89 for Class B. The teacher wants the class to decide which summary is more appropriate for describing typical performance, to identify the effect of the low score in Class A, and to recommend what the school report should include. Both classes have the same number of students and took the quiz under similar conditions.
1. Verify the means and medians. Is the first student's conclusion justified by the mean alone? Explain.
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Class A sum = 887, mean = 88.7. Sorted, the 5th and 6th scores are 94 and 96, so median = 95. Class B sum = 890, mean = 89.0; its 5th and 6th scores are 88 and 90, so median = 89. The mean difference is only 0.3 point in favor of B. But A's median is 6 points higher, and nine of A's ten scores are at or above 88, while B's top score is 98 and its bottom is 80. The mean alone does not justify saying B 'did better'; it is slightly higher because A's one very low score pulls A's mean down. The conclusion depends on the summary chosen.
2. Treat the score of 30 in Class A as an outlier. Compute its effect on Class A's mean and median. Then recompute the mean if that score were 80 instead.
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The 30 is far below the other nine scores (88-100). Mean with 30 is 88.7. If 30 is replaced by 80, the sum increases by 50, from 887 to 937, so new mean = 93.7; the mean rises by 5 points. The median is unchanged at 95: the sorted data become 80, 88, 90, 92, 94, 96, 98, 99, 100, 100, and the middle pair is still 94 and 96. This shows the median is resistant to the outlier while the mean is not.
3. Recommend how the school should report the comparison. Include at least one measure of center and one measure of spread, and explain why your choice fits the goal of describing typical performance.
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Because the goal is typical performance and Class A has an outlier, report the median as the primary measure of center, but include the mean and note the outlier. Also report a measure of spread such as the interquartile range or range. Class A: median 95; Q1 = 90, Q3 = 99, IQR = 9, range = 70. Class B: median 89; Q1 = 84, Q3 = 94, IQR = 10, range = 18. This shows that A's typical student scored higher and its middle 50% is about as consistent as B's, but A's full range is much larger because of the one low score. Ranking by mean alone would be misleading.