1992 United States presidential election in the District of Columbia

1992 United States presidential election in the District of Columbia

← 1988 November 3, 1992 1996 →

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
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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.

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]

The District of Columbia would be one of only four areas where if Bush’s and Perot’s vote had been combined, Clinton would still come out on top, along with New York, Arkansas, and Maryland.

Results[]

See also[]

• United States presidential elections in the District of Columbia

References[]