1996 AHSME Problems/Problem 4

Problem

Six numbers from a list of nine integers are and . The largest possible value of the median of all nine numbers in this list is

$\text{(A)}\ 5\qquad\text{(B)}\6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ (Error compiling LaTeX. Unknown error_msg)

Solution

First, put the six numbers we have in order, since we are concerned with the median: .

We have three more numbers to insert into the list, and the median will be the highest (and lowest) number on the list. If we top-load the list by making all three of the numbers greater than , the median will be the highest it can possibly be. Thus, the maximum median is the fifth piece of data in the list, which is , giving an answer of .

(In fact, as long as the three new integers are greater than , the median will be .)

This illustrates one important fact about medians: no matter how high the three "outlier" numbers are, the median will never be greater than . The arithmetic mean is, generally speaking, more sensitive to such outliers, while the median is resistant to a small number of data that are either too high or too low.