Lake Alfred

City in Polk County, Florida

Aerial view of Lake Alfred, Florida
Aerial view: USDA NAIP · USGS The National Map (public domain)

Lake Alfred is a city in Polk County, Florida, with 6,374 residents (2020 Census). It sits at 174 ft (53 m) above sea level. The nearest town is Auburndale, 5.3 miles W.

About Lake Alfred

Lake Alfred is a city in Polk County, Florida, United States. It is part of the Lakeland–Winter Haven Metropolitan Statistical Area. The population was approximately 6,374 at the 2020 US census.

Text: Wikipedia · CC BY-SA 4.0

Lake Alfred weather: 7-day forecast

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Source: National Weather Service (NOAA).

Live webcams in Lake Alfred

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Living in Lake Alfred

Median household income
$67,202
Median home value
$239,900
Median rent
$1,349/mo
Median age
33.8
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Map of Lake Alfred

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Lake Alfred Street View

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Lake Alfred, FL population

Lake Alfred had 6,374 residents in the 2020 Census.

2020 Census6,374
Density301 per sq mi (116 per km²)

Source: U.S. Census Bureau, Decennial Census.

Lake Alfred at a glance

StateFlorida (FL)
CountyPolk County
TypeCity
Population6,374 (2020)
Land area21.16 sq mi (54.8 km²)
Water area4.09 sq mi (10.59 km²)
Elevation174 ft (53 m)
Coordinates28.1374° N, 81.7161° W
ZIP code33850
Official websitemylakealfred.com
Census code (GEOID)1237525

Sources: U.S. Census Bureau (Gazetteer, ZCTA), USGS GNIS, Wikidata.

Towns near Lake Alfred

Straight-line distance between town centers.

Frequently asked questions about Lake Alfred

What county is Lake Alfred in?

Lake Alfred is in Polk County, Florida.

What is the population of Lake Alfred, FL?

6,374 people lived in Lake Alfred at the 2020 Census.

What is the ZIP code of Lake Alfred, FL?

The ZIP code of Lake Alfred is 33850.

What is the elevation of Lake Alfred?

174 feet (53 m) above sea level.

What towns are near Lake Alfred?

Auburndale (5.3 mi), Haines City (6.3 mi), Winter Haven (6.4 mi), Inwood (7.5 mi), Polk City (7.5 mi).

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