2011年全美数学竞赛（AMC）竞赛
如下是 Tyler 去年夏天外出捕鱼九次的数量：$$2$$，$$0$$，$$1$$，$$3$$，$$0$$，$$3$$，$$3$$，$$1$$，$$2$$．下列关于平均值，中位数，众数的描述，哪个是正确的？
美国AMC8->Knowledge Point->Counting, Probability and Statistics->Average Problems->Complex Average Problems 课内体系->知识点->统计与概率->数据的分析->平均数/加权平均数
如下是 Tyler 去年夏天外出捕鱼九次的数量：$$2$$，$$0$$，$$1$$，$$3$$，$$0$$，$$3$$，$$3$$，$$1$$，$$2$$．下列关于均值，中位数，众数的描述哪个正确？  A~ 中位数$$\textless$$均值$$\textless$$众数  B~ 均值$$\textless$$众数$$\textless$$中位数  C~ 均值数$$\textless$$中位数$$\textless$$众数  D~ 中位数$$\textless$$众数$$\textless$$均值  E~ 众数$$\textless$$中位数$$\textless$$均值  首先，把数字按增加的排列．  $$0$$，$$0$$，$$1$$，$$1$$，$$2$$，$$2$$，$$3$$，$$3$$，$$3$$  故$$\frac{0+0+1+1+2+2+3+3+3}{9}=\frac{15}{9}$$，中位数是$$2$$，众数是$$3$$．因为，$$\frac{15}{9} ~\textless{} ~2 ~\textless{} ~3$$，所以答案是mean$$ ~\textless{} ~$$median$$ ~\textless{} ~$$mode，即平均数$$ ~\textless{} ~$$中位数$$ ~\textless{} ~$$众数．  故选$$\text{C}$$．  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, $$\frac{15}{9}\textless2\textless3$$, the answer is mean $$\textless{}$$ median $$\textless{}$$ mode. 