Stout, Iowa

From Wikipedia, the free encyclopedia

Stout is a city in Grundy County, Iowa, United States. The population was 217 at the 2000 census.

[edit] Geography

Stout is located at 42°31′38″N, 92°42′41″W (42.527131, -92.711383)^GR1.

According to the United States Census Bureau, the city has a total area of 0.8 km² (0.3 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 217 people, 75 households, and 60 families residing in the city. The population density was 270.3/km² (697.3/mi²). There were 77 housing units at an average density of 95.9/km² (247.4/mi²). The racial makeup of the city was 99.08% White, and 0.92% from two or more races.

There were 75 households out of which 37.3% had children under the age of 18 living with them, 68.0% were married couples living together, 9.3% had a female householder with no husband present, and 18.7% were non-families. 14.7% 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 2.89 and the average family size was 3.23.

In the city the population was spread out with 31.3% under the age of 18, 8.8% from 18 to 24, 26.7% from 25 to 44, 18.0% from 45 to 64, and 15.2% who were 65 years of age or older. The median age was 34 years. For every 100 females there were 104.7 males. For every 100 females age 18 and over, there were 101.4 males.

The median income for a household in the city was $40,781, and the median income for a family was $41,042. Males had a median income of $28,958 versus $23,750 for females. The per capita income for the city was $12,504. About 8.5% of families and 12.5% of the population were below the poverty line, including 22.0% of those under the age of eighteen and none of those sixty five or over.