Farnhamville, Iowa

From Wikipedia, the free encyclopedia

Farnhamville is a city in Calhoun County, Iowa, United States. The population was 430 at the 2000 census.

[edit] Geography

Farnhamville is located at 42°16′35″N, 94°24′23″W (42.276421, -94.406493)^GR1.

According to the United States Census Bureau, the city has a total area of 1.7 km² (0.7 mi²). 1.7 km² (0.6 mi²) of it is land and 1.52% is water.

[edit] Demographics

As of the census^GR2 of 2000, there were 430 people, 195 households, and 130 families residing in the city. The population density was 255.4/km² (659.6/mi²). There were 213 housing units at an average density of 126.5/km² (326.7/mi²). The racial makeup of the city was 98.14% White, 0.70% Native American, 0.47% from other races, and 0.70% from two or more races. Hispanic or Latino of any race were 0.70% of the population.

There were 195 households out of which 26.7% had children under the age of 18 living with them, 54.9% were married couples living together, 9.2% had a female householder with no husband present, and 33.3% were non-families. 30.8% of all households were made up of individuals and 23.6% had someone living alone who was 65 years of age or older. The average household size was 2.21 and the average family size was 2.69.

In the city the population was spread out with 23.0% under the age of 18, 7.7% from 18 to 24, 20.5% from 25 to 44, 23.3% from 45 to 64, and 25.6% who were 65 years of age or older. The median age was 44 years. For every 100 females there were 86.1 males. For every 100 females age 18 and over, there were 87.0 males.

The median income for a household in the city was $29,107, and the median income for a family was $36,667. Males had a median income of $27,237 versus $22,813 for females. The per capita income for the city was $21,619. About 7.1% of families and 10.0% of the population were below the poverty line, including 11.2% of those under age 18 and 15.7% of those age 65 or over.