United States presidential election in Wyoming, 1992

United States presidential election in Wyoming, 1992

November 3, 1992

Nominee George H.W. Bush Bill Clinton Ross Perot Party Republican Democratic Independent Home state Texas Arkansas Texas Running mate Dan Quayle Al Gore James Stockdale Electoral vote 3 0 0 Popular vote 79,347 68,160 51,263 Percentage 39.70% 34.10% 25.65%

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 Wyoming 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.

Wyoming was won by incumbent President George H.W. Bush (R-Texas) with 39.70% of the popular vote over Governor Bill Clinton (D-Arkansas) with 34.10%. Businessman Ross Perot (I-Texas) finished in third with 25.65% of the popular vote.^[1] Clinton ultimately won the national vote, defeating incumbent President Bush. This election was one of the closest results Wyoming ever had in U.S. history, since 1948, when it narrowly voted for Harry S. Truman over Thomas E. Dewey.^[2]

Results

United States presidential election in Wyoming, 1992[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Republican	George H.W. Bush	79,347	39.70%	3
	Democratic	Bill Clinton	68,160	34.10%	0
	Independent	Ross Perot	51,263	25.65%	0
	Libertarian	Andre Marrou	844	0.42%	0
	New Alliance Party	Lenora Fulani	270	0.14%	0
Totals			199,884	100.0%	3

References

1. "1992 Presidential General Election Results - Wyoming". U.S. Election Atlas. Retrieved 11 June 2012.
2. "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 11 June 2012.