Russell, Iowa

From Wikipedia, the free encyclopedia

Russell is a city in Lucas County, Iowa, United States. The population was 559 at the 2000 census.

[edit] Geography

Russell is located at 40°58′57″N, 93°11′55″W (40.982388, -93.198642)^GR1.

According to the United States Census Bureau, the city has a total area of 2.7 km² (1.0 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 559 people, 239 households, and 144 families residing in the city. The population density was 207.5/km² (539.7/mi²). There were 265 housing units at an average density of 98.4/km² (255.8/mi²). The racial makeup of the city was 98.57% White, 0.18% Native American, 0.36% Asian, and 0.89% from two or more races. Hispanic or Latino of any race were 0.36% of the population.

There were 239 households out of which 29.3% had children under the age of 18 living with them, 50.2% were married couples living together, 6.7% had a female householder with no husband present, and 39.7% were non-families. 33.1% of all households were made up of individuals and 19.7% had someone living alone who was 65 years of age or older. The average household size was 2.34 and the average family size was 2.97.

In the city the population was spread out with 27.0% under the age of 18, 6.8% from 18 to 24, 24.0% from 25 to 44, 19.3% from 45 to 64, and 22.9% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 89.5 males. For every 100 females age 18 and over, there were 92.5 males.

The median income for a household in the city was $28,125, and the median income for a family was $29,688. Males had a median income of $26,094 versus $16,406 for females. The per capita income for the city was $13,093. About 16.5% of families and 23.6% of the population were below the poverty line, including 42.6% of those under age 18 and 9.0% of those age 65 or over.