Answer
It is appropriate to use a normal distribution because the interval $(68,92)$ lies completely within the range from 0 to 1000.
Work Step by Step
$q=1-p=1-.8=.2$
$\mu=100(.8)=80$
$\sigma=\sqrt {npq}=\sqrt {100(.8)(.2)}=4$
$\mu\pm3\sigma=80\pm3(4)=(68,92)$
