United States presidential election in Montana, 1992

United States presidential election in Montana, 1992
Montana
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 3 0 0
Popular vote 154,507 144,207 107,225
Percentage 37.63% 35.12% 26.12%

President before election

George H. W. Bush
Republican

 Elected President

Bill Clinton
Democratic

Elections in Montana

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

Montana was won by Governor Bill Clinton (D-Arkansas) with 37.63% of the popular vote over incumbent President George H.W. Bush (R-Texas) with 35.12%. Businessman Ross Perot (I-Texas) finished in third with 26.12% of the popular vote.[1] Clinton ultimately won the national vote, defeating incumbent President Bush.[2]

Results

United States presidential election in Montana, 1992[1]
Party Candidate Votes Percentage Electoral votes
Democratic Bill Clinton 154,507 37.63% 3
Republican George H.W. Bush (incumbent) 144,207 35.12% 0
Independent Ross Perot 107,225 26.12% 0
America First James "Bo" Gritz 3,658 0.89% 0
Libertarian Andre Marrou 986 0.24% 0
Totals 410,583 100.0% 3

References

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