1992 United States presidential election in Arkansas

1992 United States presidential election in Arkansas
Clinton 40–50% 50–60% 60–70% 70–80%	Bush 40–50%
President before election; George H. W. Bush; Republican	Elected President ; Bill Clinton; Democratic

The 1992 United States presidential election in Arkansas took place on November 3, 1992, as part of the 1992 United States presidential election. State voters chose six representatives, or electors to the Electoral College, who voted for president and vice president.

Arkansas was won by incumbent Governor Bill Clinton (D) with 53.21% of the popular vote over incumbent President George H. W. Bush (R-Texas) with 35.48%. Businessman Ross Perot (I-Texas) finished in third, with 10.43% of the popular vote.^[1] Clinton thus won his home state by a wide margin of 17.73%, becoming the first Democratic candidate to win the state since Jimmy Carter in 1976. Arkansas and Washington, D.C., which Clinton also won, were the only contests in 1992 in which any candidate received an absolute majority of the popular vote.

50% of white voters supported Clinton, while 40% supported Bush.^[2]

^ Cite error: The named reference results was invoked but never defined (see the help page).
^ "1992 Arkansas Exit Poll Results". CNN. Archived from the original on January 30, 2011. Retrieved December 14, 2024.

[results-1] Cite error: The named reference results was invoked but never defined (see the help page).

[2] "1992 Arkansas Exit Poll Results". CNN. Archived from the original on January 30, 2011. Retrieved December 14, 2024.

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1992 United States presidential election in Arkansas

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