Sanford, FloridaSanford is a city located in Seminole County, Florida. As of the 2000 census, the city had a total population of 38,291. It is the county seat of Seminole County6.
According to the United States Census Bureau, the city has a total area of 58.5 km² (22.6 mi²). 49.5 km² (19.1 mi²) of it is land and 9.0 km² (3.5 mi²) of it is water. The total area is 15.44% water.
As of the census of 2000, there are 38,291 people, 14,237 households, and 9,168 families residing in the city. The population density is 773.6/km² (2,004.1/mi²). There are 15,623 housing units at an average density of 315.7/km² (817.7/mi²). The racial makeup of the city is 59.73% White, 32.14% African American, 0.45% Native American, 1.05% Asian, 0.05% Pacific Islander, 4.25% from other races, and 2.31% from two or more races. 10.38% of the population are Hispanic or Latino of any race.
There are 14,237 households out of which 31.8% have children under the age of 18 living with them, 39.5% are married couples living together, 19.2% have a female householder with no husband present, and 35.6% are non-families. 27.0% of all households are made up of individuals and 9.6% have someone living alone who is 65 years of age or older. The average household size is 2.57 and the average family size is 3.13.
In the city the population is spread out with 26.8% under the age of 18, 10.7% from 18 to 24, 32.5% from 25 to 44, 19.4% from 45 to 64, and 10.5% who are 65 years of age or older. The median age is 32 years. For every 100 females there are 99.0 males. For every 100 females age 18 and over, there are 96.9 males.
The median income for a household in the city is $31,163, and the median income for a family is $36,687. Males have a median income of $28,101 versus $21,723 for females. The per capita income for the city is $15,219. 17.8% of the population and 13.2% of families are below the poverty line. Out of the total people living in poverty, 24.4% are under the age of 18 and 13.1% are 65 or older.