Answer
$p(10\leq x\leq12)=.3830$
Work Step by Step
$z=\frac{x-\mu}{\sigma}=\frac{10-11}{2}=-.5$
$z=\frac{x-\mu}{\sigma}=\frac{12-11}{2}=.5$
$p(10\leq x\leq12)=p(-.5\leq z\leq.5)=p(-.5\leq z\leq0)+p(0\lt z\leq.5)=2p(0\lt z\leq.5)=2(.1915)=.3830$
According to Table IV, page 773:
$p(0\lt z\leq.5)=.1915$