Answer:
The 95%-confidence interval for the percentage of history majors at U.S. universities who knew Abraham Lincoln's first Vice President is (10.5%, 12.1%).
Explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the z-score that has a p-value of
.
6250 students, 11.3% knew Abraham Lincoln's first vice-president.
This means that

95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

As percentages:
0.105*100% = 10.5%
0.121*100% = 12.1%
The 95%-confidence interval for the percentage of history majors at U.S. universities who knew Abraham Lincoln's first Vice President is (10.5%, 12.1%).