Rowley, Iowa

From Wikipedia, the free encyclopedia

Rowley is a city in Buchanan County, Iowa, United States. The population was 290 at the 2000 census.

1 Geography
2 Demographics
3 Constructions
4 External links

[edit] Geography

Rowley is located at 42°22′6″N, 91°50′42″W (42.368380, -91.845015)^GR1.

According to the United States Census Bureau, the city has a total area of 0.9 km² (0.4 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 290 people, 111 households, and 85 families residing in the city. The population density was 311.0/km² (806.3/mi²). There were 114 housing units at an average density of 122.3/km² (317.0/mi²). The racial makeup of the city was 99.31% White (288 people) and 0.69% Asian (2 people). None of the population is Hispanic or Latino of any race.

There were 111 households out of which 36.0% had children under the age of 18 living with them, 71.2% were married couples living together, 4.5% had a female householder with no husband present, and 23.4% were non-families. 21.6% of all households were made up of individuals and 12.6% had someone living alone who was 65 years of age or older. The average household size was 2.61 and the average family size was 3.06.

In the city the population was spread out with 26.6% under the age of 18, 7.2% from 18 to 24, 28.3% from 25 to 44, 22.1% from 45 to 64, and 15.9% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 87.1 males. For every 100 females age 18 and over, there were 91.9 males.

The median income for a household in the city was $36,563, and the median income for a family was $41,875. Males had a median income of $35,625 versus $24,375 for females. The per capita income for the city was $17,315. About 6.0% of families and 7.1% of the population were below the poverty line, including 12.0% of those under the age of eighteen and 10.0% of those sixty five or over.