Seymour is a city located in Wayne County, Iowa. Seymour has a 2026 population of 651. Seymour is currently growing at a rate of 0.46% annually and its population has increased by 2.36% since the most recent census, which recorded a population of 636 in 2020.
The median household income in Seymour is $46,071 with a poverty rate of 13.47%. The median age in Seymour is 44.5 years: 33.3 years for males, and 51.9 years for females. For every 100 females there are 99.3 males.
Data after 2023 is projected based on recent change
Overall: 44.5 years
Female: 51.9 years
Male: 33.3 years
There are 453 adults, (150 of whom are seniors) in Seymour.
Female: 298 (50.2%)
Male: 296 (49.8%)
The racial composition of Seymour includes 89.06% White, and smaller percentages for and multiracial populations.
| Race | Population ↓ | Percentage (of total) |
|---|---|---|
| White | 529 | 89.06% |
| Two or more races | 65 | 10.94% |
Seymour 's average per capita income is $41,691. Household income levels show a median of $46,071. The poverty rate stands at 13.47%.
| Name | Median ↓ | Mean |
|---|---|---|
| Married Families | $67,813 | - |
| Families | $66,667 | $72,715 |
| Households | $46,071 | $55,623 |
| Non Families | $29,167 | $33,223 |
Average Income
Median Household Income
Poverty Rate