Woodland Hills, Utah

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Woodland Hills is a city in Utah County, Utah, United States. It is part of the ProvoOrem, Utah Metropolitan Statistical Area. The population was 941 at the 2000 census. At that time Woodland Hills was a town, but it became a city at the end of 2000.[3]

According to the 2000 Census, Woodland Hills has the unique distinction of having the highest median income in the state of Utah.[4]



Woodland Hills is located at 40°0′49″N 111°39′14″W / 40.01361°N 111.65389°W / 40.01361; -111.65389 (40.013495, -111.654019).[5]

According to the United States Census Bureau, the city has a total area of 2.7 square miles (7.1 km²), all of it land.


As of the census[1] of 2000, there were 941 people, 500 households, and 207 families residing in the town. The population density was 343.3 people per square mile (132.6/km²). There were 229 housing units at an average density of 83.5/sq mi (32.3/km²). The racial makeup of the town was 97.24% White, 0.21% African American, 0.21% Native American, 0.11% Asian, 1.28% from other races, and 0.96% from two or more races. Hispanic or Latino of any race were 2.44% of the population.

There were 220 households out of which 55.5% had children under the age of 18 living with them, 90.9% were married couples living together, 2.3% had a female householder with no husband present, and 5.9% were non-families. 3.6% of all households were made up of individuals and 0.5% had someone living alone who was 65 years of age or older. The average household size was 4.28 and the average family size was 4.44.

In the town the population was spread out with 43.6% under the age of 18, 8.1% from 18 to 24, 23.0% from 25 to 44, 18.4% from 45 to 64, and 7.0% who were 65 years of age or older. The median age was 22 years. For every 100 females there were 101.1 males. For every 100 females age 18 and over, there were 97.4 males.

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