Lake Harbor

Lake Harbor is a census-designated place (CDP) in Palm Beach County, Florida, United States. The population was 195 at the 2000 census. It located along the southern banks of Lake Okeechobee, at the beginning of the Miami Canal. John Stretch Park is also located alongside the north end of Lake Harbor and the lake.GeographyLake Harbor is located at .According to the United States Census Bureau, the CDP has a total area of 3.4 km², all land.DemographicsAs of the census of 2000, there were 195 people, 65 households, and 44 families residing in the CDP. The population density was 57.5/km² (148.7/mi²). There were 118 housing units at an average density of 34.8/km² (90.0/mi²). The racial makeup of the CDP was 41.54% White (35.9% were Non-Hispanic White,) 42.56% African American, 3.08% Asian, 2.05% from other races, and 10.77% from two or more races. Hispanic or Latino of any race were 6.67% of the population.There were 65 households out of which 32.3% had children under the age of 18 living with them, 50.8% were married couples living together, 12.3% had a female householder with no husband present, and 32.3% were non-families. 24.6% of all households were made up of individuals and 4.6% had someone living alone who was 65 years of age or older. The average household size was 3.00 and the average family size was 3.55.