2019 AMC 8 第 10 题

先试着解答 2019 AMC 8 第 10 题,然后核对你的答案与视频与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2019 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

10.

图中显示了上周每个工作日参加足球训练的学生人数。教练计算了平均数和中位数后,发现星期三实际有 2121 人参加。修正后,平均数和中位数分别如何变化?
The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually 2121 participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?
平均数增加 11,中位数不变。
The mean increases by 11 and the median does not change.
平均数增加 11,中位数增加 11。
The mean increases by 11 and the median increases by 1.1.
平均数增加 11,中位数增加 55。
The mean increases by 11 and the median increases by 5.5.
平均数增加 55,中位数增加 11。
The mean increases by 55 and the median increases by 1.1.
平均数增加 55,中位数增加 55。
The mean increases by 55 and the median increases by 5.5.
答案:B
知识点:平均数中位数(数据)
小提示:
这次修正使总人数增加 55
The correction adds 55 students total
大提示:
比较把星期三的数替换前后的有序列表
Compare the ordered lists before and after replacing Wednesday’s value
视频讲解:
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文字解答:
修正后多出 55 名参与者,所以平均数增加 5÷5=15 \div 5 = 1。
原来的中位数是 2020。把 1616 换成 2121 后,中位数变为 2121。
所以中位数也增加 11。
正确答案是 B。
There are 55 more participants, which means the mean is increased by 5÷5=1.5 \div 5 = 1.
Right now, we can see that the median is 20.20. If 1616 gets replaced with 21,21, then the median becomes 21.21.
This shows that the median is also increased by 1.1.
Thus, the correct answer is B.
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