Rockville, Utah

From Wikipedia, the free encyclopedia

Rockville is a town in Washington County, Utah, United States. The population was 247 at the 2000 census.

[edit] Geography

Rockville is located at 37°9′58″N, 113°2′54″W (37.166044, -113.048304)^GR1.

According to the United States Census Bureau, the town has a total area of 21.8 km² (8.4 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 247 people, 115 households, and 64 families residing in the town. The population density was 11.3/km² (29.3/mi²). There were 142 housing units at an average density of 6.5/km² (16.9/mi²). The racial makeup of the town was 96.76% White, 0.40% Asian, 1.21% from other races, and 1.62% from two or more races. Hispanic or Latino of any race were 6.88% of the population.

There were 115 households out of which 21.7% had children under the age of 18 living with them, 46.1% were married couples living together, 7.8% had a female householder with no husband present, and 44.3% were non-families. 39.1% of all households were made up of individuals and 12.2% had someone living alone who was 65 years of age or older. The average household size was 2.15 and the average family size was 2.88.

In the town the population was spread out with 21.9% under the age of 18, 6.1% from 18 to 24, 28.3% from 25 to 44, 25.1% from 45 to 64, and 18.6% who were 65 years of age or older. The median age was 42 years. For every 100 females there were 90.0 males. For every 100 females age 18 and over, there were 91.1 males.

The median income for a household in the town was $37,917, and the median income for a family was $39,375. Males had a median income of $30,417 versus $27,917 for females. The per capita income for the town was $19,396. About 10.1% of families and 10.4% of the population were below the poverty line, including 18.2% of those under the age of eighteen and 7.4% of those sixty five or over.