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[求助]一道天山flaw题

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楼主
发表于 2008-3-21 22:02:00 | 只看该作者

[求助]一道天山flaw题

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 Meadowbrook and Parkdale 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 Meadowbrooks’ expenditures for crime prevention over the past four years compare to Parkdale’s expenditures.

答案是D,可是我怎么选也选不出D啊

沙发
发表于 2008-3-21 22:32:00 | 只看该作者
当然是D啊,因为增长的是百分比,不知道基数的话,就不知道增长了多少数量的crime,也就不知道有多少人会成为victim
板凳
发表于 2008-3-22 08:57:00 | 只看该作者
假设4年前M town只有5例犯罪,现在增长60%成了8例。而P town原来有10000例,现在11000例。。。。。。所以flaw在D。more likely to become victims of violent crime说的是实际的犯罪率,而增长只是一个相对概念。
地板
 楼主| 发表于 2008-3-22 16:40:00 | 只看该作者

看了二楼的举例,明白了

可是不用考虑人口增长问题吗

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