Question

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.

Answer 1

Answer:

97%

Step-by-step explanation:

Answer 2

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