United States presidential election in North Dakota, 1992

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United States presidential election in North Dakota, 1992

← 1988 November 3, 1992 1996 →
  43 George H.W. Bush 3x4.jpg 44 Bill Clinton 3x4.jpg RossPerotColor.jpg
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%

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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 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.

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

United States presidential election in North Dakota, 1992[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

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