Answer
$(84.8343,90.3657)$
Work Step by Step
$1-\alpha=0.95$, thus according to the table, $z_{\alpha/2}=z_{0.025}=1.96$.
Thus the confidence interval is: $\overline{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt n}$, which here is: $87.6\pm 1.96\frac{15}{\sqrt{113}}$, thus the interval is: $(84.8343,90.3657)$
