1992 United States presidential election in Maryland

1992 United States presidential election in Maryland

← 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 10 0 0 Popular vote 988,571 707,094 281,414 Percentage 49.80% 35.62% 14.18%

County Results Clinton   40–50%   50–60%   60–70%   70–80% Bush   30–40%   40–50%   50–60%

President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

Elections in Maryland
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The 1992 United States presidential election in Maryland took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose 10 representatives, or electors to the Electoral College, who voted for president and vice president.

Maryland was won by Governor Bill Clinton (D-Arkansas) with 49.80%of the popular vote over incumbent President George H. W. Bush (R-Texas) with 35.62%. Businessman Ross Perot (I-Texas) finished in third, with 14.18% of the popular vote.^[1] Clinton ultimately won the national vote, defeating incumbent President Bush.^[2] Maryland would be one of only three states, along with Washington DC, where if Bush’s and Perot’s vote had been combined, Clinton would still come out on top (albeit by an incredibly slim margin of 63 votes). The other two states are New York and Arkansas.

Results

Results by county

See also

• United States presidential elections in Maryland

References

This page was last edited on 3 February 2021, at 23:36