How much is home insurance in Sarasota?
The average cost of home insurance in Sarasota is $6,826 per year for a $300,000 dwelling policy with a $1,000 deductible. This is much higher than the national average rate of $2,584 per year for the same policy but much lower than the Florida state average of $10,675.
For a policy with a $500 deductible, the average cost of coverage is $6,868 annually. One of the biggest contributing factors to this high cost is likely Sarasota’s propensity for flooding. Flood damage can be extremely costly, so insurers have to account for this increased risk.
Cost of home insurance by dwelling coverage in Florida
Your chosen dwelling coverage level will have a direct effect on how much your policy costs. The higher your coverage level is, the higher your premiums will be, as your home insurance company is on the hook for additional damages.
Below, you can compare average annual premiums in Florida by dwelling coverage amount.
Cost of homeowners insurance by deductible amount
Your home insurance deductible is how much you have to pay out of pocket in the event of a claim. Only after you pay your deductible will your home insurance policy kick in.
Your home insurance deductible amount has a direct effect on your home insurance premiums. The higher your deductible is, the lower your premium will be, and vice versa. Compare average annual premiums by deductible amount below.
Average cost of home insurance in other cities in Florida
The cost of your home insurance policy largely depends on where you live. Homeowners pay different rates for the same policy by state, city, and ZIP code. Prices vary by location due to different weather risks, flood likelihood, vandalism frequencies, crime rates, and more.
Here are the average annual premiums in several Florida cities.
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