United States presidential election in Nebraska, 1992

United States presidential election in Nebraska, 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 5 0 0 Popular vote 344,346 217,344 174,687 Percentage 46.58% 29.40% 23.63%

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 Nebraska 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.

Nebraska was won by incumbent President George H.W. Bush (R-Texas) with 46.58% of the popular vote over Governor Bill Clinton (D-Arkansas) with 29.40%. Businessman Ross Perot (I-Texas) finished in third with 23.63% 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 Nebraska, 1992[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Republican	George H.W. Bush (incumbent)	344,346	46.58%	5
	Democratic	Bill Clinton	217,344	29.40%	0
	Independent	Ross Perot	174,687	23.63%	0
	Libertarian	Andre Marrou	1,344	0.18%	0
	New Alliance Party	Lenora Fulani	848	0.11%	0
	Natural Law	Dr. John Hagelin	714	0.10%	0
Totals			739,283	100.0%	5

References

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