1992 United States presidential election in North Dakota

The 1992 United States presidential election in North Dakota 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.

Main article: 1992 United States presidential election

1992 United States presidential election in North Dakota

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 136,244 99,168 71,084 Percentage 44.22% 32.18% 23.07%

County Results Bush   30–40%   40–50%   50–60% Clinton   30–40%   40–50%   50–60%

President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

North Dakota was won by incumbent President George H. W. Bush (R-Texas) with 44.22% of the popular vote over Governor Bill Clinton (D-Arkansas) with 32.18%. Businessman Ross Perot (I-Texas) finished in third, with 23.07% of the popular vote.[1] Clinton ultimately won the national vote, defeating incumbent President Bush and Perot.[2]

Results

1992 United States presidential election in North Dakota[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Republican	George H. W. Bush (incumbent)	136,244	44.22%	3
	Democratic	Bill Clinton	99,168	32.18%	0
	Independent	Ross Perot	71,084	23.07%	0
	Democrats for Economic Recovery	Lyndon LaRouche	642	0.21%	0
	Libertarian	Andre Marrou	416	0.14%	0
	Natural Law	Dr. John Hagelin	240	0.08%	0
	Socialist Workers Party	James Warren	193	0.06%	0
	New Alliance Party	Lenora Fulani	143	0.05%	0
	Prohibition	Earl Dodge (write-in)	3	>0.01%	0
Totals			308,133	100.0%	3

References

1. "1992 Presidential General Election Results - North Dakota". U.S. Election Atlas. Retrieved 9 June 2012.
2. "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 9 June 2012.