My Solution::
b i) no answer
b ii) no answer
b iii) no answer
You should have a table of the standard normal distribution.
Let $\displaystyle z=(t-90)/20$, where $\displaystyle t$ is the time a punter spends in the casino in minutes, then $\displaystyle z$ has a standard normal distribution.
a) That $\displaystyle t>100$ corresponds to $\displaystyle z>(100-90)/20=0.5$
If you have a table of the cumulative standard normal you can look $\displaystyle z=0.5$ up on that table to tell you $\displaystyle p(z<0.5)=0.6915$ but $\displaystyle p(z>0.5)=1-p(z<0.5)=0.3085$.
b) That $\displaystyle 80<t<130$ corresponds to $\displaystyle (80-90)/20<z<(130-90)/20$ or $\displaystyle -0.5<z<2$
If you have a table of the cumulative standard normal you can look $\displaystyle z=-0.5$ up on that table to tell you $\displaystyle p(z<-0.5)=0.3085$ and $\displaystyle z=2$ to tell you $\displaystyle p(z<2)=0.9772$, then $\displaystyle p(-0.5<z<2)=p(z<2)-p(z<-0.5)=0.9772-0.3085=0.6687$
RonL