Milford, Kosciusko County, Indiana

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Milford is a town in Kosciusko County, Indiana, United States. The population was 1,550 at the 2000 census.



Milford is located at 41°24′33″N 85°50′49″W / 41.40917°N 85.84694°W / 41.40917; -85.84694 (41.409215, -85.846866)[3].

According to the United States Census Bureau, the town has a total area of 1.1 square miles (2.8 km²), all of it land.


As of the census[1] of 2000, there were 1,550 people, 590 households, and 402 families residing in the town. The population density was 1,436.8 people per square mile (554.1/km²). There were 626 housing units at an average density of 580.3/sq mi (223.8/km²). The racial makeup of the town was 91.61% White, 0.84% African American, 0.13% Native American, 0.26% Asian, 5.29% from other races, and 1.87% from two or more races. Hispanic or Latino of any race were 11.16% of the population.

There were 590 households out of which 33.9% had children under the age of 18 living with them, 54.1% were married couples living together, 9.5% had a female householder with no husband present, and 31.7% were non-families. 28.3% of all households were made up of individuals and 13.6% had someone living alone who was 65 years of age or older. The average household size was 2.53 and the average family size was 3.11.

In the town the population was spread out with 27.2% under the age of 18, 8.3% from 18 to 24, 27.7% from 25 to 44, 19.5% from 45 to 64, and 17.4% who were 65 years of age or older. The median age was 36 years. For every 100 females there were 86.5 males. For every 100 females age 18 and over, there were 85.2 males.

The median income for a household in the town was $36,458, and the median income for a family was $43,958. Males had a median income of $32,898 versus $24,107 for females. The per capita income for the town was $17,247. About 2.4% of families and 4.0% of the population were below the poverty line, including 2.3% of those under age 18 and 6.4% of those age 65 or over.

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