Granger is a town in Sweetwater County, Wyoming, United States. The population was 146 at the 2000 census.
Geography
Granger is located at 41°35′39″N 109°58′0″W / 41.59417°N 109.96667°W / 41.59417; -109.96667 (41.594036, -109.966607)[3].
According to the United States Census Bureau, the town has a total area of 2.5 square miles (6.4 km²), all of it land.
Demographics
As of the census[1] of 2000, there were 146 people, 54 households, and 40 families residing in the town. The population density was 59.0 people per square mile (22.8/km²). There were 76 housing units at an average density of 30.7/sq mi (11.9/km²). The racial makeup of the town was 82.19% White, 8.22% from other races, and 9.59% from two or more races. Hispanic or Latino of any race were 22.60% of the population.
There were 54 households out of which 35.2% had children under the age of 18 living with them, 68.5% were married couples living together, 5.6% had a female householder with no husband present, and 25.9% were non-families. 24.1% of all households were made up of individuals and 5.6% had someone living alone who was 65 years of age or older. The average household size was 2.70 and the average family size was 3.18.
In the town the population was spread out with 28.8% under the age of 18, 10.3% from 18 to 24, 18.5% from 25 to 44, 33.6% from 45 to 64, and 8.9% who were 65 years of age or older. The median age was 38 years. For every 100 females there were 100.0 males. For every 100 females age 18 and over, there were 116.7 males.
The median income for a household in the town was $46,563, and the median income for a family was $52,083. Males had a median income of $45,750 versus $19,375 for females. The per capita income for the town was $17,764. There were 10.5% of families and 12.3% of the population living below the poverty line, including 5.7% of under eighteens and 21.1% of those over 64.
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