Cleveland, Alabama

From Wikipedia, the free encyclopedia

Cleveland is a town in Blount County, Alabama, United States. At the 2000 census the population was 1,241.

[edit] Geography

Cleveland is located at 33°59'31.326" North, 86°34'33.823" West (33.992035, -86.576062)^GR1.

According to the U.S. Census Bureau, the town has a total area of 17.9 km² (6.9 mi²). 17.8 km² (6.9 mi²) of it is land and 0.1 km² (0.1 mi²) of it (0.72%) is water.

[edit] Demographics

As of the census^GR2 of 2000, there were 1,241 people, 441 households, and 339 families residing in the town. The population density was 69.6/km² (180.5/mi²). There were 492 housing units at an average density of 27.6/km² (71.6/mi²). The racial makeup of the town was 94.52% White, 0.08% Black or African American, 0.16% Native American, 3.71% from other races, and 1.53% from two or more races. 15.79% of the population were Hispanic or Latino of any race.

There were 441 households out of which 36.7% had children under the age of 18 living with them, 67.6% were married couples living together, 5.2% had a female householder with no husband present, and 23.1% were non-families. 20.4% of all households were made up of individuals and 10.4% had someone living alone who was 65 years of age or older. The average household size was 2.76 and the average family size was 3.22.

In the town the population was spread out with 25.3% under the age of 18, 10.1% from 18 to 24, 28.0% from 25 to 44, 23.3% from 45 to 64, and 13.3% who were 65 years of age or older. The median age was 36 years. For every 100 females there were 106.1 males. For every 100 females age 18 and over, there were 105.5 males.

The median income for a household in the town was $37,037, and the median income for a family was $40,298. Males had a median income of $27,228 versus $19,667 for females. The per capita income for the town was $16,857. About 5.2% of families and 6.7% of the population were below the poverty line, including 6.3% of those under age 18 and 9.0% of those age 65 or over.