A company pays its employees an average wage of 15.90 an hour with a standard deviation of 1.50. Assume the wages are approximately normally distributed.
a) what proportion of employees receive hourly wages between 13.75 and 16.22?
b) what is the hourly wage that is exceeded by only 5% of the employees?
Here is my attempt at part a
P(13.75 < x < 16.22)
P((13.75-15.90)/1.50 < Z < (16.22-15.90)/1.50))
P(-1.433 < Z < 0.2133)
Now I am stuck, can someone explain to me what I should do next?
EDIT* I figured out how to do part a, but can someone explain to me how to do part b? I actually have no idea what part b is asking for :(.
For part b), you are asked what the 95 percentile is. You know the mean in a normal ( or symmetric) distribution is the 50 percentile. Now you need to see what z-value will give you the additional 45% of the data between the mean and the 95 percentile. By symmetry, this should be the z-value associated with the 90 percentile. By the 1-2-3 or 68-95-99 rule and by symmetry, 47.5% of the data is 2 $\sigma =3$ dviations from the mean wage. So your z-value shoulld be a bit less than $2\sigma =3$. Are you allowed tables?
I wish I knew how to draw a graph in here. Please let me know if this is not clear.