United States presidential election in West Virginia, 1992

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United States presidential election in West Virginia, 1992

← 1988 November 3, 1992 1996 →

  44 Bill Clinton 3x4 2.jpg 43 George H.W. Bush 3x4 2.jpg RossPerotColor.jpg
Nominee Bill Clinton George H.W. Bush Ross Perot
Party Democratic Republican Independent
Home state Arkansas Texas Texas
Running mate Al Gore Dan Quayle James Stockdale
Electoral vote 5 0 0
Popular vote 331,001 241,974 108,829
Percentage 48.4% 35.4% 15.9%

WV1992.jpg
County Results
  Clinton—>70%
  Clinton—60-70%
  Clinton—50-60%
  Clinton—40-50%
  Bush—40-50%
  Bush—50-60%
  Bush—60-70%
  Bush—>70%

President before election

George H. W. Bush
Republican

Elected President

Bill Clinton
Democratic

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

West Virginia was won by Governor Bill Clinton (D-Arkansas) with 48.41% of the popular vote over incumbent President George H.W. Bush (R-Texas) with 35.39%. Businessman Ross Perot (I-Texas) finished in third, with 15.92% of the popular vote.[1] Clinton ultimately won the national vote, defeating incumbent President Bush.[2]

As of the 2016 presidential election, this is the last election in which Wirt County voted for the Democratic candidate.

Results

United States presidential election in West Virginia, 1992[1]
Party Candidate Votes Percentage Electoral votes
Democratic Bill Clinton 331,001 48.41% 5
Republican George H.W. Bush (incumbent) 241,974 35.39% 0
Independent Ross Perot 108,829 15.92% 0
Libertarian Andre Marrou 1,873 0.27% 0
Totals 683,677 100.00% 5

References

  1. ^ a b "1992 Presidential General Election Results - West Virginia". U.S. Election Atlas. Retrieved 11 June 2012.
  2. ^ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 11 June 2012.
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