Kingston, Utah

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Kingston is a town in Piute County, Utah, in the United States. The population was 142 at the 2000 census.



Kingston is located at 38°12′27″N 112°11′12″W / 38.2075°N 112.18667°W / 38.2075; -112.18667 (38.207524, -112.186668).[3]

According to the United States Census Bureau, the town has a total area of 5.3 square miles (13.8 km²), all of it land.


As of the census[1] of 2000, there were 142 people, 47 households, and 38 families residing in the town. The population density was 26.7 people per square mile (10.3/km²). There were 59 housing units at an average density of 11.1/sq mi (4.3/km²). The racial makeup of the town was 98.59% White, 0.70% from other races, and 0.70% from two or more races. Hispanic or Latino of any race were 0.70% of the population.

There were 47 households out of which 31.9% had children under the age of 18 living with them, 80.9% were married couples living together, and 19.1% were non-families. 19.1% of all households were made up of individuals and 2.1% had someone living alone who was 65 years of age or older. The average household size was 3.02 and the average family size was 3.50.

In the town the population was spread out with 29.6% under the age of 18, 6.3% from 18 to 24, 22.5% from 25 to 44, 26.1% from 45 to 64, and 15.5% who were 65 years of age or older. The median age was 35 years. For every 100 females there were 108.8 males. For every 100 females age 18 and over, there were 112.8 males.

The median income for a household in the town was $23,750, and the median income for a family was $27,083. Males had a median income of $16,667 versus $16,750 for females. The per capita income for the town was $12,492. There were 22.0% of families and 32.8% of the population living below the poverty line, including 51.4% of under eighteens and none of those over 64.

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