Seymour is a city in Tennessee state in USA with a population of 15444 residents..
The poverty rate of Seymour is 8.7%, which is 38% lower than national average.
The typical household in Seymour earns $61490 a year, compared to the national median of $67,500.
Nearly 41.5% of households earn less than $50,000 a year compared to 39% national rate.
The unemployment rate of Seymour remained 5.3%, which is 13% higher than national rate. Nearly 1 times comparable to 4.7% national rate, according to the Census.
Cost of living in Seymour is 87, which is 13% lower than national average.
In Seymour, only 19% of adults have a bachelor's degree and 37% have some college but no degree.
Median home value in Seymour is $182700 , which is 20% lower than national average.
While median income in Seymour remained $61490, which is 9% lower than national average.