Seymour

Seymour is a city in Jackson County, Indiana, United States. The population was 18,101 at the 2000 census.

Geography

Seymour is located at 38°57′23″N, 85°53′24″W (38.956350, -85.890068)GR1.

According to the United States Census Bureau, the city has a total area of 10.8 square miles (28.1 km*), all land.

Demographics

As of the censusGR2 of 2000, there were 18,101 people, 7,231 households, and 4,743 families residing in the city. The population density was 644.7/km² (1,670.3/mi²). There were 7,709 housing units at an average density of 274.6/km² (711.4/mi²). The racial makeup of the city was 93.40% White, 1.02% African American, 0.30% Native American, 1.43% Asian, 0.05% Pacific Islander, 2.93% from other races, and 0.86% from two or more races. Hispanic or Latino of any race were 4.85% of the population.

There were 7,231 households out of which 32.6% had children under the age of 18 living with them, 49.4% were married couples living together, 11.5% had a female householder with no husband present, and 34.4% were non-families. 28.5% of all households were made up of individuals and 11.6% had someone living alone who was 65 years of age or older. The average household size was 2.46 and the average family size was 2.99.

In the city the population was spread out with 25.5% under the age of 18, 10.2% from 18 to 24, 31.4% from 25 to 44, 19.4% from 45 to 64, and 13.5% who were 65 years of age or older. The median age was 34 years. For every 100 females there were 95.7 males. For every 100 females age 18 and over, there were 93.0 males.

The median income for a household in the city was $36,883, and the median income for a family was $43,357. Males had a median income of $30,638 versus $22,265 for females. The per capita income for the city was $18,222. About 8.0% of families and 10.0% of the population were below the poverty line, including 9.7% of those under age 18 and 11.6% of those age 65 or over.