Rolling Hills, Wyoming

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Rolling Hills is a town in Converse County, Wyoming, United States. The population was 449 at the 2000 census.



Rolling Hills is located at 42°54′27″N 105°50′29″W / 42.9075°N 105.84139°W / 42.9075; -105.84139 (42.907620, -105.841515)[3].

According to the United States Census Bureau, the town has a total area of 0.7 square miles (1.8 km²), all of it land.


As of the census[1] of 2000, there were 449 people, 135 households, and 115 families residing in the town. The population density was 633.9 people per square mile (244.2/km²). There were 143 housing units at an average density of 201.9/sq mi (77.8/km²). The racial makeup of the town was 96.66% White, 0.22% African American, 0.45% Native American, 0.67% Asian, 0.22% from other races, and 1.78% from two or more races. Hispanic or Latino of any race were 2.90% of the population.

There were 135 households out of which 52.6% had children under the age of 18 living with them, 76.3% were married couples living together, 5.9% had a female householder with no husband present, and 14.8% were non-families. 6.7% of all households were made up of individuals and 0.7% had someone living alone who was 65 years of age or older. The average household size was 3.33 and the average family size was 3.57.

In the town the population was spread out with 35.9% under the age of 18, 8.2% from 18 to 24, 30.3% from 25 to 44, 22.9% from 45 to 64, and 2.7% who were 65 years of age or older. The median age was 31 years. For every 100 females there were 103.2 males. For every 100 females age 18 and over, there were 107.2 males.

The median income for a household in the town was $50,625, and the median income for a family was $47,917. Males had a median income of $39,250 versus $21,250 for females. The per capita income for the town was $21,767. About 1.6% of families and 3.5% of the population were below the poverty line, including 3.8% of those under age 18 and none of those age 65 or over.

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