United States presidential election in Maryland, 1988
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350px County Results
Dukakis—70-80%
Dukakis—60-70%
Dukakis—50-60%
Bush—50-60%
Bush—60-70%
Bush—70-80%
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The 1988 United States presidential election in Maryland took place on November 8, 1988, as part of the 1988 United States presidential election. Voters chose ten representatives, or electors to the Electoral College, who voted for President and Vice President.
Maryland was won by Vice President George H.W. Bush (R-Texas), with 51.11% of the popular vote, over Massachusetts Governor Mike Dukakis (D-Massachusetts) with 48.20% of the popular vote.[1] Bush ultimately won the national vote, defeating Governor Dukakis.[2]
Results
| United States presidential election in Maryland, 1988[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Republican | George H.W. Bush | 876,167 | 51.11% | 10 | |
| Democratic | Mike Dukakis | 826,304 | 48.20% | 0 | |
| Libertarian | Ron Paul | 6,748 | 0.39% | 0 | |
| New Alliance | Lenora Fulani | 5,115 | 0.29% | 0 | |
| N/A | Write-ins | 24 | <0.01% | 0 | |
| Totals | 1,714,358 | 100.0% | 10 | ||
References
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| Other 1988 elections |
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