Buck Grove, Iowa

From Wikipedia, the free encyclopedia

Buck Grove is a city in Crawford County, Iowa, United States. The population was 49 at the 2000 census.

[edit] Geography

Buck Grove is located at 41°55′7″N, 95°23′50″W (41.918646, -95.397265)^GR1.

According to the United States Census Bureau, the city has a total area of 1.0 km² (0.4 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 49 people, 20 households, and 13 families residing in the city. The population density was 51.1/km² (133.6/mi²). There were 21 housing units at an average density of 21.9/km² (57.3/mi²). The racial makeup of the city was 100.00% White.

There were 20 households out of which 25.0% had children under the age of 18 living with them, 60.0% were married couples living together, 5.0% had a female householder with no husband present, and 35.0% were non-families. 30.0% of all households were made up of individuals and 15.0% had someone living alone who was 65 years of age or older. The average household size was 2.45 and the average family size was 2.77.

In the city the population was spread out with 20.4% under the age of 18, 6.1% from 18 to 24, 34.7% from 25 to 44, 28.6% from 45 to 64, and 10.2% who were 65 years of age or older. The median age was 42 years. For every 100 females there were 122.7 males. For every 100 females age 18 and over, there were 116.7 males.

The median income for a household in the city was $25,417, and the median income for a family was $38,750. Males had a median income of $23,906 versus $18,750 for females. The per capita income for the city was $14,492. There were no families and 10.0% of the population living below the poverty line, including no under eighteens and none of those over 64.