Florida's unemployment holds steady at 3.9 percent

Posted at 11:41 AM, Apr 20, 2018

TALLAHASSEE, Fla. (AP) — Florida's unemployment rate is holding steady at 3.9 percent, slightly below the national average of 4.1 percent.

State officials released March employment figures Friday and reported that Florida added 13,800 jobs. There were 400,000 jobless Floridians out of a workforce of 10.2 million.

The March rate was 0.5 percentage point lower from a year ago.

Okaloosa and St. Johns counties had the lowest unemployment rate at 3.0 percent, followed by Monroe County at 3.1 percent.

Sumter County had the highest unemployment rate at 5.4 percent, followed by Citrus County at 5.2 percent and Hendry County at 5.1 percent.