The annual salary statistics of property, real estate managers in Sandusky, Ohio is shown in Table 1. The wage statistics of property, real estate managers in Sandusky is based on the national compensation survey conducted by the U.S. Bureau of Labor Statistics in 2019 and published in April 2020 [1].
| Percentile Bracket | Average Annual Salary |
|---|---|
| 10th Percentile Wage | $19,390 |
| 25th Percentile Wage | $33,560 |
| 50th Percentile Wage | $49,330 |
| 75th Percentile Wage | $56,630 |
| 90th Percentile Wage | $61,010 |
Table 1 shows the average annual salary for property, real estate managers in Sandusky, Ohio in 5 percentile scales. The average annual salary for the 90th percentile (the top 10 percent of the highest paid) is $61,010. The median (50th percentile) annual salary is $49,330. The average annual salary for the bottom 10 percent earners is $19,390.
The table and chart below show the trend of the median salary of property, real estate managers from 2012 to 2013. Note that there is no salary data for some years between 2012 and 2019.
| Year | Median Salary | Growth | 1-Year Growth |
|---|---|---|---|
| 2013 | $49,330 | 0.00% | 0.00% |
| 2012 | $49,330 | - | - |
From Table 3 we note that the median annual salary of $49,330 in Sandusky is in the middle of salary range for property, real estate managers in state of Ohio. In comparison, the annual salary of property, real estate managers in Sandusky is 13.2 percent (13.2%) lower than that in the highest paying Cleveland-Elyria-Mentor and 31.8 percent (31.8%) higher than that in the lowest paying Lima.