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| Elections in Idaho |
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The 1992 United States presidential election in Idaho took place on November 3, 1992, as part of the 1992 United States presidential election. State 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] As of the 2020 presidential election[update], this is the last election in which Bonner County, Clearwater County, Lewis County, and Benewah County voted for a Democratic presidential candidate.[3]
With 27.05 percent of the popular vote, Idaho would prove to be Perot's fourth strongest state after Maine, Alaska and Utah.[4]
In Franklin County, Clinton was a mere 22 votes away from placing 4th behind Independent candidate, Bo Gritz who himself scored 12.40%, his strongest county in the entire state.
- ^ Cite error: The named reference
resultswas invoked but never defined (see the help page). - ^ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved June 8, 2012.
- ^ Sullivan, Robert David; ‘How the Red and Blue Map Evolved Over the Past Century’; America Magazine in The National Catholic Review; June 29, 2016
- ^ "1992 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.