Salina, Utah

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Salina is a city in Sevier County, Utah, United States. The population was 2,393 at the 2000 census.

1 Geography
2 Demographics
3 See also
4 External links

[edit] Geography

Salina is located at 38°56′57″N, 111°51′41″W (38.949232, -111.861447)^GR1.

According to the United States Census Bureau, the city has a total area of 15.9 km² (6.2 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 2,393 people, 808 households, and 631 families residing in the city. The population density was 150.2/km² (388.9/mi²). There were 878 housing units at an average density of 55.1/km² (142.7/mi²). The racial makeup of the city was 96.66% White, 1.17% Native American, 0.13% Asian, 0.17% Pacific Islander, 0.75% from other races, and 1.13% from two or more races. Hispanic or Latino of any race were 2.80% of the population.

There were 808 households out of which 45.0% had children under the age of 18 living with them, 65.6% were married couples living together, 9.3% had a female householder with no husband present, and 21.8% were non-families. 19.1% of all households were made up of individuals and 9.3% had someone living alone who was 65 years of age or older. The average household size was 2.96 and the average family size was 3.41.

In the city the population was spread out with 35.2% under the age of 18, 10.9% from 18 to 24, 24.2% from 25 to 44, 17.3% from 45 to 64, and 12.3% who were 65 years of age or older. The median age was 28 years. For every 100 females there were 99.3 males. For every 100 females age 18 and over, there were 95.5 males.

The median income for a household in the city was $34,886, and the median income for a family was $38,816. Males had a median income of $32,204 versus $16,250 for females. The per capita income for the city was $12,967. About 10.4% of families and 13.8% of the population were below the poverty line, including 17.5% of those under age 18 and 7.7% of those age 65 or over.