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CRIME RATE

Expert replies
by pappueshwar » Wed Mar 14, 2012 7:57 am
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

OA IS D IS B IRRELEVANT? MY CHOICE WAS B HENCE ASKING
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Source: — Critical Reasoning |

by chris@magoosh » Wed Mar 14, 2012 3:04 pm
The problem with the conclusion is we do not know anything about Meadowbrook's and Parkland's crime rates four years ago. We only know about the increase. So if MB crime's rate was 10 per 1000 and it increased to 16 per 1000, overall crime would still be much lower than that of PL, if four years ago, PL had 100 per 1000. Unless the rates from four years ago are given, the conclusion cannot be made.

Therefore (D).

(B) doesn't work is we are not worried about the total population (and thus how much population has grown). We only care about the crime rate, i.e., the number of crimes per 1,000 residents).
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