Milwaukee, Wisconsin | Bribery Statistics
Milwaukee is a
city located in Wisconsin, and has a population of 597,426.
It has a rate of
0.17 (per 100,000 people) for
Bribery, which is
lower than the national average. This means that your chances of becoming a victim of Bribery in Milwaukee is
1 in 588235, if you reside there for a year.
Below are crime statistics for Bribery, according to the FBI's National Incident Based Reporting System (NIBRS), 2010-2011. For more general crime rates for even more cities, counties, and regions across the US, see
Crime Rates by City.
Bribery Statistics |
| Statistic |
Milwaukee |
USA |
| Bribery Rate |
0.17 (per 100,000)
|
0.19(per 100,000) |
| Average Victim Age |
21
|
26 |
| Typical Location |
Highway/Road/Alley
|
Residence/Home |
| Typical Time of Day |
8:00 PM |
6:00 AM |
| Typical Victim |
Acquaintance |
Acquaintance |
| Typical Weapon |
N/A |
N/A |
| Typical Victim Race |
Black
|
White |
| Typical 1st Injury |
None
|
None |
|
| Typical 2nd Injury |
None
|
None |
|
| Typical 3rd Injury |
None
|
None |
|
Note: crime statistics are estimates only and should be considered at your own risk. Data may not be representative, especially for rare crimes with small sample sizes or for lesser-populated regions. For more information on the National Incident-Based Reporting System, see the FBI's
NIBRS Technical Specification.Crime Statistics by Offence
