1996 AHSME Problems/Problem 4

Problem

Six numbers from a list of nine integers are $7,8,3,5,9$ and $5$ . The largest possible value of the median of all nine numbers in this list is

$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

Solution

First, put the six numbers we have in order, since we are concerned with the median: $3, 5, 5, 7, 8, 9$ .

We have three more numbers to insert into the list, and the median will be the $5^{th}$ highest (and $5^{th}$ lowest) number on the list. If we top-load the list by making all three of the numbers greater than $9$ , 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 $8$ , giving an answer of $\boxed{D}$ .

(In fact, as long as the three new integers are greater than $8$ , the median will be $8$ .)

This illustrates one important fact about medians: no matter how high the three "outlier" numbers are, the median will never be greater than $8$ . 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.

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1996 AHSME Problems/Problem 4

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