1992 United States presidential election in Oklahoma
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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 Oklahoma |
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 Government |
The 1992 United States presidential election in Oklahoma took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.
Oklahoma was won by incumbent President George H.W. Bush (R-Texas) with 42.65% of the popular vote over Governor Bill Clinton (D-Arkansas) with 34.02%. Businessman Ross Perot (I-Texas) finished in third with 23.01% of the popular vote.[1] Clinton ultimately won the national vote, defeating both incumbent President Bush and Perot.[2]
Results
| United States presidential election in Oklahoma, 1992[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Republican | George H.W. Bush (incumbent) | 592,929 | 42.65% | 8 | |
| Democratic | Bill Clinton | 473,066 | 34.02% | 0 | |
| Independent | Ross Perot | 319,878 | 23.01% | 0 | |
| Libertarian | Andre Marrou | 4,486 | 0.32% | 0 | |
| Totals | 1,390,359 | 100.0% | 8 | ||
References
- ^ a b "1992 Presidential General Election Results - Oklahoma". U.S. Election Atlas. Retrieved 8 June 2012.
- ^ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.
