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Compare Kendall to Sanford
This post compares Kendall, 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 | Kendall (CDP) | Sanford (city) |
|---|---|---|
| Population | 75,123 | 57,979 |
| Income (median) | $49,437 | $36,585 |
| Home Value (median) | $337,400 | $113,100 |
| White | 87% | 60% |
| Black | 3% | 26% |
| Asian | 3% | 3% |
| With Kids | 27% | 31% |
| Age (median) | 42 | 33 |
| Rentals | 36% | 47% |
Questions and Answers:
Q: Which is more populated, Kendall or Sanford?
A: Kendall has a larger population count than Sanford: 75123 compared with 57979 total people.
Q: Do incomes tend to be higher in Kendall or in Sanford?
A: Incomes are on average much higher in Kendall.
Q: Are homes more expensive in Kendall or Sanford?
A: Homes tend to be much more expensive in Kendall.
Q: Which has a higher percentage of housing rental units?
A: Sanford does. The percentage of rentals differs quite a bit: 36% for Kendall versus 47% for Sanford.
Somewhat similar in-state cities, towns, and CDPs to Kendall may include: Jupiter, Jacksonville-Beach, and Winter-Park.
The statistics and analyses on this page are based on data from the U.S. Census American Community Survey.
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