Gibson, Iowa

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Gibson is a city in Keokuk County, Iowa, United States. The population was 92 at the 2000 census.


Gibson is located at 41°28′50″N 92°23′35″W / 41.48056°N 92.39306°W / 41.48056; -92.39306 (41.480571, -92.393031)[1].

According to the United States Census Bureau, the city has a total area of 0.1 square miles (0.2 km²), all of it land.


As of the census[2] of 2000, there were 92 people, 38 households, and 19 families residing in the city. The population density was 1,389.9 people per square mile (507.4/km²). There were 41 housing units at an average density of 619.4/sq mi (226.1/km²). The racial makeup of the city was 100.00% White.

There were 38 households out of which 36.8% had children under the age of 18 living with them, 44.7% were married couples living together, 5.3% had a female householder with no husband present, and 47.4% were non-families. 44.7% of all households were made up of individuals and 23.7% had someone living alone who was 65 years of age or older. The average household size was 2.42 and the average family size was 3.60.

In the city the population was spread out with 33.7% under the age of 18, 9.8% from 18 to 24, 31.5% from 25 to 44, 6.5% from 45 to 64, and 18.5% who were 65 years of age or older. The median age was 32 years. For every 100 females there were 91.7 males. For every 100 females age 18 and over, there were 96.8 males.

The median income for a household in the city was $39,375, and the median income for a family was $45,625. Males had a median income of $21,094 versus $27,917 for females. The per capita income for the city was $12,881. There were 9.5% of families and 6.7% of the population living below the poverty line, including 7.5% of under eighteens and none of those over 64.


Delta | Gibson | Harper | Hayesville | Hedrick | Keota | Keswick | Kinross | Martinsburg | North English‡ | Ollie | Richland | Sigourney | South English | Thornburg | Webster | What Cheer

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