Sevastopol, Wisconsin

Sevastopol is a town located in Door County, Wisconsin. As of the 2000 census, the town had a total population of 2,667.

Geography

\nAccording to the United States Census Bureau, the town has a total area of 234.6 km² (90.6 mi²). 134.3 km² (51.8 mi²) of it is land and 100.3 km² (38.7 mi²) of it is water. The total area is 42.75% water.

Demographics

\nAs of the census of 2000, there are 2,667 people, 1,076 households, and 825 families residing in the town. The population density is 19.9/km² (51.4/mi²). There are 1,554 housing units at an average density of 11.6/km² (30.0/mi²). The racial makeup of the town is 98.09% White, 0.04% African American, 0.30% Native American, 0.19% Asian, 0.00% Pacific Islander, 0.75% from other races, and 0.64% from two or more races. 1.35% of the population are Hispanic or Latino of any race. There are 1,076 households out of which 27.2% have children under the age of 18 living with them, 68.5% are married couples living together, 4.6% have a female householder with no husband present, and 23.3% are non-families. 19.7% of all households are made up of individuals and 9.2% have someone living alone who is 65 years of age or older. The average household size is 2.48 and the average family size is 2.84. In the town the population is spread out with 22.8% under the age of 18, 4.9% from 18 to 24, 25.4% from 25 to 44, 29.8% from 45 to 64, and 17.1% who are 65 years of age or older. The median age is 43 years. For every 100 females there are 102.2 males. For every 100 females age 18 and over, there are 101.7 males. The median income for a household in the town is $47,227, and the median income for a family is $52,125. Males have a median income of $33,152 versus $22,632 for females. The per capita income for the town is $24,150. 7.3% of the population and 5.9% of families are below the poverty line. Out of the total people living in poverty, 13.0% are under the age of 18 and 6.0% are 65 or older.