1992 United States presidential election in the District of Columbia

The 1992 United States presidential election in the District of Columbia took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for president and vice president.

Main article: 1992 United States presidential election
1992 United States presidential election in the District of Columbia
November 3, 1992
 
Nominee Bill Clinton George H. W. Bush
Party Democratic Republican
Home state Arkansas Texas
Running mate Al Gore Dan Quayle
Electoral vote 3 0
Popular vote 192,619 20,698
Percentage 84.64% 9.10%
District Result
Clinton
  80-90%

President before election

George H. W. Bush
Republican

 Elected President

Bill Clinton
Democratic

Elections in the District of Columbia

The District of Columbia, heavily Democratic, was won in a landslide by Governor Bill Clinton (D-Arkansas) with 84.64% of the popular vote over incumbent President George H. W. Bush (R-Texas) with 9.10%. Businessman Ross Perot (I-Texas) finished in third, with 4.25% of the popular vote.[1] Clinton ultimately won the national vote, defeating incumbent President Bush and Perot.[2]

Results
1992 United States presidential election in the District of Columbia[1]
Party Candidate Votes Percentage Electoral votes
Democratic Bill Clinton 192,619 84.64% 3
Republican George H. W. Bush (incumbent) 20,698 9.10% 0
Independent Ross Perot 9,681 4.26% 0
New Alliance Lenora Fulani 1,459 0.64% 0
Independent Ronald Daniels 1,446 0.64% 0
Libertarian Andre Marrou 467 0.21% 0
N/A Write-ins 1,202 0.53% 0
Totals 227,572 100.00% 3
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See also

References
  1. "1992 Presidential General Election Results - District of Columbia". U.S. Election Atlas. Retrieved 8 June 2012.
  2. "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.
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