Granville, North Dakota

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Granville is a city in McHenry County, North Dakota in the United States. The population was 286 at the 2000 census. It is part of the Minot Micropolitan Statistical Area. Granville was founded in 1900.



Granville is located at 48°16′0″N 100°50′36″W / 48.266667°N 100.84333°W / 48.266667; -100.84333 (48.266566, -100.843355).[3]

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


As of the census[1] of 2000, there were 286 people, 120 households, and 77 families residing in the city. The population density was 1,042.5 people per square mile (409.0/km²). There were 141 housing units at an average density of 514.0/sq mi (201.6/km²). The racial makeup of the city was 96.50% White, 0.35% African American, 0.70% Native American, and 2.45% from two or more races.

There were 120 households out of which 34.2% had children under the age of 18 living with them, 55.8% were married couples living together, 6.7% had a female householder with no husband present, and 35.8% were non-families. 30.0% of all households were made up of individuals and 17.5% had someone living alone who was 65 years of age or older. The average household size was 2.38 and the average family size was 3.04.

In the city the population was spread out with 26.6% under the age of 18, 8.4% from 18 to 24, 24.5% from 25 to 44, 19.9% from 45 to 64, and 20.6% who were 65 years of age or older. The median age was 38 years. For every 100 females there were 108.8 males. For every 100 females age 18 and over, there were 100.0 males.

The median income for a household in the city was $24,444, and the median income for a family was $28,500. Males had a median income of $27,500 versus $14,464 for females. The per capita income for the city was $14,576. About 7.5% of families and 9.9% of the population were below the poverty line, including 13.3% of those under the age of eighteen and 10.5% of those sixty five or over.


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