Difference between revisions of "1996 AHSME Problems/Problem 4"
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Six numbers from a list of nine integers are <math>7,8,3,5,9</math> and <math>5</math>. The largest possible value of the median of all nine numbers in this list is | Six numbers from a list of nine integers are <math>7,8,3,5,9</math> and <math>5</math>. The largest possible value of the median of all nine numbers in this list is | ||
| − | <math> \text{(A)}\ 5\qquad\text{(B)}\6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9 </math> | + | <math> \text{(A)}\ 5\qquad\text{(B)}\6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\9 </math> |
==Solution== | ==Solution== | ||
Revision as of 12:23, 24 October 2017
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.
See also
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
