Step 1:

$X\sim N(80,10^2)$

Let $Z=\large\frac{X-\mu}{\sigma}$

$\qquad=\large\frac{X-80}{10}$ be the standard normal variate.

Step 2:

$P(65 \leq X\leq 100)$

When $X=65$

$Z=\large\frac{65-80}{10}$$=-1.5$

When $X=100,Z=2$

$\therefore P(65 \leq X \leq 100)=P(-1.5\leq Z \leq 2)$

$\qquad\qquad\qquad\quad\;\;=P(-1.5\leq Z\leq 0)+P(0\leq Z\leq 2)$

$\qquad\qquad\qquad\quad\;\;=P(0\leq Z\leq 1.5)+P(0\leq Z\leq 2)$

$\qquad\qquad\qquad\quad\;\;=0.4332+0.4772$

$\qquad\qquad\qquad\quad\;\;=0.9104$