Berwyn, Nebraska

From Wikipedia, the free encyclopedia

Berwyn is a village in Custer County, Nebraska, United States. The population was 134 at the 2000 census.

[edit] Geography

Berwyn is located at 41°21′5″N, 99°30′5″W (41.351461, -99.501294)^GR1.

According to the United States Census Bureau, the village has a total area of 0.7 km² (0.3 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 134 people, 46 households, and 33 families residing in the village. The population density was 199.0/km² (524.6/mi²). There were 50 housing units at an average density of 74.3/km² (195.8/mi²). The racial makeup of the village was 99.25% White and 0.75% Asian. Hispanic or Latino of any race were 2.24% of the population.

There were 46 households out of which 50.0% had children under the age of 18 living with them, 67.4% were married couples living together, 2.2% had a female householder with no husband present, and 26.1% were non-families. 23.9% of all households were made up of individuals and 13.0% had someone living alone who was 65 years of age or older. The average household size was 2.91 and the average family size was 3.47.

In the village the population was spread out with 37.3% under the age of 18, 6.0% from 18 to 24, 29.1% from 25 to 44, 21.6% from 45 to 64, and 6.0% who were 65 years of age or older. The median age was 29 years. For every 100 females there were 103.0 males. For every 100 females age 18 and over, there were 100.0 males.

The median income for a household in the village was $29,375, and the median income for a family was $31,750. Males had a median income of $24,000 versus $15,833 for females. The per capita income for the village was $13,288. There were 8.6% of families and 11.9% of the population living below the poverty line, including 17.2% of under eighteens and 14.3% of those over 64.