Seymour is a city located in Wayne County, Iowa. Seymour has a 2025 population of 641. Seymour is currently growing at a rate of 0.16% annually and its population has increased by 0.79% since the most recent census, which recorded a population of 636 in 2020.
The average household income in Seymour is $52,112 with a poverty rate of 18.44%.The median age in Seymour is 51.8 years: 40.5 years for males, and 57.1 years for females.
Data after 2023 is projected based on recent change
The racial composition of Seymour includes 95.88% White, and smaller percentages for and multiracial populations.
White (95.9%)
Two or more races (4.1%)
Race | Population | Percentage (of total) |
---|---|---|
White | 535 | 95.88% |
Two or more races | 23 | 4.12% |
Married
Widowed
Divorced
Separated
Never Married
Seymour's average per capita income is $40,682. Household income levels show a median of $45,000. The poverty rate stands at 18.44%.
Name | Median | Mean |
---|---|---|
Married Families | $67,857 | - |
Families | $65,833 | $67,363 |
Households | $45,000 | $52,112 |
Non Families | $35,313 | $33,172 |
Average Income
Median Household Income
Poverty Rate