1992 United States presidential election in Iowa

1992 United States presidential election in Iowa

← 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 7 0 0
Popular vote 586,353 504,891 253,468
Percentage 43.29% 37.27% 18.71%

County Results

President before election

George H. W. Bush
Republican

Elected President

Bill Clinton
Democratic

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

Iowa was won by Democratic Governor Bill Clinton of Arkansas with 43.29% of the popular vote over incumbent Republican President George H. W. Bush's 37.27%, a victory margin of 6.01%. Independent businessman Ross Perot finished in third, with 18.71% of the popular vote.[1] Clinton ultimately won the national vote, defeating incumbent President Bush and Perot.[2] Iowa was the only state which swung more Republican than it had been in 1988.

  1. ^ Cite error: The named reference results was invoked but never defined (see the help page).
  2. ^ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved June 8, 2012.

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