Sevastopol, Wisconsin
From Wikipedia, the free encyclopedia
Sevastopol is a town in Door County, Wisconsin, United States. The population was 2,667 at the 2000 census.
[edit] Geography
According 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 (42.75%) is water.
[edit] Demographics
As of the census2 of 2000, there were 2,667 people, 1,076 households, and 825 families residing in the town. The population density was 19.9/km² (51.4/mi²). There were 1,554 housing units at an average density of 11.6/km² (30.0/mi²). The racial makeup of the town was 98.09% White, 0.04% African American, 0.30% Native American, 0.19% Asian, 0.75% from other races, and 0.64% from two or more races. Hispanic or Latino of any race were 1.35% of the population.
There were 1,076 households out of which 27.2% had children under the age of 18 living with them, 68.5% were married couples living together, 4.6% had a female householder with no husband present, and 23.3% were non-families. 19.7% of all households were made up of individuals and 9.2% had someone living alone who was 65 years of age or older. The average household size was 2.48 and the average family size was 2.84.
In the town the population was 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 were 65 years of age or older. The median age was 43 years. For every 100 females there were 102.2 males. For every 100 females age 18 and over, there were 101.7 males.
The median income for a household in the town was $47,227, and the median income for a family was $52,125. Males had a median income of $33,152 versus $22,632 for females. The per capita income for the town was $24,150. About 5.9% of families and 7.3% of the population were below the poverty line, including 13.0% of those under age 18 and 6.0% of those age 65 or over.
