Rosedale, Indiana

From Wikipedia, the free encyclopedia

Rosedale is a town in Parke County, Indiana, United States. The population was 750 at the 2000 census.

[edit] Geography

Rosedale is located at 39°37′23″N, 87°16′52″W (39.623188, -87.281040)^GR1.

According to the United States Census Bureau, the town has a total area of 1.0 km² (0.4 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 750 people, 304 households, and 210 families residing in the town. The population density was 723.9/km² (1,890.6/mi²). There were 322 housing units at an average density of 310.8/km² (811.7/mi²). The racial makeup of the town was 99.60% White, 0.13% African American, 0.13% Native American and 0.13% Asian.

There were 304 households out of which 36.5% had children under the age of 18 living with them, 51.6% were married couples living together, 12.5% had a female householder with no husband present, and 30.9% were non-families. 28.9% of all households were made up of individuals and 15.8% had someone living alone who was 65 years of age or older. The average household size was 2.47 and the average family size was 2.99.

In the town the population was spread out with 28.1% under the age of 18, 9.7% from 18 to 24, 28.8% from 25 to 44, 18.5% from 45 to 64, and 14.8% who were 65 years of age or older. The median age was 33 years. For every 100 females there were 92.8 males. For every 100 females age 18 and over, there were 87.2 males.

The median income for a household in the town was $34,519, and the median income for a family was $40,556. Males had a median income of $33,000 versus $18,750 for females. The per capita income for the town was $15,301. About 11.0% of families and 14.0% of the population were below the poverty line, including 21.0% of those under age 18 and 14.8% of those age 65 or over.