An automobile insurance company claims that teenage drivers save on average $490 per year when compared to the same coverage fro

m another company, with a standard deviation of $65 per year. Natalia is a teenager and a customer of this company and her average savings was $470 per year. What is the z-value for Natalia’s savings rounded to the nearest hundredth?
Question 1 options:

a)
The z-value is -0.90.
b)
The z-value is 0.90
c)
The z-value is 0.31.
d)
The z-value is -0.31.