Aurora, Iowa

From Wikipedia, the free encyclopedia

Aurora is a city in Buchanan County, Iowa, United States. The population was 194 at the 2000 census.

[edit] Geography

Aurora is located at 42°37′5″N, 91°43′43″W (42.618061, -91.728519)^GR1.

According to the United States Census Bureau, the city has a total area of 1.5 km² (0.6 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 194 people, 78 households, and 52 families residing in the city. The population density was 131.4/km² (338.5/mi²). There were 88 housing units at an average density of 59.6/km² (153.5/mi²). The racial makeup of the city was 94.85% White, 0.52% Native American, and 4.64% from two or more races.

There were 78 households out of which 33.3% had children under the age of 18 living with them, 55.1% were married couples living together, 11.5% had a female householder with no husband present, and 32.1% were non-families. 29.5% of all households were made up of individuals and 16.7% had someone living alone who was 65 years of age or older. The average household size was 2.49 and the average family size was 3.09.

In the city the population was spread out with 28.4% under the age of 18, 7.2% from 18 to 24, 26.3% from 25 to 44, 21.6% from 45 to 64, and 16.5% who were 65 years of age or older. The median age was 37 years. For every 100 females there were 94.0 males. For every 100 females age 18 and over, there were 104.4 males.

The median income for a household in the city was $38,750, and the median income for a family was $42,188. Males had a median income of $25,000 versus $22,500 for females. The per capita income for the city was $16,254. About 15.8% of families and 19.5% of the population were below the poverty line, including 28.8% of those under the age of eighteen and 13.3% of those sixty five or over.