Lanesboro, Iowa

From Wikipedia, the free encyclopedia

Lanesboro is a city in Carroll County, Iowa, United States. The population was 152 at the 2000 census.

[edit] Geography

Lanesboro is located at 42°11′3″N, 94°41′33″W (42.184035, -94.692460)^GR1.

According to the United States Census Bureau, the city has a total area of 1.9 km² (0.8 mi²). 1.9 km² (0.7 mi²) of it is land and 1.33% is water.

[edit] Demographics

As of the census^GR2 of 2000, there were 152 people, 71 households, and 42 families residing in the city. The population density was 79.3/km² (204.5/mi²). There were 77 housing units at an average density of 40.2/km² (103.6/mi²). The racial makeup of the city was 98.68% White, 0.66% from other races, and 0.66% from two or more races. Hispanic or Latino of any race were 0.66% of the population.

There were 71 households out of which 23.9% had children under the age of 18 living with them, 53.5% were married couples living together, 2.8% had a female householder with no husband present, and 40.8% were non-families. 36.6% of all households were made up of individuals and 16.9% had someone living alone who was 65 years of age or older. The average household size was 2.14 and the average family size was 2.83.

In the city the population was spread out with 19.7% under the age of 18, 7.2% from 18 to 24, 27.0% from 25 to 44, 27.6% from 45 to 64, and 18.4% who were 65 years of age or older. The median age was 43 years. For every 100 females there were 120.3 males. For every 100 females age 18 and over, there were 117.9 males.

The median income for a household in the city was $25,750, and the median income for a family was $31,250. Males had a median income of $30,000 versus $20,625 for females. The per capita income for the city was $15,397. About 5.9% of families and 10.8% of the population were below the poverty line, including none of those under the age of eighteen and 17.6% of those sixty five or over.