Chapter 18 - Inference in Practice - Apply Your Knowledge - Page 419: 18.2a

$(1.7991,2.0409)$

$1-\alpha=0.95$, 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: $1.92\pm 1.96\frac{1.183}{\sqrt{880}}$, thus the interval is: $(1.7991,2.0409)$