Question

1. Assume that a loss history is normally distributed with a mean of $5,000 and a standard deviation of $1,500.

a. In what range, with 95 % certainty, can we expect the losses to fall this coming year? (2 points)

b. If the risk manager is willing to tolerate 1.5% chance that costs will be greater than the maximum probable loss, what is the maximum probable loss level? (1 point)

c. What’s the probability that the next loss will be greater than $5,750? Please present your final answer in percentage format, for example, 5.58%. (1.5 points)

2. Assume the probability of having a defective product is 10% and the number of loss follows a binomial distribution

a. What’s expected number of defective products for a batch of 10,000? What is the standard deviation that associates with the number of defective products? (1 point)

b. What is the probability there will be less than 1,066 defective products next year? Use the normal approximation. (1 point)

c. In what range, with 90 % certainty, can we expect the losses to fall this coming year? (2 points)

3. Assume the number of traffic accidents that occur on a particular stretch of road during a year follows a Poisson distribution with a mean of 36. For a risk manager with a risk tolerance level of 6.30%, what would be the estimated maximum probable number of traffic accidents? (1.5 point)

Answer 1

1)

a)for 95% middle values critical z =1.96

hence range of middle 95% values =mean -/+z*std deviaiton=5000-/+1.96*1500=2060 to 7940

b)

for top 1.5 % vlaues ; critical z =2.17

therefore corresponding value=mean+z*std deviation =5000+2.17*1500=8255

c)

probability that the next loss will be greater than $5,750 =P(X>5750)=P(Z>(5750-5000)/1500)

=P(Z>0.5)=0.3085

2)

a) expected number of defective products for a batch of 10,000 np=10000*0.1 =1000

std deviation =sqrt(np(1-p)) =sqrt(10000*0.1*0.9)=30

b)

probability there will be less than 1,066 defective products next year =P(X<1066)

=P(Z<(1065.5-1000)/30)=P(Z<2.18)=0.9854

c)

for middle 90% ; critical z =1.645

hence corresponding range =mean-/+ z*std deviaiton=1000-/+1.645*30=950.65 to 1049.35

3)fr poisson distribution ; mean =36

and std deviation=sqrt(36)=6

for top 6.3% ; critical z =1.53

estimated maximum probable number of traffic accidents =36+1.53*6=45.18