United States presidential election in Washington (state), 1992
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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% | |||||||||||||||||||||||||||||||||||||||
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Elections in Washington |
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The 1992 United States presidential election in Washington took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose eleven representatives, or electors to the Electoral College, who voted for President and Vice President.
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.[1] Clinton ultimately won the national vote, defeating incumbent President Bush.[2]
Results
United States presidential election in Washington, 1992[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic | Bill Clinton | 993,037 | 43.41% | 11 | |
Republican | George H.W. Bush (incumbent) | 731,234 | 31.97% | 0 | |
Independent | Ross Perot | 541,780 | 23.68% | 0 | |
Libertarian | Andre Marrou | 7,533 | 0.33% | 0 | |
America First | James "Bo" Gritz | 4,854 | 0.21% | 0 | |
Natural Law | John Hagelin | 2,456 | 0.11% | 0 | |
U.S. Taxpayers' | Howard Phillips | 2,354 | 0.10% | 0 | |
New Alliance | Lenora Fulani | 1,776 | 0.08% | 0 | |
Independent | Ronald Daniels | 1,171 | 0.05% | 0 | |
Democrats for Economic Recovery | Lyndon LaRouche | 855 | 0.04% | 0 | |
Socialist Workers | James Warren | 515 | 0.02% | 0 | |
Totals | 2,287,565 | 100.0% | 11 |
References
- 1 2 "1992 Presidential General Election Results - Washington". U.S. Election Atlas. Retrieved 11 June 2012.
- ↑ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 11 June 2012.