Seymour, Missouri

Geography Seymour is located at 37°8'52" North, 92°46'8" West (37.147671, -92.768882)¹.

According to the United States Census Bureau, the city has a total area of 6.8 km² (2.6 mi²). 6.8 km² (2.6 mi²) of it is land and none of it is covered by water.

Demographics As of the census of 2000, there are 1,834 people, 711 households, and 479 families residing in the city. The population density is 270.3/km² (699.3/mi²). There are 792 housing units at an average density of 116.7/km² (302.0/mi²). The racial makeup of the city is 96.29% White, 0.05% African American, 0.55% Native American, 0.38% Asian, 0.00% Pacific Islander, 0.71% from other races, and 2.02% from two or more races. 1.80% of the population are Hispanic or Latino of any race.

There are 711 households out of which 35.2% have children under the age of 18 living with them, 50.4% are married couples living together, 13.1% have a female householder with no husband present, and 32.5% are non-families. 29.4% of all households are made up of individuals and 18.6% have someone living alone who is 65 years of age or older. The average household size is 2.50 and the average family size is 3.08.

In the city the population is spread out with 28.5% under the age of 18, 8.1% from 18 to 24, 26.6% from 25 to 44, 19.2% from 45 to 64, and 17.7% who are 65 years of age or older. The median age is 36 years. For every 100 females there are 89.5 males. For every 100 females age 18 and over, there are 83.5 males.

The median income for a household in the city is $25,093, and the median income for a family is $30,048. Males have a median income of $25,938 versus $16,481 for females. The per capita income for the city is $12,486. 18.1% of the population and 13.7% of families are below the poverty line. Out of the total people living in poverty, 20.3% are under the age of 18 and 32.1% are 65 or older.