United States presidential election in Maryland, 1992

United States presidential election in Maryland, 1992

November 3, 1992

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 49.8% 35.6% 14.2%

County Results   Clinton—>70%   Clinton—60-70%   Clinton—50-60%   Clinton—40-50%   Bush—40-50%   Bush—50-60%   Bush—60-70%   Bush—>70%   Perot—40-50%

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 ten 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]

Results

References

State results of the 1992 U.S. presidential election
Candidates	Bill Clinton George H. W. Bush Ross Perot Lyndon LaRouche Full list
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Other 1992 elections	House Senate Gubernatorial