Woodruff, Utah

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Woodruff is a town in Rich County, Utah, United States. The population was 194 at the 2000 census.

Contents

Geography

Woodruff is located at 41°31′22″N 111°9′41″W / 41.52278°N 111.16139°W / 41.52278; -111.16139 (41.522888, -111.161383).[3]

According to the United States Census Bureau, the town has a total area of 0.5 square miles (1.4 km²), all of it land.

Demographics

As of the census[1] of 2000, there were 194 people, 60 households, and 49 families residing in the town. The population density was 361.4 people per square mile (138.7/km²). There were 75 housing units at an average density of 139.7/sq mi (53.6/km²). The racial makeup of the town was 100.00% White.

There were 60 households out of which 51.7% had children under the age of 18 living with them, 70.0% were married couples living together, 8.3% had a female householder with no husband present, and 16.7% were non-families. 13.3% of all households were made up of individuals and 1.7% had someone living alone who was 65 years of age or older. The average household size was 3.23 and the average family size was 3.62.

In the town the population was spread out with 42.8% under the age of 18, 5.2% from 18 to 24, 24.2% from 25 to 44, 15.5% from 45 to 64, and 12.4% who were 65 years of age or older. The median age was 26 years. For every 100 females there were 104.2 males. For every 100 females age 18 and over, there were 109.4 males.

The median income for a household in the town was $43,000, and the median income for a family was $42,750. Males had a median income of $38,333 versus $16,875 for females. The per capita income for the town was $13,976. About 10.5% of families and 18.7% of the population were below the poverty line, including 25.0% of those under the age of eighteen and 11.4% of those sixty five or over.

Notable residents

References

Randolph

Garden City | Laketown | Woodruff

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