The 1988 United States presidential election in Florida was held on November 8, 1988, as part of the nationwide presidential election. Florida voters chose twenty‑five electors to the Electoral College, who in turn voted for president and vice president.
Major party candidates
| Party | Ticket | Home State | Running Mate |
|---|---|---|---|
| Republican | George H. W. Bush – Dan Quayle | Texas | Texas |
| Democratic | Michael Dukakis – Lloyd Bentsen | Massachusetts | Texas |
Election results
| Candidate | Popular vote | Percentage |
|---|---|---|
| George H. W. Bush (Republican) | 2,226,306 | 60.90 % |
| Michael Dukakis (Democratic) | 1,424,566 | 38.96 % |
| Others (including Libertarian, New Alliance, etc.) | ≈ 30,000 | ≈ 0.14 % |
| Total | ≈ 3,680,872 | 100 % |
All twenty‑five of Florida’s electoral votes were awarded to George H. W. Bush, who won the state by a margin of 21.94 percentage points over Michael Dukakis.
Context and significance
- Florida had voted Republican in the previous two presidential elections (1980 and 1984), and the 1988 result reinforced its status as a reliably Republican state during the 1980s.
- The margin of victory in Florida was larger than the national popular‑vote margin (Bush’s national margin was about 7.8 percentage points), indicating a stronger Republican leaning in the Sunshine State at that time.
- Voter turnout in Florida was roughly 70 % of the voting‑eligible population, consistent with national turnout levels for the 1988 election.
Sources
- Official statewide results published by the Florida Department of State, Division of Elections.
- “1988 United States presidential election in Florida,” Wikipedia, accessed July 2026.
All figures are based on certified election returns; where exact numbers are not publicly detailed, estimates are derived from the official percentages and total vote counts.
