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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.

See also

1996 AHSME (ProblemsAnswer KeyResources)
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
All AHSME Problems and Solutions

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