United States presidential election in Mississippi, 1992

United States presidential election in Mississippi, 1992
Mississippi
November 3, 1992
 
Nominee George H.W. Bush Bill Clinton Ross Perot
Party Republican Democratic Independent
Home state Texas Arkansas Texas
Running mate Dan Quayle Al Gore James Stockdale
Electoral vote 7 0 0
Popular vote 487,793 400,258 85,626
Percentage 49.68% 40.77% 8.72%

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

Elections in Mississippi

The 1992 United States presidential election in Mississippi took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

Mississippi was won by incumbent President George H.W. Bush (R-Texas) with 49.68% of the popular vote over Governor Bill Clinton (D-Arkansas) with 40.77%. Businessman Ross Perot (I-Texas) finished in third with 8.72% of the popular vote.[1] Clinton ultimately won the national vote, defeating both incumbent President Bush and Perot.[2]

Results

United States presidential election in Mississippi, 1992[1]
Party Candidate Votes Percentage Electoral votes
Republican George H.W. Bush (incumbent) 487,793 49.68% 7
Democratic Bill Clinton 400,258 40.77% 0
Independent Ross Perot 85,626 8.72% 0
New Alliance Party Lenora Fulani 2,625 0.27% 0
Libertarian Andre Marrou 2,154 0.22% 0
Constitution Party Howard Phillips 1,652 0.17% 0
Natural Law John Hagelin 1,140 0.12% 0
Populist James "Bo" Gritz 545 0.06% 0
Totals 981,793 100.0% 7

References

  1. ↑ 1.0 1.1 "1992 Presidential General Election Results - Mississippi". U.S. Election Atlas. Retrieved 8 June 2012.
  2. ↑ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.