Milford, Nebraska

From Wikipedia, the free encyclopedia

Milford is a city in Seward County, Nebraska, United States. The population was 2,070 at the 2000 census.

1 Geography
2 Demographics
3 Images
4 External links

[edit] Geography

Milford is located at 40°46′19″N, 97°3′8″W (40.772010, -97.052311)^GR1.

According to the United States Census Bureau, the city has a total area of 1.8 km² (0.7 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 2,070 people, 722 households, and 464 families residing in the city. The population density was 1,125.7/km² (2,901.9/mi²). There were 770 housing units at an average density of 418.7/km² (1,079.5/mi²). The racial makeup of the city was 97.83% White, 0.24% African American, 0.29% Native American, 0.05% Asian, 0.05% Pacific Islander, 0.68% from other races, and 0.87% from two or more races. Hispanic or Latino of any race were 1.50% of the population.

There were 722 households out of which 30.2% had children under the age of 18 living with them, 56.1% were married couples living together, 5.7% had a female householder with no husband present, and 35.7% were non-families. 26.7% of all households were made up of individuals and 10.9% had someone living alone who was 65 years of age or older. The average household size was 2.44 and the average family size was 2.99.

In the city the population was spread out with 21.7% under the age of 18, 24.9% from 18 to 24, 21.9% from 25 to 44, 16.5% from 45 to 64, and 15.0% who were 65 years of age or older. The median age was 28 years. For every 100 females there were 132.1 males. For every 100 females age 18 and over, there were 140.5 males.

The median income for a household in the city was $37,039, and the median income for a family was $51,012. Males had a median income of $30,170 versus $22,026 for females. The per capita income for the city was $16,217. About 4.6% of families and 10.6% of the population were below the poverty line, including 6.7% of those under age 18 and 11.2% of those age 65 or over.