Shueyville, Iowa

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Shueyville is a city in Johnson County, Iowa, United States. It part of the Iowa City, Iowa Metropolitan Statistical Area. The population was 250 at the 2000 census.

[edit] Geography

Shueyville is located at 41°50′53″N, 91°38′55″W (41.848155, -91.648679)^GR1.

According to the United States Census Bureau, the city has a total area of 3.9 km² (1.5 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 250 people, 95 households, and 70 families residing in the city. The population density was 64.8/km² (167.6/mi²). There were 97 housing units at an average density of 25.1/km² (65.0/mi²). The racial makeup of the city was 99.20% White, 0.40% Asian, and 0.40% from two or more races. Hispanic or Latino of any race were 0.40% of the population.

There were 95 households out of which 35.8% had children under the age of 18 living with them, 67.4% were married couples living together, 5.3% had a female householder with no husband present, and 25.3% were non-families. 22.1% of all households were made up of individuals and 5.3% had someone living alone who was 65 years of age or older. The average household size was 2.63 and the average family size was 3.07.

In the city the population was spread out with 26.8% under the age of 18, 6.4% from 18 to 24, 33.2% from 25 to 44, 25.6% from 45 to 64, and 8.0% who were 65 years of age or older. The median age was 36 years. For every 100 females there were 101.6 males. For every 100 females age 18 and over, there were 101.1 males.

The median income for a household in the city was $61,875, and the median income for a family was $66,875. Males had a median income of $41,364 versus $26,136 for females. The per capita income for the city was $24,690. None of the families and 1.8% of the population were living below the poverty line.