Arlington, Iowa

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Arlington is a city in Fayette County, Iowa, United States. The population was 490 at the 2000 census. It was originally called Brush Creek, Iowa. [1]



Arlington is located at 42°44′53″N 91°40′18″W / 42.74806°N 91.67167°W / 42.74806; -91.67167 (42.748095, -91.671545)[1] just north of the headwaters of the Maquoketa River.

According to the United States Census Bureau, the city has a total area of 1.0 square miles (2.7 km²), all of it land.


As of the census[2] of 2000, there were 490 people, 212 households, and 140 families residing in the city. The population density was 467.4 people per square mile (180.2/km²). There were 230 housing units at an average density of 219.4/sq mi (84.6/km²). The racial makeup of the city was 99.39% White, 0.41% Native American and 0.20% Asian.

There were 212 households out of which 25.9% had children under the age of 18 living with them, 52.4% were married couples living together, 9.4% had a female householder with no husband present, and 33.5% were non-families. 30.2% of all households were made up of individuals and 17.5% had someone living alone who was 65 years of age or older. The average household size was 2.31 and the average family size was 2.87.

In the city the population was spread out with 25.9% under the age of 18, 5.3% from 18 to 24, 22.4% from 25 to 44, 24.7% from 45 to 64, and 21.6% who were 65 years of age or older. The median age was 43 years. For every 100 females there were 83.5 males. For every 100 females age 18 and over, there were 83.3 males.

The median income for a household in the city was $30,357, and the median income for a family was $37,083. Males had a median income of $25,938 versus $17,031 for females. The per capita income for the city was $14,643. About 11.9% of families and 14.9% of the population were below the poverty line, including 24.6% of those under age 18 and 12.5% of those age 65 or over.


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