Eland, Wisconsin

From Wikipedia, the free encyclopedia

Eland is a village in Shawano County, Wisconsin, in the United States. As of the 2000 census, the village population was 251.

[edit] Geography

Eland is located at 44°52′13″N, 89°12′53″W (44.870191, -89.214786)^GR1.

According to the United States Census Bureau, the village has a total area of 5.7 km² (2.2 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 251 people, 93 households, and 66 families residing in the village. The population density was 43.7/km² (113.0/mi²). There were 96 housing units at an average density of 16.7/km² (43.2/mi²). The racial makeup of the village was 92.03% White, 5.18% Native American, 1.20% from other races, and 1.59% from two or more races. Hispanic or Latino of any race were 1.20% of the population.

There were 93 households out of which 37.6% had children under the age of 18 living with them, 53.8% were married couples living together, 9.7% had a female householder with no husband present, and 28.0% were non-families. 23.7% of all households were made up of individuals and 9.7% had someone living alone who was 65 years of age or older. The average household size was 2.70 and the average family size was 3.18.

In the village the population was spread out with 33.1% under the age of 18, 3.6% from 18 to 24, 31.9% from 25 to 44, 18.7% from 45 to 64, and 12.7% who were 65 years of age or older. The median age was 36 years. For every 100 females there were 105.7 males. For every 100 females age 18 and over, there were 104.9 males.

The median income for a household in the village was $37,917, and the median income for a family was $41,607. Males had a median income of $26,750 versus $23,125 for females. The per capita income for the village was $15,909. About 10.4% of families and 7.9% of the population were below the poverty line, including 2.1% of those under the age of eighteen and 16.7% of those sixty five or over.