# 2011 AMC 8 Problems/Problem 4

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

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