United States presidential election in Minnesota, 1992

United States presidential election in Minnesota, 1992

← 1988 November 3, 1992 1996 →
Nominee Bill Clinton George H.W. Bush Ross Perot Party DFL Republican Independent Home state Arkansas Texas Texas Running mate Al Gore Dan Quayle Electoral vote 10 0 0 Popular vote 1,020,997 747,841 562,506 Percentage 43.5% 31.9% 24.0%
County Results   Clinton—50-60%   Clinton—40-50%   Clinton—<40%   Bush—<40%   Bush—40-50%
President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

George H. W. Bush
Republican

Bill Clinton
Democratic

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

Minnesota was won by Governor Bill Clinton (D-Arkansas) with 43.48% of the popular vote over incumbent President George H.W. Bush (R-Texas) who took 31.85%, a victory margin of 11.63%. Businessman Ross Perot (I-Texas) finished in third with 23.96% of the popular vote. Clinton ultimately won the national vote, defeating incumbent President Bush.

April 7, 1992, saw the first presidential primary in Minnesota since 1956. Clinton won a plurality of votes in the DFL primary and Bush won in the IR election.