Crossville, Alabama

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Crossville is a town in DeKalb County, Alabama, United States. At the 2000 census the population was 1,431. Crossville is located on top of Sand Mountain.

Contents

Geography

Crossville is located at 34°17′12″N 85°59′27″W / 34.28667°N 85.99083°W / 34.28667; -85.99083 (34.286752, -85.990814)[1].

According to the U.S. Census Bureau, the town has a total area of 6.0 square miles (15.6 km²), all of it land.

Demographics

As of the census[2] of 2000, there were 1,431 people, 539 households, and 365 families residing in the town. The population density was 237.8 people per square mile (91.8/km²). There were 593 housing units at an average density of 98.5/sq mi (38.0/km²). The racial makeup of the town was 97.69% White, 0.07% Black or African American, 0.49% Native American, 0.14% Asian, 0.07% Pacific Islander, 0.56% from other races, and 0.98% from two or more races. 2.24% of the population were Hispanic or Latino of any race.

There were 539 households out of which 33.4% had children under the age of 18 living with them, 53.2% were married couples living together, 10.0% had a female householder with no husband present, and 32.1% were non-families. 30.6% of all households were made up of individuals and 14.7% had someone living alone who was 65 years of age or older. The average household size was 2.42 and the average family size was 3.04.

In the town the population was spread out with 23.5% under the age of 18, 7.3% from 18 to 24, 27.8% from 25 to 44, 20.0% from 45 to 64, and 21.5% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 81.6 males. For every 100 females age 18 and over, there were 76.9 males.

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