Compute and interpret mean, median, mode, and range.
Free, timed SHSAT practice sets on measures of center & spread— every question includes a full worked explanation, and there's no account required.
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Each measure answers a different question, so read which one is asked.
Common traps: dividing by the wrong count, finding the "middle" of an unsorted list, adding the two middle values but forgetting to divide by 2, answering the mode with how many times it appears (2) instead of the value itself (7), and subtracting the first and last numbers as listed instead of the true maximum and minimum.
Mean = sum \(\div\) count = 96 \(\div\) 8 = 12. (Reporting the median (the middle of 2, 4, 6, 9, 14, 17, 21, 23) gives 11.5. Dividing by 7 instead of 8 (\(96 \div 7\)) gives about 13.71. Averaging only the largest and smallest scores (\((23 + 2) \div 2\)) gives 12.5.) The mean score is 12.
Sort the ages: 25, 28, 34, 36, 48, 51. With 6 values (even), the median is the average of the two middle values: (34 + 36) \(\div\) 2 = 35. (Reporting the mean (\(222 \div 6\)) gives 37. Adding the two middle values but not dividing by 2 (\(34 + 36\)) gives 70. Averaging the two middle values of the unsorted list (\(51\) and \(48\)) gives 49.5.) The median age is 35.
Range = maximum \(-\) minimum = 44 \(-\) 13 = 31. (Reporting the mean (\(135 \div 5\)) gives 27. Subtracting the first and last values as listed (\(|41 - 44|\)) instead of the largest and smallest gives 3. Using the second-smallest value, 14, as the minimum gives 30.) The range is 31\(^\circ\)F.