United States presidential election in Kansas, 1992
| |||||||||||||||||||||||||||||||||||||||
| |||||||||||||||||||||||||||||||||||||||
| |||||||||||||||||||||||||||||||||||||||
| |||||||||||||||||||||||||||||||||||||||
County Results
Clinton—>70%
Clinton—60-70%
Clinton—50-60%
Clinton—40-50%
Bush—40-50%
Bush—50-60%
Bush—60-70%
Bush—>70%
Tie (Perot-Bush)—<40%
Perot—40-50% | |||||||||||||||||||||||||||||||||||||||
|
Elections in Kansas |
---|
|
The 1992 United States presidential election in Kansas took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose six representatives, or electors to the Electoral College, who voted for President and Vice President.
Kansas was won by incumbent President George H.W. Bush (R-Texas) with 38.88% of the popular vote over Governor Bill Clinton (D-Arkansas) with 33.74%. Businessman Ross Perot (I-Texas) finished in third with 26.99% of the popular vote. Kansas was surprisingly close, likely because Ross Perot split the vote. Had Clinton won in Kansas, it would have been a major upset victory. He could have won the state had more of President Bush's voters voted for Perot instead, thus taking more votes away from him, allowing Clinton to take the lead and likely win the state's 6 electoral votes.[1] Clinton ultimately won the national vote, defeating incumbent President Bush and Perot.[2]
Results
United States presidential election in Kansas, 1992[1] | |||||
---|---|---|---|---|---|
Party | Candidate | Votes | Percentage | Electoral votes | |
Republican | George H.W. Bush (incumbent) | 449,951 | 38.88% | 6 | |
Democratic | Bill Clinton | 390,434 | 33.74% | 0 | |
Independent | Ross Perot | 312,358 | 26.99% | 0 | |
Libertarian | Andre Marrou | 4,314 | 0.37% | 0 | |
N/A | Other | 199 | 0.02% | 0 | |
Totals | 1,157,256 | 100.0% | 6 |
References
- 1 2 "1992 Presidential General Election Results - Kansas". U.S. Election Atlas. Retrieved 8 June 2012.
- ↑ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.
|
|