1988 United States presidential election in the District of Columbia
The 1988 United States presidential election in the District of Columbia took place on November 8, 1988, as part of the 1988 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for president and vice president.
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 District Result
Dukakis 80-90%
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| Elections in the District of Columbia | ||||||||
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Washington, D.C. was a landslide in favor of the Democratic candidate, Governor Michael Dukakis of Massachusetts. Vice President George H. W. Bush received 14.3% of the vote. This is the most recent election in which the Republican candidate received more than 10% of the vote in Washington, D.C. This is one of the only two areas in the country that went more Republican than that of 1984, the other being Tennessee.
Results
| 1988 United States presidential election in the District of Columbia[1] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic | Michael Dukakis | 159,407 | 82.65% | 3 | |
| Republican | George H. W. Bush | 27,590 | 14.30% | 0 | |
| New Alliance | Lenora Fulani | 2,901 | 1.50% | 0 | |
| Libertarian | Ron Paul | 554 | 0.29% | 0 | |
References
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