Milford, Utah

Milford is a city located in Beaver County, Utah. As of the 2000 census, the city had a total population of 1,451.

Geography

\nMilford is located at 38°23'41" North, 113°0'50" West (38.394752, -113.013971)¹. According to the United States Census Bureau, the city has a total area of 5.0 km² (1.9 mi²). 5.0 km² (1.9 mi²) of it is land and none of the area is covered with water.

Demographics

\nAs of the census of 2000, there are 1,451 people, 484 households, and 357 families residing in the city. The population density is 290.3/km² (753.7/mi²). There are 589 housing units at an average density of 117.8/km² (306.0/mi²). The racial makeup of the city is 90.42% White, 0.00% African American, 1.45% Native American, 1.45% Asian, 0.00% Pacific Islander, 4.07% from other races, and 2.62% from two or more races. 6.13% of the population are Hispanic or Latino of any race. There are 484 households out of which 45.0% have children under the age of 18 living with them, 61.0% are married couples living together, 9.3% have a female householder with no husband present, and 26.2% are non-families. 24.0% of all households are made up of individuals and 14.7% have someone living alone who is 65 years of age or older. The average household size is 2.95 and the average family size is 3.58. In the city the population is spread out with 37.2% under the age of 18, 9.7% from 18 to 24, 23.6% from 25 to 44, 17.2% from 45 to 64, and 12.3% who are 65 years of age or older. The median age is 28 years. For every 100 females there are 97.4 males. For every 100 females age 18 and over, there are 95.1 males. The median income for a household in the city is $35,809, and the median income for a family is $41,750. Males have a median income of $35,568 versus $19,000 for females. The per capita income for the city is $14,889. 10.8% of the population and 9.2% of families are below the poverty line. Out of the total people living in poverty, 11.1% are under the age of 18 and 17.4% are 65 or older.