AMC12 2005 B

AMC12 2005 B · Q9

AMC12 2005 B · Q9. It mainly tests Averages (mean), Logic puzzles.

On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?

在某次数学考试中，$10\%$ 的学生得 $70$ 分，$25\%$ 得 $80$ 分，$20\%$ 得 $85$ 分，$15\%$ 得 $90$ 分，其余得 $95$ 分。这次考试的平均分与中位数之差是多少？

(A) 0 0

(B) 1 1

(D) 4 4

(E) 5 5

Answer

Correct choice: (B)

正确答案：（B）

Solution

To begin, we see that the remaining $30\%$ of the students got $95$ points. Assume that there are $20$ students; we see that $2$ students got $70$ points, $5$ students got $80$ points, $4$ students got $85$ points, $3$ students got $90$ points, and $6$ students got $95$ points. The median is $85$, since the $10^{\text{th}}$ and $11^{\text{th}}$ terms are both $85$. The mean is $\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86$. The difference between the mean and median, therefore, is $\boxed{\textbf{(B)}\ 1}$.

首先可知剩下的 $30\%$ 的学生得了 $95$ 分。假设共有 $20$ 名学生，则有 $2$ 名学生得 $70$ 分，$5$ 名学生得 $80$ 分，$4$ 名学生得 $85$ 分，$3$ 名学生得 $90$ 分，$6$ 名学生得 $95$ 分。中位数为 $85$，因为第 $10^{\text{th}}$ 与第 $11^{\text{th}}$ 个分数都是 $85$。平均数为 $\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86$。因此平均数与中位数之差为 $\boxed{\textbf{(B)}\ 1}$。

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