Wilson, Wyoming

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Wilson is a census-designated place (CDP) in Teton County, Wyoming, United States. The population was 1,294 at the 2000 census.

[edit] Geography

Wilson is located at 43°29′11″N, 110°52′45″W (43.486300, -110.879279).^GR1

According to the United States Census Bureau, the CDP has a total area of 60.3 km² (23.3 mi²). 59.9 km² (23.1 mi²) of it is land and 0.4 km² (0.2 mi²) of it (0.69%) is water.

[edit] Demographics

As of the census^GR2 of 2000, there were 1,294 people, 563 households, and 305 families residing in the CDP. The population density was 21.6/km² (56.0/mi²). There were 706 housing units at an average density of 11.8/km² (30.5/mi²). The racial makeup of the CDP was 97.68% White, 0.08% African American, 0.15% Native American, 0.46% Asian, 0.54% from other races, and 1.08% from two or more races. Hispanic or Latino of any race were 1.00% of the population.

There were 563 households out of which 25.2% had children under the age of 18 living with them, 47.6% were married couples living together, 3.7% had a female householder with no husband present, and 45.8% were non-families. 28.6% of all households were made up of individuals and 2.7% had someone living alone who was 65 years of age or older. The average household size was 2.27 and the average family size was 2.86.

In the CDP the population was spread out with 20.9% under the age of 18, 4.3% from 18 to 24, 39.1% from 25 to 44, 28.3% from 45 to 64, and 7.4% who were 65 years of age or older. The median age was 37 years. For every 100 females there were 117.1 males. For every 100 females age 18 and over, there were 120.0 males.

The median income for a household in the CDP was $69,327, and the median income for a family was $93,354. Males had a median income of $30,455 versus $61,635 for females. The per capita income for the CDP was $65,489. None of the families and 6.4% of the population were living below the poverty line, including no under eighteen and none of those over 64.