1988 United States presidential election in the District of Columbia

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1988 United States presidential election in the District of Columbia
Flag of the District of Columbia.svg
  1984 November 8, 1988 1992  
  Dukakis campaign portrait 3x4.jpg VP George Bush crop.jpg
Nominee Michael Dukakis George H. W. Bush
Party Democratic Republican
Home state Massachusetts Texas
Running mate Lloyd Bentsen Dan Quayle
Electoral vote30
Popular vote159,40727,590
Percentage82.65% 14.30%

DC 1988 Presidential Election By Ward.svg
Ward Results
Dukakis
  60-70%
  70-80%
  80-90%
  90-100%

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.

Contents

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

1988 United States presidential election in the District of Columbia [1]
PartyCandidateVotesPercentageElectoral votes
Democratic Michael Dukakis 159,40782.65%3
Republican George H. W. Bush 27,59014.30%0
New Alliance Lenora Fulani 2,9011.50%0
Libertarian Ron Paul 5540.29%0

See also

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References

  1. "1988 Presidential General Election Results - District of Columbia". US Election Atlas.