Sutherland, Iowa

From Wikipedia, the free encyclopedia

Sutherland is a city in O'Brien County, Iowa, United States. The population was 707 at the 2000 census.

[edit] Geography

Sutherland is located at 42°58′21″N, 95°29′49″W (42.972364, -95.496923)^GR1.

According to the United States Census Bureau, the city has a total area of 2.3 km² (0.9 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 707 people, 315 households, and 200 families residing in the city. The population density was 313.8/km² (810.2/mi²). There were 348 housing units at an average density of 154.4/km² (398.8/mi²). The racial makeup of the city was 98.44% White, 0.14% African American, 0.71% Native American, 0.28% from other races, and 0.42% from two or more races. Hispanic or Latino of any race were 0.42% of the population.

There were 315 households out of which 22.2% had children under the age of 18 living with them, 52.4% were married couples living together, 7.0% had a female householder with no husband present, and 36.5% were non-families. 33.7% of all households were made up of individuals and 21.3% had someone living alone who was 65 years of age or older. The average household size was 2.15 and the average family size was 2.71.

In the city the population was spread out with 20.2% under the age of 18, 5.7% from 18 to 24, 20.5% from 25 to 44, 22.5% from 45 to 64, and 31.1% who were 65 years of age or older. The median age was 47 years. For every 100 females there were 87.5 males. For every 100 females age 18 and over, there were 82.5 males.

The median income for a household in the city was $31,985, and the median income for a family was $39,250. Males had a median income of $30,000 versus $19,432 for females. The per capita income for the city was $16,345. About 2.0% of families and 3.9% of the population were below the poverty line, including 1.6% of those under age 18 and 3.8% of those age 65 or over.