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

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::

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