Cleveland, Utah

From Wikipedia, the free encyclopedia

Cleveland is a town in Emery County, Utah, in the United States. As of the 2000 census, the town population was 508.

[edit] Geography

Cleveland is located at 39°20′53″N, 110°51′19″W (39.347921, -110.855178)^GR1.

According to the United States Census Bureau, the town has a total area of 2.3 km² (0.9 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 508 people, 164 households, and 139 families residing in the town. The population density was 220.4/km² (568.2/mi²). There were 173 housing units at an average density of 75.1/km² (193.5/mi²). The racial makeup of the town was 97.24% White, 0.20% Native American, 0.20% from other races, and 2.36% from two or more races. Hispanic or Latino of any race were 0.59% of the population.

There were 164 households out of which 48.2% had children under the age of 18 living with them, 76.2% were married couples living together, 6.1% had a female householder with no husband present, and 15.2% were non-families. 15.2% of all households were made up of individuals and 6.7% had someone living alone who was 65 years of age or older. The average household size was 3.10 and the average family size was 3.44.

In the town the population was spread out with 33.9% under the age of 18, 10.8% from 18 to 24, 25.0% from 25 to 44, 20.3% from 45 to 64, and 10.0% who were 65 years of age or older. The median age was 30 years. For every 100 females there were 92.4 males. For every 100 females age 18 and over, there were 95.3 males.

The median income for a household in the town was $33,500, and the median income for a family was $43,000. Males had a median income of $33,750 versus $14,286 for females. The per capita income for the town was $11,774. About 6.3% of families and 8.1% of the population were below the poverty line, including 8.5% of those under age 18 and 10.4% of those age 65 or over.