1992 United States presidential election in Minnesota

1992 United States presidential election in Minnesota

← 1988 November 3, 1992 1996 →
Turnout	73.91%[1]
Nominee Bill Clinton George H. W. Bush Ross Perot Party Democratic (DFL) Republican Independent Home state Arkansas Texas Texas Running mate Al Gore Dan Quayle James Stockdale Electoral vote 10 0 0 Popular vote 1,020,997 747,841 562,506 Percentage 43.48% 31.85% 23.96%
County Results Precinct Results Clinton   20-30%   30-40%   40–50%   50–60%   60–70%   70–80%   80–90%   90–100% Bush   20-30%   30–40%   40–50%   50–60%   60–70%   70–80%   80–90%   90–100% Perot   30–40%   40–50%   50–60%   60–70%   70–80%   90–100% Other   30–40% Tie/No Data
President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

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v t e

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.^[2] Clinton ultimately won the national vote, defeating incumbent President Bush.^[3]

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.

Clinton became the first Democrat to win the White House without carrying Lyon or Roseau Counties since Woodrow Wilson in 1912, as well as the first to do so without carrying Becker County since Woodrow Wilson in 1916.