Sand Rock, Alabama

From Wikipedia, the free encyclopedia

Sand Rock is a town in Cherokee County, Alabama, United States. At the 2000 census the population was 509.

[edit] Geography

Sand Rock is located at 34°13'57.115" North, 85°46'13.516" West (34.232532, -85.770421)^GR1.

According to the U.S. Census Bureau, the town has a total area of 11.7 km² (4.5 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 509 people, 195 households, and 154 families residing in the town. The population density was 43.4/km² (112.4/mi²). There were 215 housing units at an average density of 18.3/km² (47.5/mi²). The racial makeup of the town was 99.41% White, and 0.39% Black or African American.

There were 195 households out of which 33.8% had children under the age of 18 living with them, 68.7% were married couples living together, 7.7% had a female householder with no husband present, and 21.0% were non-families. 19.5% 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 2.61 and the average family size was 2.96.

In the town the population was spread out with 25.3% under the age of 18, 7.9% from 18 to 24, 28.9% from 25 to 44, 23.6% from 45 to 64, and 14.3% who were 65 years of age or older. The median age was 37 years. For every 100 females there were 99.6 males. For every 100 females age 18 and over, there were 97.9 males.

The median income for a household in the town was $35,179, and the median income for a family was $44,063. Males had a median income of $28,523 versus $17,375 for females. The per capita income for the town was $18,845. About 10.9% of families and 12.6% of the population were below the poverty line, including 16.5% of those under age 18 and 19.7% of those age 65 or over.