Seymour is a city in Indiana state in USA with a population of 19780 residents.. The poverty rate of Seymour is 20.2%, which is 44% higher than national average. The typical household in Seymour earns $44524 a year, compared to the national median of $67,500. Financial hardship is relatively common in the city as 1 in every 5 residents live below the poverty line, well above the national poverty rate of 14.1%. Additionally, more than 1 in every 10 area households live on an income of less than $10,000 a year. Nationwide, 5.8% of households do. Nearly 55.5% of households earn less than $50,000 a year compared to 39% national rate.

A high school education is a basic prerequisite for most jobs, and in Seymour, only 86% of adults have a high school diploma, compared to the 87.7% of adults nationwide. The unemployment rate of Seymour remained 3.5%, which is 26% lower than national rate. Higher educational attainment can lead to improved job security and reduce the likelihood of financial insecurity. In Seymour, only 19% of adults have a bachelor's degree and 29% have some college but no degree.

Cost of living in Seymour is 92, which is 8% lower than national average. Median home value in Seymour is $121400 , which is 47% lower than national average. While median income in Seymour remained $44524, which is 34% lower than national average.