Hudson, Wyoming

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Hudson is a town in Fremont County, Wyoming, United States. The population was 407 at the 2000 census.



Hudson is located at 42°54′16″N 108°34′59″W / 42.90444°N 108.58306°W / 42.90444; -108.58306 (42.904313, -108.582978)[3].

According to the United States Census Bureau, the town has a total area of 0.4 square miles (1.1 km²), all of it land.


As of the census[1] of 2000, there were 407 people, 171 households, and 112 families residing in the town. The population density was 968.4 people per square mile (374.2/km²). There were 209 housing units at an average density of 497.3/sq mi (192.1/km²). The racial makeup of the town was 92.63% White, 0.25% African American, 2.21% Native American, 1.47% from other races, and 3.44% from two or more races. Hispanic or Latino of any race were 4.91% of the population.

There were 171 households out of which 26.3% had children under the age of 18 living with them, 52.0% were married couples living together, 9.9% had a female householder with no husband present, and 34.5% were non-families. 32.2% of all households were made up of individuals and 13.5% had someone living alone who was 65 years of age or older. The average household size was 2.38 and the average family size was 3.03.

In the town the population was spread out with 25.8% under the age of 18, 6.1% from 18 to 24, 26.5% from 25 to 44, 24.8% from 45 to 64, and 16.7% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 99.5 males. For every 100 females age 18 and over, there were 93.6 males.

The median income for a household in the town was $26,563, and the median income for a family was $34,375. Males had a median income of $30,469 versus $19,125 for females. The per capita income for the town was $15,515. About 12.5% of families and 13.1% of the population were below the poverty line, including 16.1% of those under age 18 and 16.0% of those age 65 or over.


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