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Difference between revisions of "2011 AMC 8 Problems/Problem 4"

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==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?
  
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<math>\text{(A)} \text{median} < \text{mean} < \text{mode} \qquad \text{(B)} \text{mean} < \text{mode} < \text{median} \\ \\ \text{(C)} \text{mean} < \text{median} < \text{mode} \qquad \text{(D)} \text{median} < \text{mode} < \text{mean} \\ \\ \text{(E)} \text{mode} < \text{median} < \text{mean}</math>
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<math>\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}</math>
  
 
==Solution==
 
==Solution==
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First, put the numbers in increasing order.
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<cmath>0,0,1,1,2,2,3,3,3</cmath>
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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>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=3|num-a=5}}
 
{{AMC8 box|year=2011|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 01:58, 5 July 2013

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 $\boxed{\textbf{(C)}\ \text{mean} < \text{median} < \text{mode}}$

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

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