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United States presidential election in West Virginia, 1992

United States presidential election in West Virginia, 1992

November 3, 1992

 
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 48.41% 35.39% 15.92%

County Results
  Clinton—>70%
  Clinton—60-70%
  Clinton—50-60%
  Clinton—40-50%
  Bush—40-50%
  Bush—50-60%
  Bush—60-70%
  Bush—>70%
  Perot—40-50%

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.

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]

Democrat

  1. ^ a b
  2. ^

References

Totals 683,677 100.0% 5
Libertarian Andre Marrou 1,873 0.27% 0
Independent Ross Perot 108,829 15.92% 0
Republican incumbent) 241,974 35.39% 0
Democratic Bill Clinton 331,001 48.41% 5
Party Candidate Votes Percentage Electoral votes
United States presidential election in West Virginia, 1992[1]


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