Answer:
A sample of 11352 is needed.
Step-by-step 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 zscore that has a pvalue of
.
The margin of error is given by:

For this problem, we have that:
400 cards, 120 cases were correct. So

95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
How large a sample n would you need to estimate p with margin of error 0.01 and 95% confidence
We would need a sample of size n, and n is found when
. So






Rounding up
A sample of 11352 is needed.