Lu Verne, Iowa

From Wikipedia, the free encyclopedia

Lu Verne is a city in Iowa, United States. The population was 299 at the 2000 census.

[edit] Geography

Lu Verne is located at 42°54′42″N, 94°5′1″W (42.911638, -94.083618)^GR1.

According to the United States Census Bureau, the city has a total area of 5.9 km² (2.3 mi²). None of the area is covered with water.

[edit] Demographics

As of the census^GR2 of 2000, there were 299 people, 131 households, and 85 families residing in the city. The population density was 51.1/km² (132.3/mi²). There were 163 housing units at an average density of 27.8/km² (72.1/mi²). The racial makeup of the city was 99.33% White, 0.33% Native American, and 0.33% from two or more races. Hispanic or Latino of any race were 0.33% of the population.

There were 131 households out of which 28.2% had children under the age of 18 living with them, 50.4% were married couples living together, 9.9% had a female householder with no husband present, and 35.1% were non-families. 28.2% of all households were made up of individuals and 14.5% had someone living alone who was 65 years of age or older. The average household size was 2.28 and the average family size was 2.78.

In the city the population was spread out with 22.7% under the age of 18, 9.4% from 18 to 24, 26.4% from 25 to 44, 22.1% from 45 to 64, and 19.4% who were 65 years of age or older. The median age was 38 years. For every 100 females there were 104.8 males. For every 100 females age 18 and over, there were 102.6 males.

The median income for a household in the city was $30,625, and the median income for a family was $34,500. Males had a median income of $21,429 versus $20,000 for females. The per capita income for the city was $16,281. About 14.5% of families and 13.8% of the population were below the poverty line, including 23.0% of those under the age of eighteen and 6.3% of those sixty five or over.