Poplar, Montana

From Wikipedia, the free encyclopedia

Poplar is a city in Roosevelt County, Montana, United States. The population was 911 at the 2000 census.

[edit] Geography

Poplar is located at 48°6′34″N, 105°11′42″W (48.109474, -105.194891)^GR1.

According to the United States Census Bureau, the city has a total area of 0.7 km² (0.3 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 911 people, 325 households, and 206 families residing in the city. The population density was 1,302.7/km² (3,406.0/mi²). There were 350 housing units at an average density of 500.5/km² (1,308.5/mi²). The racial makeup of the city was 32.16% White, 0.11% African American, 63.67% Native American, 0.77% Asian, 0.22% from other races, and 3.07% from two or more races. Hispanic or Latino of any race were 0.88% of the population.

There were 325 households out of which 33.5% had children under the age of 18 living with them, 36.9% were married couples living together, 19.4% had a female householder with no husband present, and 36.6% were non-families. 32.3% of all households were made up of individuals and 12.6% had someone living alone who was 65 years of age or older. The average household size was 2.57 and the average family size was 3.26.

In the city the population was spread out with 30.7% under the age of 18, 8.0% from 18 to 24, 29.6% from 25 to 44, 20.2% from 45 to 64, and 11.4% who were 65 years of age or older. The median age was 33 years. For every 100 females there were 106.1 males. For every 100 females age 18 and over, there were 102.9 males.

The median income for a household in the city was $24,896, and the median income for a family was $29,688. Males had a median income of $22,250 versus $19,000 for females. The per capita income for the city was $10,579. About 23.4% of families and 31.5% of the population were below the poverty line, including 32.6% of those under age 18 and 22.2% of those age 65 or over.