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Problem Set I 1) Rick’s Toy Store faces the following probability distribution o

ID: 3354587 • Letter: P

Question

Problem Set I

1) Rick’s Toy Store faces the following probability distribution of fire losses in its store over the next year:

Probability

0.50

0.40

0.10

Loss

$0

$20,000

$70,000

Calculate the expected value and standard deviation of Rick’s losses for the year.

Assume that Rick pools his losses with Mark’s store, which has an identical loss distribution. Mark’s losses are independent of Rick’s. Rick and Mark agree to split the total losses in the pool equally. Show the revised probability distribution for the mean loss from the pool.

Calculate the expected value and standard deviation of the pooled mean losses

2) Maria is analyzing the workers’ compensation (WC) losses of the employees in the firm that occurred over a one-year period, based on the following data:

Number of WC Claims Filled/Worker

Number of Workers

Total Number of Claims

0

500

0

1

270

270

2

50

100

Use the information in the table to find the average frequency of losses per worker.

Use the information in the table to estimate a probability distribution for the frequency distribution of losses per worker in a year.

3) You are given the following table:

Range of Loss

Amount

Midpoint Dollar Amount of Loss

Number of Losses

Total $ Amt. of Losses

$1-2,000

$1,000

300

$300,000

$2,000-10,000

$6,000

15

$90,000

Greater than $12,000

NA

0

0

Use the information in the table to find the average severity per claim

Use the information in the table to estimate a probability distribution for the loss severity per claim.

Using your answers from question 3, part (a) and question 2, part( b), use convolution to find the average loss.

Probability

0.50

0.40

0.10

Loss

$0

$20,000

$70,000

Explanation / Answer

Rick’s Toy Store:

Expected Loss = 0.50 x 0 + 0.40 x 20,000 + 0.10 x 70,000 = $ 15,000

Variance of the loss = 0.50 x (0-15,000)2 + 0.40 x (20,000-15,000)2 + 0.10 x (70,000-15,000)2 = $ 425 x 106

Standard deviation of the loss = sqrt(Variance) = $ 20,615.53

For the pooled loss, the mean loss is the average of the losses of both Rick's store and Mark's store. The probability distribution of the mean loss of the pooled account is as per the below table:

Expected value of the mean loss = 0.25 x 0 + 0.2 x 10,000 + 0.05 x 35,000 + 0.2 x 10,000 + 0.16 x 20,000 + 0.04 x 45,000 + 0.05 x 35,000 + 0.04 x 45,000 + 0.01 x 70,000 = 15,000

Variance = 212.5 x 106

Standard deviation = $ 14,577.38

Rick's store Mark's store Mean Loss Probability 0 0 0 0.25 0 20000 10000 0.20 0 70000 35000 0.05 20000 0 10000 0.20 20000 20000 20000 0.16 20000 70000 45000 0.04 70000 0 35000 0.05 70000 20000 45000 0.04 70000 70000 70000 0.01
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