1992 United States presidential election in Nebraska
The 1992 United States presidential election in Nebraska took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.
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Elections in Nebraska |
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Nebraska was won by incumbent President George H. W. Bush (R-Texas) with 46.58% of the popular vote over Governor Bill Clinton (D-Arkansas) with 29.40%. Businessman Ross Perot (I-Texas) finished in third, with 23.63% of the popular vote.[1] Clinton ultimately won the national vote, defeating both incumbent President Bush and Perot.[2]
With 46.58% of the popular vote, Nebraska would prove to be Bush's fourth strongest state in the 1992 election after Mississippi, South Carolina and Alabama.[3] It was also Bush's best performance as far as margin of victory, defeating Clinton by 17.18%.
Results
1992 United States presidential election in Nebraska[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Republican | George H.W. Bush (incumbent) | 344,346 | 46.58% | 5 | |
Democratic | Bill Clinton | 217,344 | 29.40% | 0 | |
Independent | Ross Perot | 174,687 | 23.63% | 0 | |
Libertarian | Andre Marrou | 1,344 | 0.18% | 0 | |
New Alliance Party | Lenora Fulani | 848 | 0.11% | 0 | |
Natural Law | Dr. John Hagelin | 714 | 0.10% | 0 | |
Totals | 739,283 | 100.0% | 5 |
References
- "1992 Presidential General Election Results - Nebraska". U.S. Election Atlas. Retrieved 8 June 2012.
- "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.
- "1992 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.