Somerville, Indiana

From Wikipedia, the free encyclopedia

Somerville is a town in Gibson County, Indiana, United States. The population was 312 at the 2000 census.

[edit] Geography

Somerville is located at 38°16′41″N, 87°22′39″W (38.278113, -87.377426)^GR1.

According to the United States Census Bureau, the town has a total area of 0.8 km² (0.3 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 312 people, 124 households, and 81 families residing in the town. The population density was 415.4/km² (1,061.5/mi²). There were 132 housing units at an average density of 175.7/km² (449.1/mi²). The racial makeup of the town was 99.04% White, 0.64% African American, 0.32% from other races. Hispanic or Latino of any race were 0.32% of the population.

There were 124 households out of which 28.2% had children under the age of 18 living with them, 59.7% were married couples living together, 5.6% had a female householder with no husband present, and 33.9% were non-families. 30.6% of all households were made up of individuals and 16.9% had someone living alone who was 65 years of age or older. The average household size was 2.52 and the average family size was 3.18.

In the town the population was spread out with 24.7% under the age of 18, 4.8% from 18 to 24, 30.8% from 25 to 44, 24.4% from 45 to 64, and 15.4% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 89.1 males. For every 100 females age 18 and over, there were 86.5 males.

The median income for a household in the town was $34,464, and the median income for a family was $43,750. Males had a median income of $30,750 versus $22,083 for females. The per capita income for the town was $12,147. About 3.3% of families and 5.2% of the population were below the poverty line, including 4.3% of those under age 18 and 7.9% of those age 65 or over.