United States presidential election in Missouri, 1992
Main article: United States presidential election, 1992
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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% | |||||||||||||||||||||||||||||||||||||||
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Elections in Missouri | ||||||||||||||||
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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) | |
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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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