Seymour

City in Wayne County, Iowa

Aerial view of Seymour, Iowa
Aerial view: USDA NAIP · USGS The National Map (public domain)

Seymour is a city in Wayne County, Iowa, with 634 residents (2020 Census). It sits at 1,060 ft (323 m) above sea level. The nearest town is Promise City, 4.7 miles N.

About Seymour

Seymour is a city in Wayne County, Iowa, United States. The population was 634 at the time of the 2020 census.

Text: Wikipedia · CC BY-SA 4.0

Seymour weather: 7-day forecast

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Live webcams in Seymour

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Webcams: Windy.com

Living in Seymour

Median household income
$46,071
Median home value
$86,800
Median rent
$614/mo
Median age
44.5
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Map of Seymour

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Seymour, IA population

Seymour had 634 residents in the 2020 Census.

2020 Census634
Density265 per sq mi (102 per km²)

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

Seymour at a glance

StateIowa (IA)
CountyWayne County
TypeCity
Population634 (2020)
Land area2.39 sq mi (6.2 km²)
Elevation1,060 ft (323 m)
Coordinates40.6821° N, 93.1211° W
ZIP code52590
Census code (GEOID)1971760

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

Towns near Seymour

Straight-line distance between town centers.

Frequently asked questions about Seymour

What county is Seymour in?

Seymour is in Wayne County, Iowa.

What is the population of Seymour, IA?

634 people lived in Seymour at the 2020 Census.

What is the ZIP code of Seymour, IA?

The ZIP code of Seymour is 52590.

What is the elevation of Seymour?

1,060 feet (323 m) above sea level.

What towns are near Seymour?

Promise City (4.7 mi), Plano (6.4 mi), Numa (7.4 mi), Cincinnati (11.0 mi), Mystic (11.4 mi).

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