United States presidential election in Missouri, 1992

United States presidential election in Missouri, 1992
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 44.07% 33.92% 21.69%
County Results
  Clinton—>70%
  Clinton—60-70%
  Clinton—50-60%
  Clinton—40-50%
  Bush—40-50%
  Bush—50-60%
  Bush—60-70%
President before election

George H. W. Bush
Republican

 Elected President

Bill Clinton
Democratic

United States presidential election in Missouri, 1992 [1]
Party Candidate Votes Percentage Electoral votes
Democratic Bill Clinton 1,053,873 44.07% 11
Republican George H.W. Bush 811,159 33.92% 0
Independent Ross Perot 518,741 21.69% 0
Libertarian Andre Marrou 7,497 0.31% 0
Totals 2,384,522 100.0% 11

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. As of the 2016 presidential election, this is the last time that Warren County, Dallas County, Howell County, Harrison County, Adair County, Andrew County, Morgan County, Johnson County, Butler County, Atchison County, Phelps County, Platte County, Maries County, Stoddard County, Pulaski County, Dent County, Texas County, and Carter County voted for the Democratic candidate. This is also the last time that the margin of victory for a Democratic candidate for president was in double digits.

References

  1. ↑ 1992 Presidential General Election Results - Missouri
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