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Compare Arvin to Lawndale
This post compares Arvin, California to Lawndale, California across various dimensions, including population, median income, median age, percentage of housing that is rental, percent of households with kids, and other demographics.
| Dimension | Arvin (city) | Lawndale (city) |
|---|---|---|
| Population | 20,826 | 33,191 |
| Income (median) | $21,815 | $37,317 |
| Home Value (median) | $132,300 | $417,600 |
| White | 87% | 42% |
| Black | 1% | 10% |
| Asian | 0% | 10% |
| With Kids | 67% | 39% |
| Age (median) | 24 | 35 |
| Rentals | 54% | 67% |
Questions and Answers:
Q: Which is more populated, Arvin or Lawndale?
A: Arvin has a smaller population count than Lawndale: 20826 compared with 33191 total people.
Q: Do incomes tend to be higher in Arvin or in Lawndale?
A: Incomes are on average much lower in Arvin.
Q: Are homes more expensive in Arvin or Lawndale?
A: Homes tend to be much less expensive in Arvin.
Q: Which has a higher percentage of housing rental units?
A: Lawndale does. The percentage of rentals differs quite a bit: 54% for Arvin versus 67% for Lawndale.
Somewhat similar in-state cities, towns, and CDPs to Arvin may include: Wasco, Dinuba, and Madera.
The statistics and analyses on this page are based on data from the U.S. Census American Community Survey.
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