United States presidential election in Oklahoma, 1992

United States presidential election in Oklahoma, 1992
Oklahoma
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 8 0 0
Popular vote 592,929 473,066 319,878
Percentage 42.65% 34.02% 23.01%

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

Elections in Oklahoma

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

Oklahoma was won by incumbent President George H.W. Bush (R-Texas) with 42.65% of the popular vote over Governor Bill Clinton (D-Arkansas) with 34.02%. Businessman Ross Perot (I-Texas) finished in third with 23.01% 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 Oklahoma, 1992[1]
Party Candidate Votes Percentage Electoral votes
Republican George H.W. Bush (incumbent) 592,929 42.65% 8
Democratic Bill Clinton 473,066 34.02% 0
Independent Ross Perot 319,878 23.01% 0
Libertarian Andre Marrou 4,486 0.32% 0
Totals 1,390,359 100.0% 8

References

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