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

Revision as of 00:58, 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 $\boxed{\textbf{(C)}\ \text{mean} < \text{median} < \text{mode}}$

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AJHSME/AMC 8 Problems and Solutions

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

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 ==See Also==
 {{AMC8 box|year=2011|num-b=3|num-a=5}}
+{{MAA Notice}}

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

Revision as of 00:58, 5 July 2013

Problem

Solution

See Also