United States presidential election in Missouri, 1992

United States presidential election in Missouri, 1992
Missouri
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 11 0 0
Popular vote 1,053,873 811,159 518,741
Percentage 44.1% 33.9% 21.7%

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

This article describes the United States presidential election, 1992, in Missouri. Since 1904, Missouri has voted for the eventual winner of the election in a presidential election, with the exceptions of the 1956, 2008, and 2012 elections.

Presidential Candidate Running Mate Party Electoral Vote (EV) Popular Vote (PV)
Bill Clinton of Arkansas Al Gore of Tennessee Democratic 11[1] 1,053,873 44.07%
George H. W. Bush Dan Quayle Republican 0 811,159 33.92%
Ross Perot James Stockdale Independent Party 0 518,741 21.69%
Andre Marrou Nancy Lord Libertarian Party 0 7,497 0.31%

Statewide winner in bold. See main article : U.S. presidential election, 1992.

References

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