Little America, Wyoming

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Little America is a census-designated place (CDP) in Sweetwater County, Wyoming, United States. The population was 56 at the 2000 census.

[edit] Geography

Little America is located at 41°32′43″N, 109°52′20″W (41.545398, -109.872112)^GR1.

According to the United States Census Bureau, the CDP has a total area of 20.1 km² (7.8 mi²). 20.1 km² (7.8 mi²) of it is land and 0.13% water.

[edit] Demographics

As of the census^GR2 of 2000, there were 56 people, 24 households, and 13 families residing in the CDP. The population density was 2.8/km² (7.2/mi²). There were 48 housing units at an average density of 2.4/km² (6.2/mi²). The racial makeup of the CDP was 71.43% White, 28.57% from other races. Hispanic or Latino of any race were 44.64% of the population.

There were 24 households out of which 45.8% had children under the age of 18 living with them, 33.3% were married couples living together, 20.8% had a female householder with no husband present, and 45.8% were non-families. 41.7% of all households were made up of individuals and none had someone living alone who was 65 years of age or older. The average household size was 2.33 and the average family size was 3.23.

In the CDP the population was spread out with 32.1% under the age of 18, 28.6% from 18 to 24, 28.6% from 25 to 44, 10.7% from 45 to 64, . The median age was 21 years. For every 100 females there were 100.0 males. For every 100 females age 18 and over, there were 123.5 males.

The median income for a household in the CDP was $18,125, and the median income for a family was $18,750. Males had a median income of $12,250 versus $8,750 for females. The per capita income for the CDP was $7,407. None of the population or the families were below the poverty line.