Answer:
Normal distribution table
Step-by-step explanation:
We need to define all the variables that are presented to us in the problem, like this:
Data point in question = xi = 36000
The mean = μ = 32000
The standar variation = s = 3000
calculate the z-score for the left bound
x = 36000:
[tex]z = \frac{x_{i}-\mu}{\frac{s}{\sqrt{n}} }[/tex]
where μ = 32000, s = 3000, n = 1. (There is only one
teacher in the sample.)
z-score for 36000 is calculated thusly:
[tex]z = \frac{36000-32000}{3000} =1.3333[/tex]
It is necessary to look in the normalized table for this value, and look in the area on the left that for 1.33 the result is 0.9082
So our probability is 0.9082 that he or she makes more than $36000
