Question 1170269
405/732=0.553
half-interval for 95% CI is z*sqrt(p*(1-p)/n)=1.96* sqrt (0.553*0.447/732)
=0.036
95CI is (0.517, 0.589)
We don't know the true proportion of judges who are introverts, but we are highly confident the proportion lies within the interval.
If we constructed 100 such intervals with this sample size, 95 of them would contain the true proportion.  We wouldn't know, however, which 95.
np and n(1-p) are both greater than 10. So normal distribution may be used.