Palmyra, Utah

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Palmyra is a census-designated place (CDP) in Utah County, Utah, United States. The population was 485 at the 2000 census.

[edit] Geography

Palmyra is located at 40°7′36″N, 111°41′7″W (40.126648, -111.685291)^GR1.

According to the United States Census Bureau, the CDP has a total area of 30.4 km² (11.7 mi²). 29.8 km² (11.5 mi²) of it is land and 0.6 km² (0.2 mi²) of it (2.05%) is water.

[edit] Demographics

As of the census^GR2 of 2000, there were 485 people, 142 households, and 123 families residing in the CDP. The population density was 16.3/km² (42.2/mi²). There were 148 housing units at an average density of 5.0/km² (12.9/mi²). The racial makeup of the CDP was 97.53% White, 0.21% Native American, 2.06% Asian, 0.21% from other races. Hispanic or Latino of any race were 3.71% of the population.

There were 142 households out of which 40.8% had children under the age of 18 living with them, 80.3% were married couples living together, 4.2% had a female householder with no husband present, and 12.7% were non-families. 12.0% of all households were made up of individuals and 4.2% had someone living alone who was 65 years of age or older. The average household size was 3.42 and the average family size was 3.72.

In the CDP the population was spread out with 34.0% under the age of 18, 10.1% from 18 to 24, 23.5% from 25 to 44, 21.6% from 45 to 64, and 10.7% who were 65 years of age or older. The median age was 31 years. For every 100 females there were 119.5 males. For every 100 females age 18 and over, there were 114.8 males.

The median income for a household in the CDP was $58,750, and the median income for a family was $90,122. Males had a median income of $42,917 versus $24,531 for females. The per capita income for the CDP was $18,967. About 5.8% of families and 8.5% of the population were below the poverty line, including 17.0% of those under age 18 and none of those age 65 or over.