Rosewood Heights, Illinois

Rosewood Heights is located at 38°53′29″N 90°4′19″W (38.891414, -90.071813).[2]

According to the United States Census Bureau, the CDP has a total area of 2.1 square miles (5.4 km²), all land.

At the 2000 census there were 4,262 people, 1,688 households, and 1,299 families living in the CDP. The population density was 1,988.0 people per square mile (769.0/km²). There were 1,754 housing units at an average density of 818.1/sq mi (316.5/km²). The racial makeup of the CDP was 98.78% White, 0.12% African American, 0.23% Native American, 0.21% Asian, 0.21% from other races, and 0.45% from two or more races. Hispanic or Latino of any race were 0.82%.[3]

Of the 1,688 households 29.7% had children under the age of 18 living with them, 64.5% were married couples living together, 9.1% had a female householder with no husband present, and 23.0% were non-families. 20.0% of households were one person and 10.4% were one person aged 65 or older. The average household size was 2.52 and the average family size was 2.88.

The age distribution was 22.2% under the age of 18, 7.6% from 18 to 24, 26.1% from 25 to 44, 26.0% from 45 to 64, and 18.1% 65 or older. The median age was 41 years. For every 100 females, there were 90.6 males. For every 100 females age 18 and over, there were 90.5 males.

The median household income was $46,701 and the median family income was $53,365. Males had a median income of $44,111 versus $28,176 for females. The per capita income for the CDP was $20,527. About 7.2% of families and 7.8% of the population were below the poverty line, including 8.4% of those under age 18 and 3.2% of those age 65 or over.