gun death stats

[Bulzeye]

This is more political than technical, but I though it would be good to share here where the most gun owners could see it. Regardless of your position on the Iracwar, I feel this puts things in perspective, if indeed the numbers are accurate.

[blockquote]If you consider that there has been an average of 160,000 troops in the Iraq theatre of operations during the last 22 months, and a total of 2,112 deaths, that gives a firearm death rate of 60 per 100,000 soldiers.

The firearm death rate in Washington D.C. is 80.6 per 100,000 for the same period.

That means that you are about 25% more likely to be shot and killed in the U.S. Capitol, which has some of the strictest gun control laws in the nation, than you are in Iraq.

Conclusion: The U.S. should pull out of Washington[/blockquote]

[Californiadoctor]

Let us redo the math on this one and this time do it correctly;

160,000 divided by 100,000 equals 1.6. Now divide the number of fatalities, 2112 by 1.6.

2112/1.6 = 1,320fatalitiesper 100,000 soldiers.

Now convert that intoan annual rate by dividing 22 by 12 months.

12/22 = 0.5454

Now calculate the annual fatality rate

1,320 x 0.5454 = 720 deaths per 100,000 soldiers per year.

This is almost 10 times the D.C. rate and D.C. is a pretty dangerous place, I have lived there.

Caldoc

[tj_cubin]

yeah... those original numbers are pretty messed up

californiadoctor, you wouldnt have to find the annual rate because according to this highly reliable source, the rate for DC was for the same period, so you would be comparing the 1320 per 100,000 in Iraq to 80 per 100,000 in DC. both those numbers are for a 22 month period.

this means that being killed in iraq is 16.5 times MORE likely than in DC.

[biscuit jake]

Actually you loose more people in combat due to accidents than direct fire.

[#40Fan]

ORIGINAL: Californiadoctor

Let us redo the math on this one and this time do it correctly;

160,000 divided by 100,000 equals 1.6. Now divide the number of fatalities, 2112 by 1.6.

2112/1.6 = 1,320fatalitiesper 100,000 soldiers.

Now convert that intoan annual rate by dividing 22 by 12 months.

12/22 = 0.5454

Now calculate the annual fatality rate

1,320 x 0.5454 = 720 deaths per 100,000 soldiers per year. Correct up to here.

This is almost 10 times the D.C. rate and D.C. is a pretty dangerous place, I have lived there.

Caldoc

If there is 720/100,000 in Iraq and only 80.6/100,000, there is a 89% greater chance of being killed in Iraq.

[Ideaman]

100% of the persons killed inIraq are heros, if the cause is just or not they died trying to do the right thing.

A large % of the persons killed in DC died because nobody there tried to do anything.

[Californiadoctor]

#40Fan, I think you are off by one decimal place.

(720/100,000)/(80.6/100,000) = 8.93 or 893% greater risk of death in Iraq


Caldoc

[onebadf250]

Not on the exact subject...but I read in a national geographic a few weeks back while at the doc's office that you are more likely to be killed riding your bicycle than you are to being killed by a gun. I thought that was pretty funny. Maybe we should stiffin up our bicycling laws.

[#40Fan]

The way that I did the math was to divide 80.9 into 720 which gave me .112361 or 11.2%. 100%-11.2% is about 89%.

To confirm it, you can divide 80.9 by 11.2361 and then multiply that by 100.