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 "2011 AMC 8 Problems/Problem 4"
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 +
==Problem==
 
Here is a list of the numbers of fish that Tyler caught in nine outings last summer: <cmath>2,0,1,3,0,3,3,1,2.</cmath> Which statement about the mean, median, and mode is true?
 
Here is a list of the numbers of fish that Tyler caught in nine outings last summer: <cmath>2,0,1,3,0,3,3,1,2.</cmath> Which statement about the mean, median, and mode is true?
  
Line 10: Line 11:
  
 
The mean is <math>\frac{0+0+1+1+2+2+3+3+3}{9} = \frac{15}{9},</math> the median is <math>2,</math> and the mode is <math>3.</math> Because, <math>\frac{15}{9} < 2 < 3,</math> the answer is <math>\boxed{\textbf{(C)}\ \text{mean} < \text{median} < \text{mode}}</math>
 
The mean is <math>\frac{0+0+1+1+2+2+3+3+3}{9} = \frac{15}{9},</math> the median is <math>2,</math> and the mode is <math>3.</math> Because, <math>\frac{15}{9} < 2 < 3,</math> the answer is <math>\boxed{\textbf{(C)}\ \text{mean} < \text{median} < \text{mode}}</math>
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/T9V6PXgaheI
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=3|num-a=5}}
 
{{AMC8 box|year=2011|num-b=3|num-a=5}}
 +
{{MAA Notice}}

Latest revision as of 10:02, 18 November 2024

Problem

Here is a list of the numbers of fish that Tyler caught in nine outings last summer: \[2,0,1,3,0,3,3,1,2.\] Which statement about the mean, median, and mode is true?

$\textbf{(A)}\ \text{median} < \text{mean} < \text{mode} \qquad \textbf{(B)}\ \text{mean} < \text{mode} < \text{median} \\ \\ \textbf{(C)}\ \text{mean} < \text{median} < \text{mode} \qquad \textbf{(D)}\ \text{median} < \text{mode} < \text{mean} \\ \\ \textbf{(E)}\ \text{mode} < \text{median} < \text{mean}$

Solution

First, put the numbers in increasing order.

\[0,0,1,1,2,2,3,3,3\]

The mean is $\frac{0+0+1+1+2+2+3+3+3}{9} = \frac{15}{9},$ the median is $2,$ and the mode is $3.$ Because, $\frac{15}{9} < 2 < 3,$ the answer is $\boxed{\textbf{(C)}\ \text{mean} < \text{median} < \text{mode}}$

Video Solution by WhyMath

https://youtu.be/T9V6PXgaheI

See Also

2011 AMC 8 (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
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_4&oldid=234878"