1990 AMC 8 第 20 题

先试着解答 1990 AMC 8 第 20 题,然后核对你的答案与精心整理的解答,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 1990 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。

20.

1,0001{,}000 个家庭的年收入从 $8200\$8200 到 $98,000\$98{,}000 不等。由于错误,最高收入在电脑中被输入为 $980,000\$980{,}000。错误数据的平均数与实际数据的平均数相差多少?
The annual incomes of 1,0001{,}000 families range from $8200\$8200 to $98,000.\$98{,}000. In error, the largest income was entered on the computer as $980,000.\$980{,}000. The difference between the mean of the incorrect data and the mean of the actual data is
$882\$882
$980\$980
$1078\$1078
$482,000\$482{,}000
$882,000\$882{,}000
答案:A
知识点:平均数
小提示:
只有一个数据变了,所以只有总和改变;数据个数仍是 10001000。
Only one value changed, so only the total sum changes; the count stays at 10001000
大提示:
先求这个错误输入让总和多了多少,再除以 10001000。
Find how much the one wrong entry inflated the total, then divide that by 10001000
解答:
只有一个数据被改错。错误总和比实际总和多 $980,000−$98,000=$882,000\$980{,}000 - \$98{,}000 = \$882{,}000。
这多出的金额分摊到 10001000 个家庭上,平均数相差 $882,0001000=$882\dfrac{\$882{,}000}{1000} = \$882。
所以正确答案是 A。
Only one entry changed. The incorrect total exceeds the actual total by $980,000−$98,000=$882,000.\$980{,}000 - \$98{,}000 = \$882{,}000.
Since this extra amount is spread over 10001000 families, the means differ by $882,0001000=$882.\dfrac{\$882{,}000}{1000} = \$882.
Thus, the correct answer is A .
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