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Compare Jacksonville to Sanford
This post compares Jacksonville, Florida to Sanford, Florida across various dimensions, including population, median income, median age, percentage of housing that is rental, percent of households with kids, and other demographics.
| Dimension | Jacksonville (city) | Sanford (city) |
|---|---|---|
| Population | 867,313 | 57,979 |
| Income (median) | $40,449 | $36,585 |
| Home Value (median) | $150,200 | $113,100 |
| White | 59% | 60% |
| Black | 31% | 26% |
| Asian | 4% | 3% |
| With Kids | 30% | 31% |
| Age (median) | 35 | 33 |
| Rentals | 42% | 47% |
Questions and Answers:
Q: Which is more populated, Jacksonville or Sanford?
A: Jacksonville has a much larger population count than Sanford: 867313 compared with 57979 total people.
Q: Do incomes tend to be higher in Jacksonville or in Sanford?
A: Incomes are on average higher in Jacksonville.
Q: Are homes more expensive in Jacksonville or Sanford?
A: Homes tend to be more expensive in Jacksonville.
Q: Which has a higher percentage of housing rental units?
A: Sanford does. The percentage of rentals is quite similar: 42% for Jacksonville versus 47% for Sanford.
Somewhat similar in-state cities, towns, and CDPs to Jacksonville may include: Tampa, Miami, and Orlando.
The statistics and analyses on this page are based on data from the U.S. Census American Community Survey.
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