Question

Suppose that an insurance company sells a policy whose losses are distributed exponentially with mean $1500. Further suppose that the company sells a large number of claims. An actuary wishes to analyze the performance of the product and takes a random sample of 100 policies for which there was a claim filed from this population.

a. What are the mean and variance of an individual insurance policy?

b. What are the mean and variance of the total claim amount T of these 100 selected policies?

c. What are the mean and variance of the average claim amount ?̅ = ?/100 of these 100 policies?

d. Calculate the approximate probability that the total claim amount T from the sample is between $160,000 and $170,000?

e. Calculate the approximate probability that the average claim ?̅ from the sample exceeds $1700.

f. What average claim amount will 18.67% of sample means exceed based on the information above?

Answer 1

a)for exponential distribution

mean =1500

and variance=1500² =2250000

b)

mean of total claim of 100 policies =1500*100 =150000

and variance =2250000*100=225000000

c)

mean of average claim of 100 policies =1500

and variance =2250000/100=22500

d)

for std deviation of total amount =sqrt(225000000)=15000

therefore from normal approximation:

probability that the total claim amount T from the sample is between $160,000 and $170,000 =P(160000<T<170000)

=P((160000-150000)/15000<Z<(170000-150000)/15000)=P(0.67 <Z <1.33)=0.9082-0.7486 =0.1596

e) std deviation of average amount=sqrt(22500)=150

P(Xbar>1700)=P(Z>(1700-1500)/150)=P(Z>1.33)=0.0918

f)

for 18.67% top values; crtiical z =0.89

therefore corresponding amount =mean+z*std deviation =1500+0.89*150=1633.5