Lynndyl, Utah

From Wikipedia, the free encyclopedia

Lynndyl is a town in Millard County, Utah, United States. The population was 134 at the 2000 census.

[edit] Geography

Lynndyl is located at 39°31′3″N, 112°22′35″W (39.517406, -112.376357)^GR1.

According to the United States Census Bureau, the town has a total area of 9.1 km² (3.5 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 134 people, 45 households, and 39 families residing in the town. The population density was 14.7/km² (38.2/mi²). There were 55 housing units at an average density of 6.1/km² (15.7/mi²). The racial makeup of the town was 89.55% White, 2.24% Native American, 5.97% from other races, and 2.24% from two or more races. Hispanic or Latino of any race were 13.43% of the population.

There were 45 households out of which 42.2% had children under the age of 18 living with them, 71.1% were married couples living together, 6.7% had a female householder with no husband present, and 13.3% were non-families. 13.3% of all households were made up of individuals and 4.4% had someone living alone who was 65 years of age or older. The average household size was 2.98 and the average family size was 3.26.

In the town the population was spread out with 32.8% under the age of 18, 4.5% from 18 to 24, 23.9% from 25 to 44, 22.4% from 45 to 64, and 16.4% who were 65 years of age or older. The median age was 35 years. For every 100 females there were 100.0 males. For every 100 females age 18 and over, there were 91.5 males.

The median income for a household in the town was $35,625, and the median income for a family was $40,278. Males had a median income of $31,250 versus $21,250 for females. The per capita income for the town was $11,738. There were 11.1% of families and 7.5% of the population living below the poverty line, including 13.0% of under eighteens and none of those over 64.