Difference between revisions of "1996 AHSME Problems/Problem 4"
Talkinaway (talk | contribs) (Created page with "==See also== {{AHSME box|year=1996|num-b=3|num-a=5}}") |
Talkinaway (talk | contribs) |
||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | First, put the six numbers we have in order, since we are concerned with the median: <math>3, 5, 5, 7, 8, 9</math>. | ||
+ | |||
+ | We have three more numbers to insert into the list, and the median will be the <math>5^{th}</math> highest (and <math>5^{th}</math> lowest) number on the list. If we top-load the list by making all three of the numbers greater than <math>9</math>, 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 <math>8</math>, giving an answer of <math>\boxed{D}</math>. | ||
+ | |||
+ | (In fact, as long as the three new integers are greater than <math>8</math>, the median will be <math>8</math>.) | ||
+ | |||
+ | This illustrates one important fact about medians: no matter how high the three "outlier" numbers are, the median will never be greater than <math>8</math>. 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== | ==See also== | ||
{{AHSME box|year=1996|num-b=3|num-a=5}} | {{AHSME box|year=1996|num-b=3|num-a=5}} |
Revision as of 20:26, 18 August 2011
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 |