Wilhoit, Arizona

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Wilhoit is a census-designated place (CDP) in Yavapai County, Arizona, United States. The population was 664 at the 2000 census.

Geography

Wilhoit is located at 34°25′8″N 112°36′16″W / 34.41889°N 112.60444°W / 34.41889; -112.60444 (34.418777, -112.604547).[1]

According to the United States Census Bureau, the CDP has a total area of 15.7 square miles (40.6 km²), all of it land.

Demographics

As of the census[2] of 2000, there were 664 people, 298 households, and 187 families residing in the CDP. The population density was 42.4 people per square mile (16.4/km²). There were 363 housing units at an average density of 23.2/sq mi (8.9/km²). The racial makeup of the CDP was 95.63% White, 0.90% Native American, 0.30% Asian, 1.81% from other races, and 1.36% from two or more races. 5.72% of the population were Hispanic or Latino of any race.

There were 298 households out of which 17.1% had children under the age of 18 living with them, 55.0% were married couples living together, 6.7% had a female householder with no husband present, and 37.2% were non-families. 32.2% of all households were made up of individuals and 16.1% had someone living alone who was 65 years of age or older. The average household size was 2.23 and the average family size was 2.81.

In the CDP the population was spread out with 19.1% under the age of 18, 4.5% from 18 to 24, 17.6% from 25 to 44, 33.7% from 45 to 64, and 25.0% who were 65 years of age or older. The median age was 50 years. For every 100 females there were 90.3 males. For every 100 females age 18 and over, there were 95.3 males.

The median income for a household in the CDP was $23,500, and the median income for a family was $30,938. Males had a median income of $27,222 versus $25,938 for females. The per capita income for the CDP was $14,466. About 10.1% of families and 16.8% of the population were below the poverty line, including 23.1% of those under age 18 and 9.7% of those age 65 or over.

References

Cottonwood | Peoria‡ | Prescott | Sedona

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