According to the current Commissioners Standard Ordinary mor

According to the current Commissioners\' Standard Ordinary mortality table, adopted by state insurance regulators in December 2002, a 25-year-old man has these probabilities of dying during the next five years:

Age at death

25

26

27

28

29

Probability

0.00039

0.00044

0.00051

0.00057

0.00060

a. What is the probability that the man does not die in the next five years?

b. An online insurance site offers a term insurance policy that will pay $100,000 if a 25-year-old man dies within the next 5 years. The cost is $175 per year. So the insurance company will take in $875 from this policy if the man does not die within five years. If he does die, the company must pay $100,000. Its loss depends on how many premiums were paid, as follows:

What is the insurance company\'s mean cash intake from such polices? Hint: These losses should be considered as negative numbers whereas the payments are positive.

c. Lab2204-142data.xls contains a simulation of insuring 1000 25-year-old men with the probabilities in this problem.

To run the simulation, type a 0 in one of the blank cells in column H and notice that the number alive in row 3 changes. Copy and paste the first four rows, columns A-F, into your lab report.

Repeat this 4 more times.

Answer the questions below:

i.In these 5 trials, what was the maximum number of men who died?

ii.If the insurance company had to pay out $100,000 for a death, how much money did the insurance company pay out for the trial with the most deaths?

iii.How many deaths will cause the company to take a loss?

Solution

a. What is the probability that the man does not die in the next five years?

Probability of not die = 1 - probability of die

probability of die = 0.00039 + 0.00044 + 0.00051 + 0.00057 + 0.00060 = 0.00251

Probability of not die = 1 - 0.00251 = 0.99749

b)

expected value

-100,000* 0.00251 + 875* 0.99749 = 621.80375

c)

Need more data ( the Lab2204-142data.xls) is not a link you need to copy and paste the information to this site or you can put a link

I can gladly help you but you should post it in a new question