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Sandusky, Iowa
Sandusky is an unincorporated community and census-designated place (CDP) in Lee County, Iowa, United States. It is in the southeast part of the county, on the west bank of the Mississippi River 4 miles (6 km) north of Keokuk, the county seat, and 6 miles (10 km) south of Montrose. Sandusky was first listed as a CDP prior to the 2020 census. As of the 2020 census, its population was 297.[3] Demographics
2020 censusAs of the census of 2020,[5] there were 297 people, 146 households, and 124 families residing in the community. The population density was 208.4 inhabitants per square mile (80.4/km2). There were 149 housing units at an average density of 104.5 per square mile (40.4/km2). The racial makeup of the community was 92.9% White, 0.0% Black or African American, 0.3% Native American, 0.7% Asian, 0.0% Pacific Islander, 1.0% from other races and 5.1% from two or more races. Hispanic or Latino persons of any race comprised 1.7% of the population. Of the 146 households, 26.0% of which had children under the age of 18 living with them, 65.1% were married couples living together, 9.6% were cohabitating couples, 17.8% had a female householder with no spouse or partner present and 7.5% had a male householder with no spouse or partner present. 15.1% of all households were non-families. 8.2% of all households were made up of individuals, 2.1% had someone living alone who was 65 years old or older. The median age in the community was 49.9 years. 16.5% of the residents were under the age of 20; 3.7% were between the ages of 20 and 24; 24.6% were from 25 and 44; 31.3% were from 45 and 64; and 23.9% were 65 years of age or older. The gender makeup of the community was 60.6% male and 39.4% female. HistoryFounded in the 1800s, Sandusky's population was 85 in 1902,[6] and 60 in 1925.[7] References
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