A national restaurant chain claims that their servers make an average of $12.85 in tips per hour, with a standard deviation of $2.15. The servers at the restaurant chain's location in Dallas make an average of $8.65 in tips per hour. Given that the data is approximately normal, find the probability that a server, chosen at random, will make more than $8.65 in tips per hour.

We need to calculate the probability:
[tex]P(N\geq 8.65)[/tex]
wherein N is the normal law. 
[tex]P(\dfrac{N-12.85}{2.15}\geq \dfrac{8.65-12.85}{2.15})[/tex]
Simplifying the right hand side of the inequality we get:
[tex]P(\dfrac{N-12.85}{2.15}\geq -1.95[/tex]
The above formula can be written in the form:
[tex]P(N(0,1)\geq -1.95)[/tex]
[tex]N(0,1)[/tex] is the normal law with mean 0 and standard deviation 1. 
Using the table of the values we find the probability 
[tex]P(N(0,1)\geq -1.95)=P(Z\leq1.95)=0.97[/tex]