Marbleton, Wyoming

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Marbleton is a town in Sublette County, Wyoming, United States. The population was 720 at the 2000 census.



Marbleton is located at 42°33′26″N 110°6′22″W / 42.55722°N 110.10611°W / 42.55722; -110.10611 (42.557108, -110.106196)[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 720 people, 268 households, and 201 families residing in the town. The population density was 1,060.8 people per square mile (408.8/km²). There were 310 housing units at an average density of 456.7/sq mi (176.0/km²). The racial makeup of the town was 96.25% White, 0.42% African American, 0.69% Native American, 0.42% Asian, 0.14% Pacific Islander, 0.56% from other races, and 1.53% from two or more races. Hispanic or Latino of any race were 1.67% of the population.

There were 268 households out of which 42.5% had children under the age of 18 living with them, 64.2% were married couples living together, 6.0% had a female householder with no husband present, and 25.0% were non-families. 20.1% of all households were made up of individuals and 2.2% had someone living alone who was 65 years of age or older. The average household size was 2.69 and the average family size was 3.09.

In the town the population was spread out with 31.1% under the age of 18, 8.1% from 18 to 24, 33.1% from 25 to 44, 22.6% from 45 to 64, and 5.1% who were 65 years of age or older. The median age was 34 years. For every 100 females there were 106.3 males. For every 100 females age 18 and over, there were 105.8 males.

The median income for a household in the town was $41,406, and the median income for a family was $46,250. Males had a median income of $38,250 versus $17,500 for females. The per capita income for the town was $18,446. About 3.2% of families and 4.2% of the population were below the poverty line, including 5.8% of those under age 18 and none of those age 65 or over.

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