United States presidential election in Mississippi, 1992

United States presidential election in Mississippi, 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 7 0 0 Popular vote 487,793 400,258 85,626 Percentage 49.7% 40.8% 8.7%

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 Mississippi took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

Mississippi was won by incumbent President George H.W. Bush (R-Texas) with 49.68% of the popular vote over Governor Bill Clinton (D-Arkansas) with 40.77%. Businessman Ross Perot (I-Texas) finished in third with 8.72% of the popular vote.^[1] Clinton ultimately won the national vote, defeating both incumbent President Bush and Perot.^[2]

Results

United States presidential election in Mississippi, 1992[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Republican	George H.W. Bush (incumbent)	487,793	49.68%	7
	Democratic	Bill Clinton	400,258	40.77%	0
	Independent	Ross Perot	85,626	8.72%	0
	New Alliance Party	Lenora Fulani	2,625	0.27%	0
	Libertarian	Andre Marrou	2,154	0.22%	0
	Constitution Party	Howard Phillips	1,652	0.17%	0
	Natural Law	John Hagelin	1,140	0.12%	0
	Populist	James "Bo" Gritz	545	0.06%	0
Totals			981,793	100.0%	7

References

1. "1992 Presidential General Election Results - Mississippi". U.S. Election Atlas. Retrieved 8 June 2012.
2. "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.