Franklin, Iowa

From Wikipedia, the free encyclopedia

Franklin is a city in Lee County, Iowa, United States. The population was 136 at the 2000 census.

[edit] Geography

Franklin is located at 40°39′59″N, 91°30′43″W (40.666460, -91.511828)^GR1.

According to the United States Census Bureau, the city has a total area of 0.4 km² (0.2 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 136 people, 57 households, and 36 families residing in the city. The population density was 328.2/km² (851.3/mi²). There were 58 housing units at an average density of 140.0/km² (363.1/mi²). The racial makeup of the city was 97.79% White, 0.74% Native American, and 1.47% from two or more races. Hispanic or Latino of any race were 2.21% of the population.

There were 57 households out of which 33.3% had children under the age of 18 living with them, 52.6% were married couples living together, 7.0% had a female householder with no husband present, and 36.8% were non-families. 35.1% of all households were made up of individuals and 12.3% had someone living alone who was 65 years of age or older. The average household size was 2.39 and the average family size was 3.11.

In the city the population was spread out with 25.0% under the age of 18, 7.4% from 18 to 24, 26.5% from 25 to 44, 31.6% from 45 to 64, and 9.6% who were 65 years of age or older. The median age was 41 years. For every 100 females there were 103.0 males. For every 100 females age 18 and over, there were 100.0 males.

The median income for a household in the city was $33,125, and the median income for a family was $41,250. Males had a median income of $31,042 versus $14,107 for females. The per capita income for the city was $18,129. There were 6.1% of families and 5.5% of the population living below the poverty line, including 13.0% of under eighteens and none of those over 64.