United States presidential election in Idaho, 1992

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United States presidential election in Idaho, 1992
Idaho
1988 ←
November 3, 1992
→ 1996

  43 George H.W. Bush 3x4.jpg 44 Bill Clinton 3x4.jpg RossPerotColor.jpg
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 4 0 0
Popular vote 202,645 137,013 130,395
Percentage 42.03% 28.42% 27.05%

ID1992.jpg

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

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

Idaho was won by incumbent President George H.W. Bush (R-Texas) with 42.03% of the popular vote over Governor Bill Clinton (D-Arkansas) with 28.42%. Businessman Ross Perot (I-Texas) finished in a close third with 27.05% of the popular vote.[1] Clinton ultimately won the national vote, defeating both incumbent President Bush and Perot.[2]

Results[edit]

United States presidential election in Idaho, 1992[1]
Party Candidate Votes Percentage Electoral votes
Republican George H.W. Bush (incumbent) 202,645 42.03% 4
Democratic Bill Clinton 137,013 28.42% 0
Independent Ross Perot 130,395 27.05% 0
America First James "Bo" Gritz 10,281 2.13% 0
Libertarian Andre Marrou 1,167 0.24% 0
New Alliance Party Lenora Fulani 613 0.13% 0
Totals 482,114 100.0% 4

References[edit]

  1. ^ a b "1992 Presidential General Election Results - Idaho". U.S. Election Atlas. Retrieved 8 June 2012. 
  2. ^ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.