AMC8 2011

AMC8 2011 · Q4

AMC8 2011 · Q4. It mainly tests Averages (mean), Casework.

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, mode, and median of these numbers is true?

以下是 Tyler 去年夏天九次出游捕鱼的数量列表：2, 0, 1, 3, 0, 3, 3, 1, 2。关于这些数字的均值、众数和中位数的哪个陈述是正确的？

(A) median < mean < mode 中位数 < 均值 < 众数

(B) mean < mode < median 均值 < 众数 < 中位数

(D) median < mode < mean 中位数 < 众数 < 均值

(E) mode < median < mean 众数 < 中位数 < 均值

Answer

Correct choice: (C)

正确答案：（C）

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

首先，将数字按升序排列。 \[0,0,1,1,2,2,3,3,3\] 均值是 $\frac{0+0+1+1+2+2+3+3+3}{9} = \frac{15}{9}$，中位数是 $2$，众数是 $3$。因为 $\frac{15}{9} < 2 < 3$，答案是 $\boxed{\textbf{(C)}\ \text{mean} < \text{median} < \text{mode}}$

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