# Thread: need help on probability

1. ## need help on probability

My Solution::

2. Originally Posted by littlemissy
From the long experience, a staff in casino found that the duration a visitor would spend in the casino per visit is normally distributed with the mean 90 minutes and standard deviation 20mins

1)If one visitor was selected randomly, find the probability that he would spend
a) more than 100minutes
b) between 80 minutes and 130 minutes
...
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

3. Quention 2 is hard for me....sorry cant help