Answer
$p(7.8\leq x\leq12.6)=.7333$
Work Step by Step
$z=\frac{x-\mu}{\sigma}=\frac{7.8-11}{2}=-1.6$
$z=\frac{x-\mu}{\sigma}=\frac{12.6-11}{2}=.8$
$p(7.8\leq x\leq12.6)=p(-1.6\leq z\leq.8)=p(-1.6\leq z\leq0)+p(0\lt z\leq.8)=p(0\lt z\leq1.6)+p(0\lt z\leq.8)=.4452+.2881=.7333$
According to Table IV, page 773:
$p(0\lt z\leq.8)=.2881$
$p(0\lt z\leq1.6)=.4452$