Brown Deer, Wisconsin

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Brown Deer is a village in Milwaukee County, Wisconsin, United States. It is a suburb of Milwaukee. The population was 12,170 at the 2000 census. The village incorporated on January 20, 1955 from a portion of the former town of Granville.[3]



Brown Deer is located at 43°10′22″N 87°58′40″W / 43.17278°N 87.97778°W / 43.17278; -87.97778 (43.172858, -87.977899)[4].

According to the United States Census Bureau, the village has a total area of 4.4 square miles (11.4 km²), all of it land.


As of the census[2] of 2000, there were 12,170 people, 5,134 households, and 3,249 families residing in the village. The population density was 2,768.1 people per square mile (1,067.9/km²). There were 5,335 housing units at an average density of 1,213.5/sq mi (468.1/km²). The racial makeup of the village was 82.04% White, 12.51% African American, 0.25% Native American, 2.62% Asian, 0.04% Pacific Islander, 0.66% from other races, and 1.88% from two or more races. Hispanic or Latino of any race were 2.14% of the population.

Brown Deer is one of only two Milwaukee suburbs (Racine being the other) where African-Americans make up at least 10% of the general population.

There were 5,134 households out of which 25.6% had children under the age of 18 living with them, 52.8% were married couples living together, 7.8% had a female householder with no husband present, and 36.7% were non-families. 31.2% of all households were made up of individuals and 11.4% had someone living alone who was 65 years of age or older. The average household size was 2.27 and the average family size was 2.87.

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