United States presidential election in Kansas, 1992

United States presidential election in Kansas, 1992

November 3, 1992

Nominee George H.W. Bush Bill Clinton Ross Perot Party Republican Democratic Independent Home state Texas Arkansas Texas Running mate Dan Quayle Al Gore James Stockdale Electoral vote 6 0 0 Popular vote 446,951 390,434 312,358 Percentage 38.9% 33.7% 27.0%

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%

President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

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.^[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. "1992 Presidential General Election Results - Kansas". U.S. Election Atlas. Retrieved 8 June 2012.
2. "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.