1988 United States presidential election in the District of Columbia
This article relies largely or entirely on a single source. (April 2016) |
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Ward Results
Dukakis 60-70% 70-80% 80-90% 90-100%
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Elections in the District of Columbia |
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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.
Washington, D.C. overwhelmingly voted for Governor Michael Dukakis of Massachusetts, the Democratic candidate. 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 the District of Columbia, and it was one of only two areas that leaned more Republican than in the presidential election of 1984, which had resulted in a Republican landslide, the other being Tennessee.
Results
[edit]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 |
See also
[edit]References
[edit]- ^ "1988 Presidential General Election Results - District of Columbia". US Election Atlas.