Answer
See below.
Work Step by Step
$\hat{p}=\frac{x}{n}=\frac{82}{84}\approx0.9762$
The z-value belonging to the $95\%$ confidence interval according to the table is $z=1.96$, thus the confidence interval is: $\hat{p}\pm z\sqrt{\frac{p(1-p)}{n}}$, which here is: $0.9762\pm 1.96\sqrt{\frac{0.9762\cdot(1-0.9762)}{84}}$, thus the confidence interval is $(0.9436,1]$
