Lee, Florida

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Lee is a town in Madison County, Florida, United States. The population was 352 at the 2000 census. As of 2004, the population recorded by the U.S. Census Bureau is 357 [1].



Lee is located at 30°25′10″N 83°17′59″W / 30.419355°N 83.299631°W / 30.419355; -83.299631.[3]

According to the United States Census Bureau, the town has a total area of 1.2 square miles (3.2 km²), all of it land.


As of the census[1] of 2000, there were 352 people, 130 households, and 92 families residing in the town. The population density was 287.8 people per square mile (111.4/km²). There were 154 housing units at an average density of 125.9/sq mi (48.7/km²). The racial makeup of the town was 95.45% White, 2.84% African American, 0.28% Native American, 0.28% Asian, and 1.14% from two or more races. Hispanic or Latino of any race were 2.56% of the population.

There were 130 households out of which 34.6% had children under the age of 18 living with them, 52.3% were married couples living together, 16.2% had a female householder with no husband present, and 28.5% were non-families. 26.2% of all households were made up of individuals and 12.3% had someone living alone who was 65 years of age or older. The average household size was 2.71 and the average family size was 3.13.

In the town the population was spread out with 30.7% under the age of 18, 5.4% from 18 to 24, 28.7% from 25 to 44, 22.2% from 45 to 64, and 13.1% who were 65 years of age or older. The median age was 37 years. For every 100 females there were 96.6 males. For every 100 females age 18 and over, there were 101.7 males.

The median income for a household in the town was $26,250, and the median income for a family was $26,875. Males had a median income of $33,750 versus $21,875 for females. The per capita income for the town was $13,242. About 19.8% of families and 26.5% of the population were below the poverty line, including 46.2% of those under age 18 and 12.8% of those age 65 or over.

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