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The violent crime rate (number of violent crimes per 1,000 residents) in Meadowbrook is 60 percent higher now than it was four years ago. The corresponding increase for Parkdale is only 10 percent. These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.

The argument above is flawed because it fails to take into account

正确答案: D

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T-4-Q20旧题新问 看了几个帖子还是迷茫 求大神解惑

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楼主
发表于 2014-10-11 08:49:40 | 只看该作者 回帖奖励 |倒序浏览 |阅读模式
The violent crime rate (number of violent crimes per 1,000 residents) in Meadowbrook is 60 percent higher now than it was four years ago.  The corresponding increase for Parkdale is only 10 percent.  These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.
The argument above is flawed because it fails to take into account

A. changes in the population density of both Parkdale and Meadowbrook over the past four years
B. how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale
C. the ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale
D. the violent crime rates in Meadowbrook and Parkdale four years ago
E. how Meadowbrook's expenditures for crime prevention over the past four years compare to Parkdale's expenditures

我选的是B,我认为既然犯罪增长率说的是每1000个居民中所占犯罪人的比率那么人口基数就会对结果产生影响啊。比如说M地过去的四年里面增长了1000个人,那么自然会有相应的犯罪的人的增加,那么必然造成有更多的人受害。===
绕不出来了 请大神踢醒我~~谢谢
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沙发
 楼主| 发表于 2014-10-11 22:27:14 | 只看该作者
DDDDDDDDDDD
板凳
发表于 2014-10-13 09:41:00 | 只看该作者
我觉得crime rate是一个比率(就好像出生率、死亡率一样),跟人口数的关系就是犯罪数比上人口数等于犯罪率。但是比较两个地方犯罪可能性的大小要用rate而非人数,所以跟人口增长数没有直接和绝对的关系,因为还涉及犯罪人数,所以最终的rate才是唯一衡量标准。

另外我认为本题的关键是考基数和增量的区别,比如M之前只有5%,增加60%才65%,P之前就有70%,增加10%就80%。所以怎么比较都是D更好。
地板
 楼主| 发表于 2014-10-13 11:55:07 | 只看该作者
patricialiu 发表于 2014-10-13 09:41
我觉得crime rate是一个比率(就好像出生率、死亡率一样),跟人口数的关系就是犯罪数比上人口数等于犯罪率 ...

我想说最后的这句话These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.的意思是说居民是否更加容易成为受害者也就是说在M地和在P地的犯罪人的多少,如果罪犯多了,那么居民就更容易受害。我的理解是这样的,虽然M地犯罪率上升的多但是如果M地的人口增长的少,那么罪犯人数=总人口*犯罪率,也不会增长。所以进行了削弱。

我可能理解的有点偏了,但是还是有点纠结,望你指正。谢谢。
5#
发表于 2014-10-13 13:41:48 | 只看该作者
sunyuwei 发表于 2014-10-13 11:55
我想说最后的这句话These figures support the conclusion that residents of Meadowbrook are more like ...

举个例子,如果一个城市100万人,有10个人是犯罪的,则每个人受害的概率是10/100万,如果人口是200万,犯罪人20个,则每个人受害的概率是20/200万。衡量可能性要用概率或者比率等相对值,不能用绝对数字。
6#
 楼主| 发表于 2014-10-13 15:09:47 | 只看该作者
patricialiu 发表于 2014-10-13 13:41
举个例子,如果一个城市100万人,有10个人是犯罪的,则每个人受害的概率是10/100万,如果人口是200万,犯 ...

我明白了~你的意思是说最后一句的结论These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.比较的受害可能性也是一个比率问题,所以这部分也是要按照比率折算的~

谢谢指正
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