Assume the cost of an extended 100,000 mile warranty for a particular SUV follow

Question

Assume the cost of an extended 100,000 mile warranty for a particular SUV follows the normal distribution with a mean of $1,330 and a standard deviation of $90.

Complete parts (a) through (d) below.

a) Determine the interval of warranty costs from various companies that are one standard deviation around the mean.

The interval of warranty costs that are one standard deviation around the mean ranges from $_____?____to $___?_____

(Type integers or decimals. Use ascending order.)

b) Determine the interval of warranty costs from various companies that are two standard deviations around the mean.

The interval of warranty costs that are two standard deviations around the mean ranges from $___?____to $___?_____

(Type integers or decimals. Use ascending order.)

c) Determine the interval of warranty costs from various companies that are three standard deviations around the mean.

The interval of warranty costs that are three standard deviations around the mean ranges from $__?___ to $ ___?__

(Type integers or decimals. Use ascending order.)

d) An extended 100,000 mile warranty for this type of vehicle is advertised at $1690

Based on the previous results, what conclusions can you make? Choose the right answer :

A.The $1690 cost of this warranty is slightly higher than average due to the fact that it is more than three standard deviations above the mean.

B.The $1690 cost of this warranty must be an error in the advertisement because a data value cannot be more than three standard deviations from the mean.

C.The $1690 cost of this warranty is much higher than average due to the fact that it is more than three standard deviations above the mean.

D.This warranty is better quality than warranties offered by competing companies due to the fact that it is more than three standard deviations above the mean.

Explanation / Answer

Answer to the question :

a. Interval one deviation around the mean will have a Z of 1.
So, interval is : Mean +/- Z*Stdev = 1330+/-1*90 = $1240 to $1420

b.Interval two deviations around the mean will have a Z of 2.
So, interval is : Mean +/- Z*Stdev = 1330+/-2*90 = $1150 to $1510

c.Interval three deviations around the mean will have a Z of 3.
So, interval is : Mean +/- Z*Stdev = 1330+/-2*90 = $1060 to $1600

d. C is right. 1690 is 4 deviations above mean (1330+360). So, the price paid is very high compared
to the average.

D - we can't comment on quality of warranty, as no data is given
B - technically not an error, a possibility for sure. A data can be more than 3 deviations from mean
A - It is not 'slightly' higher. It is much higher.

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