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.

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]

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	100.0%	5

References

1. "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.