Humboldt, Wisconsin

Humboldt is a town located in Brown County, Wisconsin. As of the 2000 census, the town had a total population of 1,338.

Geography

\nAccording to the United States Census Bureau, the town has a total area of 62.0 km² (24.0 mi²). 62.0 km² (24.0 mi²) of it is land and none of the area is covered with water.

Demographics

\nAs of the census of 2000, there are 1,338 people, 453 households, and 370 families residing in the town. The population density is 21.6/km² (55.9/mi²). There are 460 housing units at an average density of 7.4/km² (19.2/mi²). The racial makeup of the town is 99.25% White, 0.00% African American, 0.45% Native American, 0.00% Asian, 0.00% Pacific Islander, 0.15% from other races, and 0.15% from two or more races. 0.22% of the population are Hispanic or Latino of any race. There are 453 households out of which 39.7% have children under the age of 18 living with them, 73.3% are married couples living together, 4.9% have a female householder with no husband present, and 18.3% are non-families. 15.7% of all households are made up of individuals and 6.6% have someone living alone who is 65 years of age or older. The average household size is 2.95 and the average family size is 3.31. In the town the population is spread out with 28.3% under the age of 18, 9.0% from 18 to 24, 29.4% from 25 to 44, 23.8% from 45 to 64, and 9.6% who are 65 years of age or older. The median age is 36 years. For every 100 females there are 110.7 males. For every 100 females age 18 and over, there are 108.7 males. The median income for a household in the town is $54,821, and the median income for a family is $60,063. Males have a median income of $38,889 versus $24,489 for females. The per capita income for the town is $19,813. 2.2% of the population and 2.3% of families are below the poverty line. Out of the total people living in poverty, 2.5% are under the age of 18 and 4.4% are 65 or older.