Jonestown, Texas

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Jonestown is a city in Travis County, Texas, United States. The population was 1,681 at the 2000 census.

Contents

Geography

Jonestown is located at 30°29′20″N 97°55′27″W / 30.488872°N 97.924208°W / 30.488872; -97.924208 (30.488872, -97.924208).[3] This is at the north end of Lake Travis, 19 miles (31 km) northwest of Austin.

According to the United States Census Bureau, the city has a total area of 5.3 square miles (13.7 km2), of which, 4.7 square miles (12.1 km2) of it is land and 0.6 square miles (1.6 km2) of it (11.34%) is water.

Demographics

As of the census[1] of 2000, there were 1,681 people, 699 households, and 434 families residing in the city. The population density was 358.0 people per square mile (138.4/km2). There were 770 housing units at an average density of 164.0/sq mi (63.4/km2). The racial makeup of the city was 92.27% White, 0.83% African American, 0.42% Native American, 0.42% Asian, 3.57% from other races, and 2.50% from two or more races. Hispanic or Latino of any race were 9.58% of the population.

There were 699 households out of which 29.6% had children under the age of 18 living with them, 46.9% were married couples living together, 9.7% had a female householder with no husband present, and 37.8% were non-families. 29.5% of all households were made up of individuals and 5.0% had someone living alone who was 65 years of age or older. The average household size was 2.40 and the average family size was 2.98.

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