United States presidential election in Maryland, 1992

United States presidential election in Maryland, 1992
Maryland
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 49.80% 35.62% 14.18%

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

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

United States presidential election in Maryland, 1992[1]
Party Candidate Votes Percentage Electoral votes
Democratic Bill Clinton 988,571 49.80% 10
Republican George H.W. Bush 707,094 35.62% 0
Independent Ross Perot 281,414 14.18% 0
Libertarian Andre Marrou 4,715 0.24% 0
New Alliance Lenora Fulani 2,786 0.14% 0
N/A Write-ins 466 0.02% 0
Totals 1,985,046 100.0% 10

References

  1. ↑ 1.0 1.1 "1992 Presidential General Election Results - Maryland". U.S. Election Atlas. Retrieved 8 June 2012.
  2. ↑ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.