Question

Assume the cost of an extended​ 100,000 mile warranty for a particular SUV follows the normal...

Assume the cost of an extended​ 100,000 mile warranty for a particular SUV follows the normal distribution with a mean of ​$1,450 and a standard deviation of ​$75. 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 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.)

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 deviation 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 ​$1,750. Based on the previous​ results, what conclusions can you​ make?

A. 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.

B.The $1,750 cost of this warranty is slightly higher than average due to the fact that it is more than three standard deviations above the mean.

C.The ​$1,750 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 ​$1,750 cost of this warranty is much higher than average due to the fact that it is more than three standard deviations above the mean.

Homework Answers

Answer #1

Dear student , please like it.

Thanks.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Assume the cost of an extended 100,000 mile warranty for a particular SUV follows the normal...
Assume the cost of an extended 100,000 mile warranty for a particular SUV follows the normal distribution with a mean of $1200 and a standard deviation of $60. Complete parts. 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 (Type integers or decimals. Use ascending order.) b) Determine the interval of warranty costs from various companies...
Assume the cost of an extended? 100,000 mile warranty for a particular SUV follows the normal...
Assume the cost of an extended? 100,000 mile warranty for a particular SUV follows the normal distribution with a mean of ?$1,320 and a standard deviation of ?$105 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 ___ ?b) Determine the interval of warranty costs from various...
A credit score measures a​ person's creditworthiness. Assume the average credit score for Americans is 709....
A credit score measures a​ person's creditworthiness. Assume the average credit score for Americans is 709. Assume the scores are normally distributed with a standard deviation of 53. ​a) Determine the interval of credit scores that are one standard deviation around the mean. ​b) Determine the interval of credit scores that are two standard deviations around the mean. ​c) Determine the interval of credit scores that are three standard deviations around the mean. ​a)The interval of credit scores that are...
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for...
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 909 people age 15 or​ older, the mean amount of time spent eating or drinking per day is 1.59 hours with a standard deviation of 0.57 hour. Determine and interpret a 90​% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day. Select the correct choice below and fill in the...
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for...
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1025 people age 15 or​ older, the mean amount of time spent eating or drinking per day is 1.46 hours with a standard deviation of 0.66 hour. Complete parts ​(a) through ​(d) below. Select the correct choice below and fill in the answer​ boxes, if​ applicable, in your choice. ​(Type integers or decimals rounded to three decimal places as needed....
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for...
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1005 people age 15 or​ older, the mean amount of time spent eating or drinking per day is 1.03 hours with a standard deviation of 0.65 hour. Complete parts ​(a) through ​(d) below. ​(c) Determine and interpret a 9595​% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day. Select the...
1. Assume a normal distribution and find the following probabilities. (Round the values of z to...
1. Assume a normal distribution and find the following probabilities. (Round the values of z to 2 decimal places. Round your answers to 4 decimal places.) (a) P(x < 24 | μ = 27 and σ = 4) enter the probability of fewer than 24 outcomes if the mean is 27 and the standard deviation is 4 (b) P(x ≥ 63 | μ = 50 and σ = 7) enter the probability of 63 or more outcomes if the mean...
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for...
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 949 people age 15 or​ older, the mean amount of time spent eating or drinking per day is 1.65 hours with a standard deviation of 0.54 hour. Complete parts ​(a) through ​(d) below. c) Determine and interpret a 95 % confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day. Select...
Question 1 Suppose that the distance of fly balls hit to the outfield (in baseball) is...
Question 1 Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 248 feet and a standard deviation of 54 feet. We randomly sample 49 fly balls. a) If X = average distance in feet for 49 fly balls, then give the distribution of X. Round your standard deviation to two decimal places. X -N ( ?, ?) b)What is the probability that the 49 balls traveled an average of...
The reading speed of second grade students in a large city is approximately​ normal, with a...
The reading speed of second grade students in a large city is approximately​ normal, with a mean of 89 words per minute​ (wpm) and a standard deviation of 10 wpm. Complete parts​ (a) through​ (f). ​(a) What is the probability a randomly selected student in the city will read more than 94 words per​ minute? The probability is ___ ​(Round to four decimal places as​ needed.) ​(b) What is the probability that a random sample of 13 second grade students...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT