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United States presidential election in Arkansas, 1992

United States presidential election in Arkansas, 1992
Arkansas
1988 ←
November 3, 1992 → 1996

  44 Bill Clinton 3x4.jpg 43 George H.W. Bush 3x4.jpg RossPerotColor.jpg
Nominee Bill Clinton George H.W. Bush Ross Perot
Party Democratic Republican Independent
Home state Arkansas Texas Texas
Running mate Al Gore Dan Quayle
Electoral vote 6 0 0
Popular vote 505,823 337,324 99,132
Percentage 53.2% 35.5% 10.4%

AR1992.jpg

County Results
  Clinton—70-80%
  Clinton—60-70%
  Clinton—50-60%
  Clinton—40-50%
  Bush—40-50%

President before election

George H. W. Bush
Republican

Elected President

Bill Clinton
Democratic


AR1992.jpg

George H. W. Bush
Republican

Bill Clinton
Democratic

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

Arkansas was won by the state's Governor Bill Clinton (D) with 53.21% of the popular vote over incumbent President George H.W. Bush (R-Texas) with 35.48%. Businessman Ross Perot (I-Texas) finished in third with 10.43% of the popular vote. Clinton ultimately won the national vote, defeating incumbent President Bush. Clinton was Governor of Arkansas at the time, and as he was popular within the state, he easily won by a margin of 17.73%, making it the first time that Arkansas had voted Democratic since 1976 when it voted for Jimmy Carter. Arkansas and Washington, D.C. were the only contests in which Clinton, or any candidate, received over 50% of the popular vote.


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