Ferdinand, Indiana
Ferdinand is a town located in
Dubois County, Indiana. As of the
2000 census, the town had a total population of 2,277.
Geography
\nFerdinand is located at 38°13'31" North, 86°51'40" West (38.225392, -86.860996)1.
According to the
United States Census Bureau, the town has a total area of 5.8
km² (2.2
mi²). 5.8 km² (2.2 mi²) of it is land and none of it is covered by water.
Demographics
\nAs of the census of
2000, there are 2,277 people, 808 households, and 569 families residing in the town. The
population density is 392.5/km² (1,017.3/mi²). There are 845 housing units at an average density of 145.7/km² (377.5/mi²). The racial makeup of the town is 99.03% White, 0.04%
African American, 0.09%
Native American, 0.18%
Asian, 0.09% Pacific Islander, 0.31% from other races, and 0.26% from two or more races. 0.61% of the population are
Hispanic or
Latino of any race.
There are 808 households out of which 35.3% have children under the age of 18 living with them, 60.9% are married couples living together, 7.7% have a female householder with no husband present, and 29.5% are non-families. 27.1% of all households are made up of individuals and 14.0% have someone living alone who is 65 years of age or older. The average household size is 2.51 and the average family size is 3.09.
In the town the population is spread out with 24.8% under the age of 18, 5.7% from 18 to 24, 28.6% from 25 to 44, 20.9% from 45 to 64, and 20.1% who are 65 years of age or older. The median age is 39 years. For every 100 females there are 78.9 males. For every 100 females age 18 and over, there are 74.3 males.
The median income for a household in the town is $41,326, and the median income for a family is $52,065. Males have a median income of $34,205 versus $23,646 for females. The per capita income for the town is $18,335. 9.9% of the population and 3.4% of families are below the poverty line. Out of the total people living in poverty, 4.2% are under the age of 18 and 16.1% are 65 or older.