United States presidential election in Oregon, 1992

United States presidential election in Oregon, 1992
Oregon
November 3, 1992
 
Nominee Bill Clinton George H.W. Bush Ross Perot
Party Democratic Republican Independent
Home state Arkansas Texas Texas
Running mate Al Gore Dan Quayle James Stockdale
Electoral vote 7 0 0
Popular vote 621,314 475,757 354,091
Percentage 42.48% 32.53% 24.21%

County Results
  Clinton—50-60%
  Clinton—40-50%
  Clinton—<40%
  Bush—<40%
  Bush—40-50%
  Bush—50-60%
President before election

George H. W. Bush
Republican

 Elected President

Bill Clinton
Democratic

The 1992 United States presidential election in Oregon 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.

Oregon was won by Governor Bill Clinton (D-Arkansas) with 42.48% of the popular vote over incumbent President George H.W. Bush (R-Texas) with 32.53%. Businessman Ross Perot (I-Texas) finished in third with 24.21% of the popular vote.[1] Clinton ultimately won the national vote, defeating incumbent President Bush.[2]

Results

United States presidential election in Oregon, 1992[1]
Party Candidate Votes Percentage Electoral votes
Democratic Bill Clinton 621,314 42.48% 7
Republican George H.W. Bush (incumbent) 475,757 32.53% 0
Independent Ross Perot 354,091 24.21% 0
Libertarian Andre Marrou 4,277 0.29% 0
New Alliance Lenora Fulani 3,030 0.21% 0
Other write-ins 2,609 0.18% 0
America First James "Bo" Gritz (write-in) 1,470 0.10% 0
Natural Law Dr. John Hagelin (write-in) 91 0.01% 0
Independent J. Quinn Brisben (write-in) 4 0.00% 0
Totals 1,462,643 100.0% 7

References

  1. ↑ 1.0 1.1 "1992 Presidential General Election Results - Oregon". U.S. Election Atlas. Retrieved 9 June 2012.
  2. ↑ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 9 June 2012.