Answer
$p(92\leq x\leq 96)=.1498$
Work Step by Step
$z_0=\frac{x-\mu}{\sigma}=\frac{92-100}{8}=-1$
$z_0=\frac{x-\mu}{\sigma}=\frac{96-100}{8}=-.5$
According to table IV, page 773:
$p(0\lt z\lt 1)=.3413$
$p(0\lt z\lt .5)=.1915$
$p(92\leq x\leq 96)=p(-1\leq z\leq -.5)=p(-1\leq z\lt0)-p(-.5\leq z\leq0)=p(0\lt z\leq1)-p(0\leq z\leq.5)=.3413-.1915=.1498$