Alexander, Iowa

From Wikipedia, the free encyclopedia

Alexander is a city in Franklin County, Iowa, United States. The population was 165 at the 2000 census.

[edit] Geography

Alexander is located at 42°48′17″N, 93°28′39″W (42.804760, -93.477395)^GR1.

According to the United States Census Bureau, the city has a total area of 11.1 km² (4.3 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 165 people, 79 households, and 48 families residing in the city. The population density was 14.9/km² (38.6/mi²). There were 88 housing units at an average density of 7.9/km² (20.6/mi²). The racial makeup of the city was 96.36% White, 3.64% from other races. Hispanic or Latino of any race were 3.64% of the population.

There were 79 households out of which 24.1% had children under the age of 18 living with them, 55.7% were married couples living together, 3.8% had a female householder with no husband present, and 38.0% were non-families. 35.4% of all households were made up of individuals and 27.8% had someone living alone who was 65 years of age or older. The average household size was 2.09 and the average family size was 2.59.

In the city the population was spread out with 17.0% under the age of 18, 7.3% from 18 to 24, 24.2% from 25 to 44, 24.8% from 45 to 64, and 26.7% who were 65 years of age or older. The median age was 46 years. For every 100 females there were 87.5 males. For every 100 females age 18 and over, there were 95.7 males.

The median income for a household in the city was $31,250, and the median income for a family was $33,750. Males had a median income of $27,344 versus $20,833 for females. The per capita income for the city was $14,995. About 6.3% of families and 9.3% of the population were below the poverty line, including none of those under the age of eighteen and 11.3% of those sixty five or over.