2022 AMC 8 第 19 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

19.

Ramos 老师给 2020 名学生进行了一次测试。下面的点图显示了成绩分布。

后来 Ramos 老师发现某题评分有误。他重新评分,给一些学生额外加了 55 分,使得测试成绩的中位数提高到 8585。至少有多少名学生得到了额外分数?

(注意,若 2020 个成绩按从小到大排列,中位数等于中间 22 个成绩的平均数。)

Mr. Ramos gave a test to his class of 2020 students. The dot plot below shows the distribution of test scores.

Later Mr. Ramos discovered that there was a scoring error on one of the questions. He regraded the tests, awarding some of the students 55 extra points, which increased the median test score to 85.85. What is the minimum number of students who received extra points?

(Note that the median test score equals the average of the 22 scores in the middle if the 2020 test scores are arranged in increasing order.)

22

33

44

55

66

答案:C
知识点:中位数(数据)数据与图表解读
难度评级:1370
解答:

所有成绩都是 55 的倍数,加上 55 分后仍然如此。如果第 1010 和第 1111 个成绩的平均数是 8585,但两者不都是 8585,那么第 1111 个成绩至少为 9090。这就要求至少有 1010 个成绩不低于 9090

可是重新评分后,只有原来不低于 8585 的成绩才可能达到 9090。点图中这样的成绩只有 77 个,所以中间两个成绩必须都是 8585

点图中原来有 77 个不低于 8585 的成绩。要使第 1010 和第 1111 个成绩都为 8585,至少要有 1111 个成绩不低于 8585。把四个 8080 分提高到 8585 分既是必要的,也是充分的。

所以正确答案是 C

All scores are multiples of 5,5, and adding 55 preserves that fact. If the 1010th and 1111th scores average 8585 but are not both 85,85, then the 1111th score must be at least 90.90. That would require at least 1010 scores of 9090 or more.

However, even after regrading, only the original scores of 8585 or more can reach 90.90. The plot has only 77 such scores, so the two middle scores must both be 85.85.

The plot initially has 77 scores of at least 85.85. To make the 1010th and 1111th scores both 85,85, there must be at least 1111 scores of at least 85.85. Raising four of the 8080's to 8585 is both necessary and sufficient.

Thus, the correct answer is C.

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