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%   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

1. 1992 Presidential General Election Results - Missouri