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== | ||
| Line 17: | Line 17: | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=3|num-a=5}} | {{AHSME box|year=1996|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:27, 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
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. 
