United States presidential election in Wyoming, 1992

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United States presidential election in Wyoming, 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 79,347 68,160 51,263
Percentage 39.70% 34.10% 25.65%

WY1992.jpg
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 Wyoming 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.

Wyoming was won by incumbent President George H.W. Bush (R-Texas) with 39.70% of the popular vote over Governor Bill Clinton (D-Arkansas) with 34.10%. Businessman Ross Perot (I-Texas) finished in third with 25.65% of the popular vote.[1] Clinton ultimately won the national vote, defeating incumbent President Bush. This election was one of the closest results Wyoming ever had in U.S. history, since 1948, when it narrowly voted for Harry S. Truman over Thomas E. Dewey.[2]

Results

United States presidential election in Wyoming, 1992[1]
Party Candidate Votes Percentage Electoral votes
Republican George H.W. Bush 79,347 39.70% 3
Democratic Bill Clinton 68,160 34.10% 0
Independent Ross Perot 51,263 25.65% 0
Libertarian Andre Marrou 844 0.42% 0
New Alliance Party Lenora Fulani 270 0.14% 0
Totals 199,884 100.0% 3

References

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