United States presidential election in Missouri, 1992

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United States presidential election in Missouri, 1992

← 1988 November 3, 1992 1996 →
  44 Bill Clinton 3x4.jpg 43 George H.W. Bush 3x4.jpg Ross Perot.jpg
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 44.07% 33.92% 21.69%

MO1992.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

Lua error in package.lua at line 80: module 'strict' not found. 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