Problem:
Melanie computes the mean , the median , and the modes of the values that are the dates in the months of . Thus her data consist of 12 1s, 12 2s,..., 12 28s, 11 29s, 11 30s, and 7 31s. Let be the median of the modes. Which of the following statements is true?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
First of all, obviously has to be smaller than , since when calculating , we must take into account the s, s, and . So we can eliminate choices and . Since there are total entries, the median, , must be the rd one, at which point we note that is , so has to be the median (because is between and ). Now, the mean, , must be smaller than , since there are many fewer , s, and s. is less than , because when calculating , we would include , and . Thus the answer is
(E) .
OR
As in Solution , we find that the median is . Then, looking at the modes , we realize that even if we were to have of each, their median would remain the same, being . As for the mean, we note that the mean of the first is simply the same as the median of them, which is . Hence, since we in fact have 's, 's, and 's, the mean has to be higher than . On the other hand, since there are fewer 's, 's, and 's than the rest of the numbers, the mean has to be lower than 16 (the median). By comparing these values, the answer is .
We can solve this problem simply by carefully calculating each of the values, which turn out to be , , and . Thus the answer is .
The problems on this page are the property of the MAA's American Mathematics Competitions