Troy, Idaho

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Troy is a city in Latah County, Idaho, United States. The population was 798 at the 2000 census.

Contents

Geography

Troy is located at 46°44′19″N 116°46′15″W / 46.73861°N 116.77083°W / 46.73861; -116.77083 (46.738571, -116.770785).[1]

According to the United States Census Bureau, the city has a total area of 0.8 square miles (2.0 km²), all of it land.

Demographics

As of the census[2] of 2000, there were 798 people, 309 households, and 224 families residing in the city. The population density was 1,004.8 people per square mile (390.0/km²). There were 341 housing units at an average density of 429.4/sq mi (166.7/km²). The racial makeup of the city was 96.12% White, 0.50% Native American, 0.38% Asian, and 3.01% from two or more races. Hispanic or Latino of any race were 0.13% of the population.

There were 309 households out of which 38.5% had children under the age of 18 living with them, 61.2% were married couples living together, 7.8% had a female householder with no husband present, and 27.2% were non-families. 20.7% of all households were made up of individuals and 7.8% had someone living alone who was 65 years of age or older. The average household size was 2.58 and the average family size was 3.02.

In the city the population was spread out with 29.8% under the age of 18, 7.1% from 18 to 24, 29.2% from 25 to 44, 22.4% from 45 to 64, and 11.4% who were 65 years of age or older. The median age was 34 years. For every 100 females there were 99.5 males. For every 100 females age 18 and over, there were 92.4 males.

The median income for a household in the city was $36,250, and the median income for a family was $42,031. Males had a median income of $33,194 versus $23,295 for females. The per capita income for the city was $16,557. About 11.9% of families and 12.1% of the population were below the poverty line, including 14.7% of those under age 18 and 3.1% of those age 65 or over.

Population history

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