1992 United States presidential election in Washington (state)
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The 1992 United States presidential election in Washington took place on 3 November 1992, as part of the 1992 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for president and vice president.
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Turnout | 82.60% 5.66%[1] | ||||||||||||||||||||||||||||||||
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County Results
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Washington was won by Governor Bill Clinton (D-Arkansas) with 43.41% of the popular vote over incumbent President George H. W. Bush (R–Texas) with 31.97%. Businessman Ross Perot (I-Texas) finished in third, with 23.68% of the popular vote.[2] Clinton ultimately won the national vote, defeating incumbent President Bush.[3]
As of the 2020 presidential election[update], this is the last election in which Okanogan County has voted for a Democratic Presidential nominee.[4] It was also the first occasion since Franklin D. Roosevelt’s 1936 landslide that Whitman County had supported a Democratic presidential candidate, as well as the first time that King County was the most Democratic in the state, a trend that has continued in every presidential election since.[5]