Difference between revisions of "2011 AMC 8 Problems/Problem 4"
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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> | 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}} | {{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: ![\[2,0,1,3,0,3,3,1,2.\]](http://latex.artofproblemsolving.com/6/b/7/6b7acd21d15b3e6744a3b750c1c6fbd688b75635.png)
Which statement about the mean, median, and mode is true?
Solution
First, put the numbers in increasing order.
![\[0,0,1,1,2,2,3,3,3\]](http://latex.artofproblemsolving.com/3/d/7/3d76e96d7a46a5eacd2956de74936b866cd03a45.png)
The mean is 
the median is 
and the mode is 
Because, 
the answer is 
Video Solution by WhyMath
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. 
