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Florala, Alabama - Wikipedia

Florala is a town in Covington County, Alabama, United States. At the 2020 census, the population was 1,923.

Florala, Alabama
Location of Florala in Covington County, Alabama.
Location of Florala in Covington County, Alabama.
Coordinates: 31°0′27″N 86°19′29″W / 31.00750°N 86.32472°W / 31.00750; -86.32472
CountryUnited States
StateAlabama
CountyCovington
Area
 • Total10.98 sq mi (28.44 km2)
 • Land10.53 sq mi (27.28 km2)
 • Water0.45 sq mi (1.16 km2)
Elevation
262 ft (80 m)
Population
 (2020)
 • Total1,923
 • Density182.59/sq mi (70.49/km2)
Time zoneUTC−6 (Central (CST))
 • Summer (DST)UTC−5 (CDT)
ZIP code
36442
Area code334
FIPS code01-26848
GNIS feature ID0159617

Geography

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Florala is located along the Alabama–Florida state line at 31°0′28″N 86°19′30″W / 31.00778°N 86.32500°W / 31.00778; -86.32500 (31.007712, -86.324957).[2] It is bordered by the town of Lockhart to the west and the town of Paxton, Florida, to the south.

According to the U.S. Census Bureau, the city has a total area of 11.0 square miles (28.4 km2), of which 10.5 square miles (27.3 km2) is land and 0.46 square miles (1.2 km2), or 4.07%, is water.[3] Lake Jackson lies on the state line, half in Florala. Florala City Park occupies all of the lake's shoreline in Alabama.

Climate

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Climate data for Florala, Alabama (Florala Municipal Airport), 1991–2020 normals, extremes 2006–2020
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Year
Record high °F (°C) 83
(28)
84
(29)
89
(32)
94
(34)
99
(37)
104
(40)
102
(39)
103
(39)
103
(39)
100
(38)
91
(33)
81
(27)
104
(40)
Mean daily maximum °F (°C) 62.2
(16.8)
66.2
(19.0)
73.1
(22.8)
79.4
(26.3)
86.6
(30.3)
90.3
(32.4)
91.8
(33.2)
91.7
(33.2)
88.5
(31.4)
80.0
(26.7)
71.3
(21.8)
63.6
(17.6)
78.7
(26.0)
Daily mean °F (°C) 51.3
(10.7)
54.7
(12.6)
61.0
(16.1)
67.0
(19.4)
74.7
(23.7)
80.1
(26.7)
82.1
(27.8)
82.0
(27.8)
78.4
(25.8)
68.4
(20.2)
59.1
(15.1)
52.9
(11.6)
67.6
(19.8)
Mean daily minimum °F (°C) 40.5
(4.7)
43.3
(6.3)
48.8
(9.3)
54.6
(12.6)
62.7
(17.1)
70.0
(21.1)
72.4
(22.4)
72.4
(22.4)
68.3
(20.2)
56.7
(13.7)
46.9
(8.3)
42.3
(5.7)
56.6
(13.7)
Record low °F (°C) 16
(−9)
20
(−7)
28
(−2)
35
(2)
43
(6)
58
(14)
59
(15)
62
(17)
50
(10)
34
(1)
19
(−7)
22
(−6)
16
(−9)
Average precipitation inches (mm) 5.28
(134)
5.00
(127)
5.28
(134)
5.00
(127)
3.78
(96)
6.01
(153)
6.35
(161)
5.17
(131)
5.18
(132)
3.61
(92)
4.31
(109)
4.85
(123)
59.82
(1,519)
Source 1: NOAA[4]
Source 2: XMACIS2[5]

Demographics

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Historical population
CensusPop.Note
19102,439
19202,6338.0%
19302,580−2.0%
19402,99916.2%
19502,713−9.5%
19603,01111.0%
19702,701−10.3%
19802,165−19.8%
19902,075−4.2%
20001,964−5.3%
20101,9800.8%
20201,923−2.9%
U.S. Decennial Census[6]
2013 Estimate[7]

2020 Census data

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Florala racial composition[8]
Race Num. Perc.
White (non-Hispanic) 1,481 77.02%
Black or African American (non-Hispanic) 296 15.39%
Native American 3 0.16%
Asian 11 0.57%
Other/Mixed 70 3.64%
Hispanic or Latino 62 3.22%

As of the 2020 United States census, there were 1,923 people, 646 households, and 352 families residing in the town.

2010 Census data

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As of the census[9] of 2010, there were 1,980 people, 839 households, and 514 families living in the city. The population density was 180.2 inhabitants per square mile (69.6/km2). There were 1,107 housing units at an average density of 105.0 per square mile (40.5/km2). The racial makeup of the city was 80.1% White, 15.8% Black or African American, 0.8% Native American, 0.4% Asian, 0.7% from other races, and 2.3% from two or more races. 2.8% of the population were Hispanic or Latino of any race.

There were 839 households, out of which 23.5% had children under the age of 18 living with them, 38.1% were married couples living together, 17.6% had a female householder with no husband present, and 38.7% were non-families. 34.7% of all households were made up of individuals, and 17.8% had someone living alone who was 65 years of age or older. The average household size was 2.26 and the average family size was 2.90.

In the city, the population was 21.6% under the age of 18, 7.9% from 18 to 24, 18.9% from 25 to 44, 27.7% from 45 to 64, and 23.8% who were 65 years of age or older. The median age was 46.1 years. For every 100 females, there were 88.0 males. For every 100 females age 18 and over, there were 92.9 males.

The median income for a household in the city was $30,833, and the median income for a family was $36,435. Males had a median income of $24,000 versus $20,100 for females. The per capita income for the city was $16,344. About 15.7% of families and 20.2% of the population were below the poverty line, including 28.5% of those under age 18 and 15.4% of those age 65 or over.

2000 Census data

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As of the census[10] of 2000, there were 1,964 people, 898 households, and 527 families living in the city. The population density was 187.0 inhabitants per square mile (72.2/km2). There were 1,103 housing units at an average density of 105.0 per square mile (40.5/km2). The racial makeup of the city was 80.75% White, 15.68% Black or African American, 0.87% Native American, 0.61% Asian, 0.66% from other races, and 1.43% from two or more races. 2.65% of the population were Hispanic or Latino of any race.

There were 898 households, out of which 23.9% had children under the age of 18 living with them, 40.5% were married couples living together, 14.3% had a female householder with no husband present, and 41.3% were non-families. 36.4% of all households were made up of individuals, and 21.6% had someone living alone who was 65 years of age or older. The average household size was 2.19 and the average family size was 2.86.

In the city, the population was 21.8% under the age of 18, 7.8% from 18 to 24, 24.6% from 25 to 44, 24.2% from 45 to 64, and 21.5% who were 65 years of age or older. The median age was 42 years. For every 100 females, there were 82.5 males. For every 100 females age 18 and over, there were 76.3 males. The median income for a household in the city was $17,377, and the median income for a family was $21,176. Males had a median income of $27,569 versus $15,625 for females. The per capita income for the city was $11,495. About 29.3% of families and 32.4% of the population were below the poverty line, including 38.8% of those under age 18 and 26.6% of those age 65 or over.

Etymology

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The name is a portmanteau of Florida and Alabama.

History

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In 1818, Andrew Jackson stopped at the lake with his soldiers, and thus Lake Jackson is named after him. Since 1870, Florala has served as the home of the world's oldest consecutive annual Masonic Day celebration, through Florala's Fidelity Masonic Lodge #685 (beginning with Chapel Hill and Lake City Lodge #377), and Chapter #441 of the Order of the Eastern Star. The celebration is in honor of St. John's Day, June 24, 1717, when the first Grand Masonic Lodge in England was established. The celebration is held each year on the Friday before June 24 and concluding on the Saturday after the 24th.

Notable people

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References

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  1. ^ "2020 U.S. Gazetteer Files". United States Census Bureau. Retrieved October 29, 2021.
  2. ^ "US Gazetteer files: 2010, 2000, and 1990". United States Census Bureau. February 12, 2011. Retrieved April 23, 2011.
  3. ^ "Geographic Identifiers: 2010 Demographic Profile Data (G001): Florala town, Alabama". U.S. Census Bureau, American Factfinder. Archived from the original on February 12, 2020. Retrieved June 10, 2014.
  4. ^ "U.S. Climate Normals Quick Access – Station: Florala MUNI AP, AL". National Oceanic and Atmospheric Administration. Retrieved March 4, 2023.
  5. ^ "xmACIS2". National Oceanic and Atmospheric Administration. Retrieved March 4, 2023.
  6. ^ "U.S. Decennial Census". Census.gov. Retrieved June 6, 2013.
  7. ^ "Annual Estimates of the Resident Population: April 1, 2010 to July 1, 2013". Archived from the original on May 22, 2014. Retrieved June 3, 2014.
  8. ^ "Explore Census Data". data.census.gov. Retrieved December 17, 2021.
  9. ^ "U.S. Census website". United States Census Bureau. Retrieved July 20, 2015.
  10. ^ "U.S. Census website". United States Census Bureau. Retrieved January 31, 2008.
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31°00′28″N 86°19′30″W / 31.007712°N 86.324957°W / 31.007712; -86.324957