Osceola (town), Wisconsin

Osceola is a town located in Polk County, Wisconsin. As of the 2000 census, the town had a total population of 2,085.


\nAccording to the
United States Census Bureau, the town has a total area of 95.2 km² (36.7 mi²). 90.5 km² (35.0 mi²) of it is land and 4.6 km² (1.8 mi²) of it is water. The total area is 4.87% water.


\nAs of the
census of 2000, there are 2,085 people, 744 households, and 596 families residing in the town. The population density is 23.0/km² (59.7/mi²). There are 829 housing units at an average density of 9.2 persons/km² (23.7 persons/mi²). The racial makeup of the town is 98.80% White, 0.14% African American, 0.10% Native American, 0.19% Asian, 0.00% Pacific Islander, 0.05% from other races, and 0.72% from two or more races. 0.48% of the population are Hispanic or Latino of any race. There are 744 households out of which 41.8% have children under the age of 18 living with them, 69.4% are married couples living together, 7.1% have a woman whose husband does not live with her, and 19.8% are non-families. 15.3% of all households are made up of individuals and 3.4% have someone living alone who is 65 years of age or older. The average household size is 2.80 and the average family size is 3.13. In the town the population is spread out with 30.3% under the age of 18, 5.5% from 18 to 24, 33.7% from 25 to 44, 22.3% from 45 to 64, and 8.3% who are 65 years of age or older. The median age is 36 years. For every 100 females there are 106.2 males. For every 100 females age 18 and over, there are 104.1 males. The median income for a household in the town is $55,509, and the median income for a family is $59,688. Males have a median income of $41,200 versus $28,693 for females. The per capita income for the town is $21,865. 2.2% of the population and 1.5% of families are below the poverty line. Out of the total people living in poverty, 1.2% are under the age of 18 and 0.0% are 65 or older.

copyright 2004 FactsAbout.com