| Bay Harbor Islands, Florida | |
|---|---|
| — Town — | |
| Location in Miami-Dade County and the state of Florida | |
| U.S. Census Bureau map showing town boundaries | |
| Coordinates: | |
| Country | United States |
| State | Florida |
| County | Miami-Dade |
| Incorporated | 1947 |
| Area | |
| • Total | 0.6 sq mi (1.6 km2) |
| • Land | 0.4 sq mi (1 km2) |
| • Water | 0.2 sq mi (0.6 km2) |
| Elevation | 7 ft (2 m) |
| Population (2009)[1] | |
| • Total | 5,131 |
| • Density | 13,835/sq mi (5,131/km2) |
| Time zone | Eastern (EST) (UTC-5) |
| • Summer (DST) | EDT (UTC-4) |
| ZIP code | 33154 |
| Area code(s) | 305 |
| FIPS code | 12-03975[2] |
| GNIS feature ID | 0278201[3] |
| Website | http://www.bayharborislands.org |
Bay Harbor Islands is a town in Miami-Dade County, Florida, United States. The population was 5,146 at the 2000 census. It is separated from the mainland by Biscayne Bay, with which it is connected via the Broad Causeway. On the mainland side, BHI is bordered by the city of North Miami, while to the east it borders the villages of Bal Harbour and Surfside. To the south lies the exclusive Indian Creek island. BHI is about 20 minutes driving distance away from Miami International Airport, and is situated between the larger centers of Aventura and Miami Beach.
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It was founded by Shepard Broad in April 1947.
Bay Harbor Islands is located at (25.887522, -80.135739).[4] It consists of two islands. The west island contains exclusively single family homes and the east island contains a business district and multi-family housing.
According to the United States Census Bureau, the town has a total area of 0.6 square miles (1.6 km2). 0.4 square miles (1.0 km2) of it is land and 0.2 square miles (0.52 km2) of it (39.34%) is water.
As of the census[2] of 2000, there were 5,146 people, 2,612 households, and 1,295 families residing in the town. The population density was 13,875.4 inhabitants per square mile (5,370.0/km²). There were 3,103 housing units at an average density of 8,366.8 per square mile (3,238.0/km²). The racial makeup of the town is 91.22% White (60.1% were Non-Hispanic White),[5] 1.79% African American, 0.08% Native American, 1.22% Asian, 0.06% Pacific Islander, 2.84% from other races, and 2.80% from two or more races. 35.29% of the population are Hispanic or Latino of any race.
There were 2,612 households out of which 20.3% had children under the age of 18 living with them, 35.6% were married couples living together, 10.7% had a female householder with no husband present, and 50.4% were non-families. 43.1% of all households were made up of individuals and 18.6% had someone living alone who was 65 years of age or older. The average household size was 1.97 and the average family size was 2.71.
In the town the population was spread out with 18.0% under the age of 18, 5.0% from 18 to 24, 32.1% from 25 to 44, 21.7% from 45 to 64, and 23.3% who were 65 years of age or older. The median age was 42 years. For every 100 females there were 80.1 males. For every 100 females age 18 and over, there were 75.3 males.
The median income for a household in the town was $38,514, and the median income for a family was $43,939. Males had a median income of $38,750 versus $31,044 for females. The per capita income for the town was $29,261. About 8.0% of families and 13.1% of the population were below the poverty line, including 20.7% of those under age 18 and 6.8% of those age 65 or over.
As of 2000, speakers of English as a first language consisted of 52.04%, while Spanish accounted for 43.90%, Portuguese was at 1.65%, Hebrew at 1.29%, and French made up 1.10% of the population.[6]
As of 2000, Bay Harbor Islands had the twenty-second highest percentage of Colombian residents in the US, with 4.72% of the US populace.[7] It had the fifty-first highest percentage of Cuban residents in the US, at 8.51% of the town's population.[8] It also had the ninth most Peruvians in the US, at 3.11%,[9] while it had the twentieth highest percentage of Romanians, at 1.10% of all residents (tied with Hollywood, Florida and many other areas in the US.)[10] Bay Harbor Islands' Turkish community had the seventh highest percentage of residents, which was at 1.20% (tying with five other US areas, such as Highland Beach.)[11] It's also home to the seventeenth highest percentage of Venezuelan residents in the US, at 1.15% of the population.[12]
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