Question 1201062
You are missing some information, but here is the approach.
error = z*sqrt(p*(1-p))/n
suppose z=1.96 for 95% CI
and suppose p=0.4 and n=50
then error=1.96*(sqrt(.24/50)
=0.1358 
np and n(1-p) are both more than 10 so yes can assume normality.

note that the error for probability around 50% is about 1/sqrt(n).
This comes from using z=2 and p=0.5,  Very useful to know this.