Question

Assume the cost of an extended​ 100,000 mile warranty for a particular SUV follows the normal distribution with a mean of ​$1710 and a standard deviation of ​$95. Complete parts ​(a) through ​(d) below. ​a) Determine the interval of warranty costs from various companies that are one standard deviation around the mean. The interval of warranty costs that are one standard deviation around the mean ranges from ​$ nothing to ​$ nothing. ​(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 ​$ nothing to ​$ nothing. ​(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 ​$ nothing to ​$ nothing. ​(Type integers or decimals. Use ascending​ order.) ​d) An extended​ 100,000 mile warranty for this type of vehicle is advertised at ​$2 comma 090. Based on the previous​ results, what conclusions can you​ make? A. The ​$2090 cost of this warranty is much higher than average due to the fact that it is more than three standard deviations above the mean. B. 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. C. The ​$2090 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. D. The ​$2090 cost of this warranty is slightly higher than average due to the fact that it is more than three standard deviations above the mean.

Answer 1

mean = 1710 , s = 95

a)

Interval = mean +/- 1 s
= 1710 +/- 95
= ( 1615 , 1805)

b)

Interval = 1710 +/- 2s
= 1710 +/- 2 *95
= (1520,1900)

c)

Interval = 1710 +/- 3s
= 1710 +/- 3 *95
= (1425,1995)

d)

. D. The ​$2090 cost of this warranty is slightly higher than average due to the fact that it is more than three standard deviations above the mean.