Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1996 AHSME Problems/Problem 4"

(Created page with "==See also== {{AHSME box|year=1996|num-b=3|num-a=5}}")
 
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 19:26, 18 August 2011

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

$\text{(A)}\ 5\qquad\text{(B)}\6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ (Error compiling LaTeX. Unknown error_msg)

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 (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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1996_AHSME_Problems/Problem_4&oldid=41632"