Question

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

Answer 1

d) The z-value is -0.31

Explanation:

-Given that Natalia's saving is $470, mean claim amount is $490 and the standard deviation is $65

-Let Natalia's savings be X

#The z-value is calculated using the formula:

`z=(X-\mu)/(\sigma)\\\\=(470-490)/(65)\\\\=-0.30\approx -0.31`

Hence, Natalia's z-value is -0.31