Waverly, Washington

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Waverly is a town in Spokane County, Washington, United States. The population was 121 at the 2000 census.

1 History
2 Geography
3 Demographics
4 External links

[edit] History

Waverly was officially incorporated on May 15, 1907.

[edit] Geography

Waverly is located at 47°20′23″N, 117°13′43″W (47.339852, -117.228643)^GR1.

According to the United States Census Bureau, the town has a total area of 1.0 km² (0.4 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 121 people, 46 households, and 31 families residing in the town. The population density was 116.8/km² (299.1/mi²). There were 49 housing units at an average density of 47.3/km² (121.1/mi²). The racial makeup of the town was 95.87% White, 2.48% Native American, and 1.65% from two or more races.

There were 46 households out of which 41.3% had children under the age of 18 living with them, 60.9% were married couples living together, 2.2% had a female householder with no husband present, and 32.6% were non-families. 28.3% of all households were made up of individuals and 6.5% had someone living alone who was 65 years of age or older. The average household size was 2.63 and the average family size was 3.29.

In the town the population was spread out with 30.6% under the age of 18, 4.1% from 18 to 24, 32.2% from 25 to 44, 23.1% from 45 to 64, and 9.9% who were 65 years of age or older. The median age was 36 years. For every 100 females there were 101.7 males. For every 100 females age 18 and over, there were 100.0 males.

The median income for a household in the town was $38,125, and the median income for a family was $43,750. Males had a median income of $31,042 versus $20,625 for females. The per capita income for the town was $15,072. There were 5.9% of families and 8.4% of the population living below the poverty line, including 14.6% of under eighteens and none of those over 64.