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 1 has annual cost less than $550?

a. Between 10% and 13%

b. Between 23% and 29%

c. Between 55% and 70%

d. None of these

14. Which of the three types of cars is most likely to cost more than $1000?

a. Type 1

b. Type 2

c. Type 3

15. For which of the three types we have the highest average cost?

a. Type 1

b. Type 2

c. Type 3

Answer 1

Given Datos- Mean: $500 Cara Mem- $720 can Mean = $1970 ; Standard ; Strmdard ; Standard deviation deviation deviation a = $110 T: $170 or $80 110 13. P(x, <650) = pſz_ 550-1620 7. P[720.27277 A from z table PLZ 40 2727] = 0.6064 -60.6W' Lies betweon 55% and 70 % Hence option "C" is correct 1 14. car Type I Plx< 1000) =P ( 22 1000-520 ) - P122436) P ETIT 01. from a table Pize 4.36) = 0.99999 P(X<1000) - 0.9999 PX, >1000) = 1-0.9999 - 0-0000! Car type zi PCX 71000) = pſz > 1000 17001 - [z> x647] = v=P[z<1.00] from z table P(X) >\Occ) - - 06 - 0.05 car Type 3 :- P(xg > 1000) = pſ2> 1000-47c] - P[776625] = 1 - P[z261698] P(X3 >1000) 1-1 0 Option "B" type 2 15. Pupe ? option 'B' is correct