Harris, Iowa

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Harris is a city in Osceola County, Iowa, United States. The population was 200 at the 2000 census.


Harris is located at 43°26′44″N 95°25′59″W / 43.44556°N 95.43306°W / 43.44556; -95.43306 (43.445575, -95.433168)[1].

According to the United States Census Bureau, the city has a total area of 0.8 square miles (2.0 km²), all of it land.


As of the census[2] of 2000, there were 200 people, 86 households, and 47 families residing in the city. The population density was 252.6 people per square mile (97.7/km²). There were 91 housing units at an average density of 114.9/sq mi (44.5/km²). The racial makeup of the city was 95.00% White, 0.50% Native American, 1.50% from other races, and 3.00% from two or more races. Hispanic or Latino of any race were 4.00% of the population.

There were 86 households out of which 25.6% had children under the age of 18 living with them, 45.3% were married couples living together, 7.0% had a female householder with no husband present, and 44.2% were non-families. 37.2% 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 2.33 and the average family size was 3.19.

In the city the population was spread out with 25.5% under the age of 18, 10.5% from 18 to 24, 33.0% from 25 to 44, 21.0% from 45 to 64, and 10.0% who were 65 years of age or older. The median age was 35 years. For every 100 females there were 94.2 males. For every 100 females age 18 and over, there were 106.9 males.

The median income for a household in the city was $35,625, and the median income for a family was $46,250. Males had a median income of $27,500 versus $23,571 for females. The per capita income for the city was $15,788. About 13.7% of families and 14.1% of the population were below the poverty line, including 29.1% of those under the age of eighteen and 26.1% of those sixty five or over.


Ashton | Harris | Melvin | Ocheyedan | Sibley

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