Laketown, Utah

From Wikipedia, the free encyclopedia

Laketown is a town in Rich County, Utah, United States. The population was 188 at the 2000 census.

[edit] Geography

Laketown is located at 41°49'28" North, 111°19'17" West (41.824327, -111.321339)^GR1.

According to the United States Census Bureau, the town has a total area of 2.6 km² (1.0 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 188 people, 60 households, and 51 families residing in the town. The population density was 71.9/km² (186.3/mi²). There were 89 housing units at an average density of 34.0/km² (88.2/mi²). The racial makeup of the town was 96.28% White, 3.19% Asian, and 0.53% from two or more races.

There were 60 households out of which 48.3% had children under the age of 18 living with them, 80.0% were married couples living together, 1.7% had a female householder with no husband present, and 15.0% were non-families. 15.0% of all households were made up of individuals and 8.3% had someone living alone who was 65 years of age or older. The average household size was 3.13 and the average family size was 3.51.

In the town the population was spread out with 32.4% under the age of 18, 8.5% from 18 to 24, 18.6% from 25 to 44, 27.7% from 45 to 64, and 12.8% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 97.9 males. For every 100 females age 18 and over, there were 111.7 males.

The median income for a household in the town was $60,893, and the median income for a family was $65,000. Males had a median income of $40,972 versus $31,875 for females. The per capita income for the town was $23,519. About 3.6% of families and 7.0% of the population were below the poverty line, including 13.3% of those under the age of eighteen and 10.5% of those sixty five or over.