Ely, Iowa

From Wikipedia, the free encyclopedia

Ely is a city in Linn County, Iowa, United States. The population was 1,149 at the 2000 census.

[edit] Geography

Ely is located at 41°52′32″N, 91°34′59″W (41.875431, -91.583061)^GR1.

According to the United States Census Bureau, the city has a total area of 3.5 km² (1.4 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 1,149 people, 424 households, and 315 families residing in the city. The population density was 328.6/km² (848.1/mi²). There were 434 housing units at an average density of 124.1/km² (320.3/mi²). The racial makeup of the city was 98.09% White, 0.35% African American, 0.26% Native American, 0.17% Asian, 0.09% from other races, and 1.04% from two or more races. Hispanic or Latino of any race were 0.96% of the population.

There were 424 households out of which 42.2% had children under the age of 18 living with them, 61.8% were married couples living together, 9.4% had a female householder with no husband present, and 25.7% were non-families. 17.9% of all households were made up of individuals and 6.4% had someone living alone who was 65 years of age or older. The average household size was 2.71 and the average family size was 3.10.

In the city the population was spread out with 29.2% under the age of 18, 9.6% from 18 to 24, 34.0% from 25 to 44, 20.5% from 45 to 64, and 6.7% who were 65 years of age or older. The median age was 32 years. For every 100 females there were 98.8 males. For every 100 females age 18 and over, there were 91.3 males.

The median income for a household in the city was $57,250, and the median income for a family was $62,500. Males had a median income of $41,292 versus $29,286 for females. The per capita income for the city was $20,936. About 2.4% of families and 5.8% of the population were below the poverty line, including 7.0% of those under age 18 and 8.5% of those age 65 or over.