Question

Your insurance company has converage for three types of cars. The annual cost for each type...

Your insurance company has converage for three types of cars. The annual cost for each type of car can be modeled using Gaussian (Normal) distribution, with the following parameters:

  • Car Type 1: Mean=$520 and Standard Deviation=$110
  • Car Type 2: Mean=$720 and Standard Deviation=$170
  • Car Type 3: Mean=$470 and Standard Deviation=$80

Use the Random number generator and simulate 1000-long columns, for each of the three cases. Example: for the Car Type 1, use Number of variables=1, Number of random numbers=1000, Distribution=Normal, Mean=520 and Standard deviation=110, and leave random Seed empty.

Next, use either sorting to construct the appropriate histogram or rule of thumb to answer the questions:

13. What is approximate probability that Car Type 1 has an annual cost less than $350?

  • a. Between 1% and 3%
  • b. Between 3% and 9%
  • c. Between 10% and 15%
  • d. None of these

14. Which of the three types of cars is least likely to cost less than $350?

  • a. Type 1
  • b. Type 2
  • c. Type 3

15. For which of the three types we expect that (approximately) 95% of cases will be between $300 and $740?

  • a. Type 1
  • b. Type 2
  • c. Type 3

Homework Answers

Answer #1

Answer:

13. What is approximate probability that Car Type 1 has an annual cost less than $350?

The following information has been provided:

We need to compute. The corresponding z-value needed to be computed:

Therefore,

b. Between 3% and 9%

14. Which of the three types of cars is least likely to cost less than $350?

The car with maximum mean and standard deviation is least likely to cost less than $350

The correct answer is type 2

15. For which of the three types we expect that (approximately) 95% of cases will be between $300 and $740?

Simplest way to check is to obtain the middle value of the interval, the car with middle value of the interval = mean will be the answer

( 300 + 740 )/2= 1140/2 = 520

a. Type 1

NOTE::

I HOPE YOUR HAPPY WITH MY ANSWER....***PLEASE SUPPORT ME WITH YOUR RATING...

***PLEASE GIVE ME "LIKE"...ITS VERY IMPORTANT FOR ME NOW....PLEASE SUPPORT ME ....THANK YOU

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
Your insurance company has converged for three types of cars. The annual cost for each type...
Your insurance company has converged for three types of cars. The annual cost for each type of cars can be modeled using Gaussian (Normal) distribution, with the following parameters: (Discussions allowed!) Car type 1 Mean=$520 and Standard Deviation=$110 Car type 2 Mean=$720 and Standard Deviation=$170 Car type 3 Mean=$470 and Standard Deviation=$80 Use Random number generator and simulate 1000 long columns, for each of the three cases. Example: for the Car type 1, use Number of variables=1, Number of random...
Your health insurance company has converge for three types of medical treatment. The annual cost for...
Your health insurance company has converge for three types of medical treatment. The annual cost for each type of treatment can be modeled using Gaussian (Normal) distribution, with the following parameters: Treatment 1: Mean=$1600 and Standard Deviation=$120 Treatment 2: Mean=$1500 and Standard Deviation=$95 Treatment 3: Mean=$1200 and Standard Deviation=$70 Use Random number generator and simulate 1000 long columns, for each of the three cases. Example: for the Treatment 1, use Number of variables=1, Number of random numbers=1000, Distribution=Normal, Mean=1600 and...
CNNBC recently reported that the mean annual cost of auto insurance is 1008 dollars. Assume the...
CNNBC recently reported that the mean annual cost of auto insurance is 1008 dollars. Assume the standard deviation is 130 dollars. You will use a simple random sample of 110 auto insurance policies. Find the probability that a single randomly selected policy has a mean value between 998.1 and 1041.5 dollars. P(998.1 < X < 1041.5) = Find the probability that a random sample of size n=110n=110 has a mean value between 998.1 and 1041.5 dollars. P(998.1 < M <...
A company has a policy of retiring company cars; this policy looks at number of miles...
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 35 months and a standard deviation of 3 months. Using the 68-95-99.7 rule, what is the approximate percentage of cars that remain in service between 38 and 44 months? Do not enter the...
A company has a policy of retiring company cars; this policy looks at number of miles...
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 58 months and a standard deviation of 9 months. Using the empirical rule, what is the approximate percentage of cars that remain in service between 31 and 49 months?
A company has a policy of retiring company cars; this policy looks at number of miles...
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 47 months and a standard deviation of 9 months. Using the 68-95-99.7 rule, what is the approximate percentage of cars that remain in service between 20 and 38 months?
A company has a policy of retiring company cars; this policy looks at number of miles...
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 65 months and a standard deviation of 10 months. Using the empirical rule (as presented in the book), what is the approximate percentage of cars that remain in service between 35 and 45...
A company has a policy of retiring company cars; this policy looks at number of miles...
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 40 months and a standard deviation of 6 months. Using the 68-95-99.7 rule, what is the approximate percentage of cars that remain in service between 28 and 34 months?
A company has a policy of retiring company cars; this policy looks at number of miles...
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 60 months and a standard deviation of 7 months. Using the 68-95-99.7 rule, what is the approximate percentage of cars that remain in service between 67 and 81 months?
A company has a policy of retiring company cars; this policy looks at number of miles...
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 36 months and a standard deviation of 5 months. Using the empirical rule (as presented in the book), what is the approximate percentage of cars that remain in service between 21 and 31...