Albin, Wyoming

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Albin is a town in Laramie County, Wyoming, United States. It is part of the Cheyenne, Wyoming Metropolitan Statistical Area. The population was 120 at the 2000 census.



Albin is located at 41°25′1″N 104°6′1″W / 41.41694°N 104.10028°W / 41.41694; -104.10028 (41.417082, -104.100166)[3], about 34 miles (55 km) ENE of Cheyenne.

According to the United States Census Bureau, the town has a total area of 0.1 square miles (0.4 km²), all of it land.


As of the census[1] of 2000, there were 120 people, 54 households, and 32 families residing in the town. The population density was 850.6 people per square mile (330.9/km²). There were 70 housing units at an average density of 496.2/sq mi (193.1/km²). The racial makeup of the town was 95.83% White, 0.83% Native American, 0.83% Asian, 2.50% from other races. Hispanic or Latino of any race were 9.17% of the population.

There were 54 households out of which 24.1% had children under the age of 18 living with them, 51.9% were married couples living together, 3.7% had a female householder with no husband present, and 38.9% were non-families. 37.0% of all households were made up of individuals and 22.2% had someone living alone who was 65 years of age or older. The average household size was 2.22 and the average family size was 2.91.

In the town the population was spread out with 22.5% under the age of 18, 7.5% from 18 to 24, 20.0% from 25 to 44, 27.5% from 45 to 64, and 22.5% who were 65 years of age or older. The median age was 46 years. For every 100 females there were 87.5 males. For every 100 females age 18 and over, there were 89.8 males.

The median income for a household in the town was $25,625, and the median income for a family was $38,750. Males had a median income of $26,000 versus $18,750 for females. The per capita income for the town was $13,174. There were 6.7% of families and 12.5% of the population living below the poverty line, including 19.0% of under eighteens and 31.3% of those over 64.

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