Melvin, Iowa

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Melvin is a city in Osceola County, Iowa, United States. The population was 243 at the 2000 census.

Geography

Melvin is located at 43°17′12″N 95°36′27″W / 43.28667°N 95.6075°W / 43.28667; -95.6075 (43.286659, -95.607462)[1].

According to the United States Census Bureau, the city has a total area of 0.2 square miles (0.4 km²), all of it land.

Demographics

As of the census[2] of 2000, there were 243 people, 117 households, and 67 families residing in the city. The population density was 1,403.5 people per square mile (551.9/km²). There were 134 housing units at an average density of 774.0/sq mi (304.3/km²). The racial makeup of the city was 99.18% White and 0.82% Native American.

There were 117 households out of which 20.5% had children under the age of 18 living with them, 53.0% were married couples living together, 3.4% had a female householder with no husband present, and 42.7% were non-families. 39.3% of all households were made up of individuals and 27.4% had someone living alone who was 65 years of age or older. The average household size was 2.08 and the average family size was 2.82.

In the city the population was spread out with 19.3% under the age of 18, 4.9% from 18 to 24, 22.2% from 25 to 44, 19.8% from 45 to 64, and 33.7% who were 65 years of age or older. The median age was 47 years. For every 100 females there were 96.0 males. For every 100 females age 18 and over, there were 84.9 males.

The median income for a household in the city was $27,750, and the median income for a family was $36,250. Males had a median income of $25,250 versus $18,250 for females. The per capita income for the city was $17,827. About 2.9% of families and 8.5% of the population were below the poverty line, including 18.4% of those under the age of eighteen and 7.9% of those sixty five or over.

References

Ashton | Harris | Melvin | Ocheyedan | Sibley

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