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:


The z-value is -0.90.


The z-value is 0.90


The z-value is 0.31.


The z-value is -0.31.


d) The z-value is -0.31

Step-by-step 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=\frac{X-\mu}{\sigma}\\\\=\frac{470-490}{65}\\\\=-0.30\approx -0.31

Hence, Natalia's z-value is -0.31

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