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Farragut, Iowa

Farragut is a city located in Fremont County, Iowa. As of the 2000 census, the city had a total population of 509.

Geography Farragut is located at 40°43'18" North, 95°28'49" West (40.721645, -95.480226)1.

According to the United States Census Bureau, the city has a total area of 1.0 km² (0.4 mi²). 1.0 km² (0.4 mi²) of it is land and none of it is covered by water.

Demographics As of the census of 2000, there are 509 people, 221 households, and 144 families residing in the city. The population density is 531.2/km² (1,390.2/mi²). There are 234 housing units at an average density of 244.2/km² (639.1/mi²). The racial makeup of the city is 97.64% White, 0.00% African American, 0.00% Native American, 0.39% Asian, 0.00% Pacific Islander, 0.98% from other races, and 0.98% from two or more races. 1.57% of the population are Hispanic or Latino of any race.

There are 221 households out of which 27.6% have children under the age of 18 living with them, 55.7% are married couples living together, 7.2% have a female householder with no husband present, and 34.8% are non-families. 32.1% of all households are made up of individuals and 22.2% have someone living alone who is 65 years of age or older. The average household size is 2.30 and the average family size is 2.85.

In the city the population is spread out with 25.9% under the age of 18, 5.1% from 18 to 24, 23.2% from 25 to 44, 22.0% from 45 to 64, and 23.8% who are 65 years of age or older. The median age is 42 years. For every 100 females there are 85.1 males. For every 100 females age 18 and over, there are 80.4 males.

The median income for a household in the city is $34,250, and the median income for a family is $47,813. Males have a median income of $31,397 versus $21,000 for females. The per capita income for the city is $16,667. 7.5% of the population and 3.9% of families are below the poverty line. Out of the total people living in poverty, 0.0% are under the age of 18 and 10.9% are 65 or older.



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