Question

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 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 3 has annual cost less than $550?

• a. Between 1% and 3%
• b. Between 27% and 39%
• c. Between 75% and 90%
• d. None of these

14. Which of the three types of cars is most likely to cost less than $400?

• a. Type 1
• b. Type 2
• c. Type 3

15. For which of the three types we have the highest probability that it will cost between $500 and $700?

• a. Type 1
• b. Type 2
• c. Type 3

Answer 1

Answer:

13)

Given,

For car type 3 :

Mean = $470

Standard Deviation = $80

To determine the probability that Car Type 3 has annual cost less than $550

i.e.,

P(CT3 < 550) = P(Z < (x-)/)

substitute values

P(CT3 < 550) = P(Z < (550 - 470)/80)

= P(Z < 80/80)

= P(Z < 1)

From the standard normal distribution table

P(CT3 < 550) = 0.8413447

= 0.8413

So option C is right answer i.e., between 75% & 90%

14)

Car type 1:

Mean = $520

Standard Deviation = $110

P(CT1 < 400) = P(Z < (x-)/)

substitute values

= P(Z < (400 - 520)/110)

= P(Z < - 1.091)

from the z table

= 0.1376364

P(CT1 < 400) = 0.1376

Car type 2:

Mean = $720

Standard Deviation = $170

P(CT2 < 400) = P(Z < (x-)/)

substitute values

= P(Z < (400 - 720)/170)

= P(Z < - 1.8824)

from z table

P(CT2 < 400) = 0.0299

Car type 3:

Mean = $470

Standard Deviation = $80

P(CT3 < 400) = P(Z < (x-)/)

substitute values

= P(Z < (400 - 470)/80)

= P(Z < - 0.875)

from z table

P(CT3 < 400) = 0.1908

Option C is right answer. i.e., type 3 is most likely

15)

Type 3