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Find Health Insurance Agents Near You in Woodbury, NY
Compare health insurance agents in Woodbury. Review contact information and insurance categorys before you decide.
5 agents listed
Updated May 2026
Woodbury, NY
Important: Listings are compiled from publicly available sources and have not been independently verified by BeforeInsuranceUSA.com. We do not endorse any agent or guarantee outcomes. Verify licensing and disciplinary status with the Department of Insurance of New York before hiring.
Showing 5 health insurance agents in Woodbury, NY
Listings are displayed in no particular order and are not ranked by BeforeInsuranceUSA. Order does not constitute a recommendation or endorsement of any agent.
SterlingRisk Insurance | Part of a Nationally Ranked Top 15 Insurance Brokerage
135 Crossways Park Dr # 300, Woodbury, NY 11797
(516) 487-0300
sterlingrisk.com
NBP Insurance
7600 Jericho Tpke # 100, Woodbury, NY 11797
(516) 626-6271
nbpinsurance.com
Acrisure Woodbury, NY (Bender Insurance Agency)
365 Crossways Park Dr, Woodbury, NY 11797
(516) 799-8222
acrisure.com
FNA Insurance Services
1000 Woodbury Rd #403, Woodbury, NY 11797
(516) 352-7000
fnainsurance.com
Peoples Choice Insurance Agency
1000 Woodbury Rd, Woodbury, NY 11797
(516) 709-2000
peopleschoiceinsure.com
Health insurance agents in Woodbury, New York help individuals and families find coverage through the NY State of Health marketplace or private plans. These agents are licensed by the New York State Department of Financial Services and must follow state regulations. They can explain options like Qualified Health Plans, Essential Plans, and subsidies available under New York law.
What Does a Health Insurance Agent in Woodbury Cost?
In New York, health insurance agents typically do not charge clients directly because they earn commissions from insurance carriers. Commissions can range from 2% to 5% of the annual premium, but this does not affect your premium. Some agents may charge a flat fee for specialized services, often between 50 and 200 dollars. Costs vary by situation and plan type. This is general information, not insurance advice.
* Cost estimates are general ranges based on publicly available data and compiled using automated research tools. Actual fees vary by agent and case complexity. This is not insurance advice — consult directly with an agent for fee specifics.
Frequently Asked Questions
What does a health insurance agent in Woodbury do?
A health insurance agent helps you compare plans, apply for subsidies, and enroll in coverage. They are licensed and can assist with both marketplace and off-marketplace plans. In New York, agents must complete continuing education to stay current.
How do I choose a health insurance agent in Woodbury?
Look for an agent licensed in New York with experience in individual or family plans. You can verify their license through the New York State Department of Financial Services website. Ask if they work with multiple carriers to offer a range of options.
Is there a fee for using a health insurance agent in Woodbury?
Most health insurance agents in New York are paid by the insurance company, not by you. This means their services are usually free to you. However, some agents may charge a consulting fee for complex cases, so ask before you start.
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Notice
BeforeInsuranceUSA.com is not an insurance agency, agent referral service, or insurance advice provider. No agent-client relationship is formed by submitting a form or using this site. We do not endorse, recommend, or vouch for the qualifications of any agent. Past results do not guarantee future outcomes. Each situation is different. Verify agent credentials with your state department of insurance before hiring.
About Our Listings
Agents listed on BeforeInsuranceUSA.com are compiled from publicly available directories, state department of insurance records, and business listings. Inclusion does not constitute endorsement. We do not verify agent licensing, disciplinary status, or qualifications. Verify each agent's current standing with the Department of Insurance of New York.