Harcourt, Iowa

From Wikipedia, the free encyclopedia

Harcourt is a city in Webster County, Iowa, United States. The population was 340 at the 2000 census.

[edit] Geography

Harcourt is located at 42°15′43″N, 94°10′36″W (42.262056, -94.176589)^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 340 people, 137 households, and 86 families residing in the city. The population density was 230.3/km² (592.5/mi²). There were 149 housing units at an average density of 100.9/km² (259.7/mi²). The racial makeup of the city was 98.53% White, 0.29% African American, 0.88% Native American and 0.29% Asian. Hispanic or Latino of any race were 0.88% of the population.

There were 137 households out of which 29.9% had children under the age of 18 living with them, 59.1% were married couples living together, 3.6% had a female householder with no husband present, and 36.5% were non-families. 35.0% of all households were made up of individuals and 18.2% had someone living alone who was 65 years of age or older. The average household size was 2.48 and the average family size was 3.24.

In the city the population was spread out with 26.2% under the age of 18, 9.7% from 18 to 24, 26.8% from 25 to 44, 19.1% from 45 to 64, and 18.2% who were 65 years of age or older. The median age was 38 years. For every 100 females there were 98.8 males. For every 100 females age 18 and over, there were 90.2 males.

The median income for a household in the city was $37,212, and the median income for a family was $42,500. Males had a median income of $25,625 versus $18,854 for females. The per capita income for the city was $22,461. None of the families and 3.3% of the population were living below the poverty line, including no under eighteens and 4.5% of those over 64.