1992 United States presidential election in Wisconsin

1992 United States presidential election in Wisconsin

← 1988 November 3, 1992 1996 →
 
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 1,041,066 930,855 544,479
Percentage 41.13% 36.78% 21.51%


President before election

George H. W. Bush
Republican

Elected President

Bill Clinton
Democratic

The 1992 United States presidential election in Wisconsin took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for president and vice president.

Wisconsin was won by Governor Bill Clinton (D-Arkansas) with 41.13 percent of the popular vote over incumbent President George H. W. Bush (R-Texas) with 36.78 percent. Businessman Ross Perot (I-Texas) finished in third, with 21.51 percent of the popular vote.[1] Clinton ultimately won the national vote, defeating incumbent President Bush.[2]

As of the 2020 presidential election, this is the last election in which Florence County voted for a Democratic presidential candidate, and the last time that Door County backed a losing candidate.[3] Despite Clinton's decently solid victory in Wisconsin, it marks his narrowest one in a state that Michael Dukakis won in 1988.

  1. ^ Cite error: The named reference results was invoked but never defined (see the help page).
  2. ^ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved June 11, 2012.
  3. ^ Sullivan, Robert David; ‘How the Red and Blue Map Evolved Over the Past Century’; America Magazine in The National Catholic Review; June 29, 2016

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