Sanborn, Iowa

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Sanborn is a city in O'Brien County, Iowa, United States. The population was 1,353 at the 2000 census.


Sanborn is located at 43°10′59″N 95°39′24″W / 43.18306°N 95.65667°W / 43.18306; -95.65667 (43.183112, -95.656666)[1].

According to the United States Census Bureau, the city has a total area of 1.8 square miles (4.7 km²), all of it land.


As of the census[2] of 2000, there were 1,353 people, 558 households, and 372 families residing in the city. The population density was 747.1 people per square mile (288.6/km²). There were 593 housing units at an average density of 327.5/sq mi (126.5/km²). The racial makeup of the city was 99.41% White, 0.07% African American, 0.07% Native American, 0.15% from other races, and 0.30% from two or more races. Hispanic or Latino of any race were 0.44% of the population.

There were 558 households out of which 23.8% had children under the age of 18 living with them, 61.6% were married couples living together, 3.9% had a female householder with no husband present, and 33.3% were non-families. 30.8% of all households were made up of individuals and 18.8% had someone living alone who was 65 years of age or older. The average household size was 2.28 and the average family size was 2.84.

In the city the population was spread out with 20.7% under the age of 18, 6.8% from 18 to 24, 20.3% from 25 to 44, 21.4% from 45 to 64, and 30.9% who were 65 years of age or older. The median age was 46 years. For every 100 females there were 86.9 males. For every 100 females age 18 and over, there were 82.8 males.

The median income for a household in the city was $34,250, and the median income for a family was $42,500. Males had a median income of $31,792 versus $19,750 for females. The per capita income for the city was $18,189. About 2.8% of families and 4.8% of the population were below the poverty line, including 4.7% of those under age 18 and 3.2% of those age 65 or over.


Archer | Calumet | Hartley | Paullina | Primghar | Sanborn | Sheldon‡ | Sutherland

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