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Alford is a town in Jackson County, Florida in the Florida Panhandle near Marianna. The population was 466 at the 2000 census. According to the 2004 U.S. Census Bureau's Estimates, the town slightly declined to 464.
GeographyAlford is located at (30.694794, -85.393119)[3]. According to the United States Census Bureau, the town has a total area of 3.4 km² (1.3 mi²). 3.3 km² (1.3 mi²) of it is land and 0.1 km² (0.04 mi²) of it (1.53%) is water. DemographicsAs of the census[1] of 2000, there were 466 people, 196 households, and 133 families residing in the town. The population density was 140.6/km² (362.9/mi²). There were 235 housing units at an average density of 70.9/km² (183.0/mi²). The racial makeup of the town was 96.35% White, 1.50% African American, 0.86% Native American, 0.21% from other races, and 1.07% from two or more races. Hispanic or Latino of any race were 2.36% of the population. There were 196 households out of which 30.1% had children under the age of 18 living with them, 48.0% were married couples living together, 14.3% had a female householder with no husband present, and 32.1% were non-families. 29.6% of all households were made up of individuals and 15.8% had someone living alone who was 65 years of age or older. The average household size was 2.38 and the average family size was 2.93. In the town the population was spread out with 27.7% under the age of 18, 5.6% from 18 to 24, 27.0% from 25 to 44, 23.0% from 45 to 64, and 16.7% who were 65 years of age or older. The median age was 37 years. For every 100 females there were 82.0 males. For every 100 females age 18 and over, there were 82.2 males. The median income for a household in the town was $19,250, and the median income for a family was $24,375. Males had a median income of $28,333 versus $16,250 for females. The per capita income for the town was $14,689. About 27.9% of families and 36.9% of the population were below the poverty line, including 52.4% of those under age 18 and 15.5% of those age 65 or over. References
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