St. Paul, Indiana

St. Paul is a small town on the border of Decatur and Shelby counties in the U.S. state of Indiana. The population was 1,031 at the 2010 census.

Geography

St. Paul is located at (39.428227, -85.629222)^[3], along the Flatrock River.^[4]

According to the United States Census Bureau, the town has a total area of 0.3 square miles (0.78 km²), all of it land.

Demographics

As of the census^[1] of 2000, there were 1,022 people, 372 households, and 286 families residing in the town. The population density was 3,314.7 people per square mile (1,272.9/km²). There were 394 housing units at an average density of 1,277.9 per square mile (490.7/km²). The racial makeup of the town was 97.95% White, 0.10% African American, 0.29% Native American, 0.20% Asian, 0.10% Pacific Islander, and 1.37% from two or more races. Hispanic or Latino of any race were 0.78% of the population.

There were 372 households out of which 44.4% had children under the age of 18 living with them, 58.6% were married couples living together, 12.1% had a female householder with no husband present, and 23.1% were non-families. 18.0% of all households were made up of individuals and 4.8% had someone living alone who was 65 years of age or older. The average household size was 2.75 and the average family size was 3.09.

In the town the population was spread out with 32.0% under the age of 18, 8.5% from 18 to 24, 31.9% from 25 to 44, 19.4% from 45 to 64, and 8.2% who were 65 years of age or older. The median age was 32 years. For every 100 females there were 94.7 males. For every 100 females age 18 and over, there were 98.6 males.

The median income for a household in the town was $39,079, and the median income for a family was $42,650. Males had a median income of $31,806 versus $20,670 for females. The per capita income for the town was $14,819. About 5.3% of families and 9.1% of the population were below the poverty line, including 10.9% of those under age 18 and 8.1% of those age 65 or over.

References