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1988 United States presidential election in the District of Columbia

  (Redirected from United States presidential election in the District of Columbia, 1988)

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.

United States presidential election in the District of Columbia, 1988

← 1984 November 8, 1988 1992 →
  1988 Dukakis.jpg 1988 Bush.jpg
Nominee Michael Dukakis George H. W. Bush
Party Democratic Republican
Home state Massachusetts Texas
Running mate Lloyd Bentsen Dan Quayle
Electoral vote 3 0
Popular vote 159,407 27,590
Percentage 82.65% 14.30%

President before election

Ronald Reagan
Republican

Elected President

George H. W. Bush
Republican

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.

ResultsEdit

United States presidential election in the District of Columbia, 1988[1]
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

ReferencesEdit