Jefferson City, Montana

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Jefferson City is a census-designated place (CDP) in Jefferson County, Montana, United States. The population was 295 at the 2000 census. It is part of the Helena Micropolitan Statistical Area.



Jefferson City is located at 46°23′12″N 112°02′28″W / 46.386685°N 112.041154°W / 46.386685; -112.041154 (46.386685, -112.041154).[1]

According to the United States Census Bureau, the CDP has a total area of 12.1 square miles (31.2 km2), all of it land.


As of the census[2] of 2000, there were 295 people, 110 households, and 85 families residing in the CDP. The population density was 24.5 people per square mile (9.4/km2). There were 115 housing units at an average density of 9.5/sq mi (3.7/km2). The racial makeup of the CDP was 95.59% White, 2.03% Native American, 1.36% Asian, 0.34% from other races, and 0.68% from two or more races. Hispanic or Latino of any race were 0.34% of the population.

There were 110 households out of which 33.6% had children under the age of 18 living with them, 69.1% were married couples living together, 4.5% had a female householder with no husband present, and 22.7% were non-families. 20.0% of all households were made up of individuals and 4.5% had someone living alone who was 65 years of age or older. The average household size was 2.68 and the average family size was 3.11.

In the CDP the population was spread out with 25.4% under the age of 18, 5.8% from 18 to 24, 30.8% from 25 to 44, 27.5% from 45 to 64, and 10.5% who were 65 years of age or older. The median age was 40 years. For every 100 females there were 99.3 males. For every 100 females age 18 and over, there were 105.6 males.

The median income for a household in the CDP was $53,125, and the median income for a family was $52,917. Males had a median income of $28,125 versus $32,500 for females. The per capita income for the CDP was $21,953. None of the families and 1.4% of the population were living below the poverty line.

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