question archive A midsize corporation is considering purchase of Cyber Liability Insurance Policy
A midsize corporation is considering purchase of Cyber Liability Insurance Policy
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A midsize corporation is considering purchase of Cyber Liability Insurance Policy. The following table provides the probability distribution of yearly cost coverage associated with different level of Cyber Security IT network interruptions for this corporation.
Type of Risks Probability Cost Coverage
None 0.90 $0
Minor: 0.050 $70,000
Moderate: 0.030 $500,000
Major 0.019 $2,000,000
Catastrophic: 0.001 $10,000,000
a- What is the expected cost of this policy for the insurance company? What is the Standard deviation of cost? Hint: Use Excel.
b- If the cost of purchasing a network interruption insurance is $100,000, calculate the expected gain or loss for a corporation that purchase this insurance. Would you recommend purchasing this insurance? Why?
c- Would you a buy this insurance policy at cost of $50,000? Why?
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Answer:
a.
expected cost $66,500.00
standard deviation is $423,465.17
b.
Buying the insurance policy at $100,000 whereas its expected cost is $66,500 means the purchase results in expected loss of $33,500, I would not be in a position to recommend the insurance policy because its fair price should have been $66,500, such a decision would mean the cost of the insurance policy would outweigh its benefits
c.
Buying the insurance policy at $50,000 results in cash savings of $16,500 and since the priority of the company would be to save costs and improve profitability, spending less to have access to the same quality of insurance coverage is a step in the right direction
Step-by-step explanation
The expected cost of the policy is the weighted average of cost which is the sum of the costs multiplied by probabilities of cost occurring in each case.
The expected cost considers the likely costs that potential insurance policyholder would pay for each type of risk alternatives and their associated likelihood of occurrence as shown summarized in the formula below:
expected cost=(prob of None Risk *cost of None Risk)+(prob of Minor Risk*cost of Minor Risk)+(prob of Moderate Risk*cost of Moderate Risk)+(prob of Major Risk*cost of Major Risk)+(Prob of Catastrophic Risk*cost of Catastrophic Risk)
expected cost=(0.90*$0)+(0.050*$70,000)+(0.030*$500,000)+( 0.019*$2,000,000)+(0.001*$10,000,000)
expected cost=$66,500
| x | P(x) | x *P(x) | (x - μ)2 *P(x) |
| $0 | 0.9000 | - | 3,980,025,000.00 |
| $70,000 | 0.0500 | 3,500.00 | 612,500.00 |
| $500,000 | 0.0300 | 15,000.00 | 5,637,667,500.00 |
| $2,000,000 | 0.0190 | 38,000.00 | 71,030,022,750.00 |
| $10,000,000 | 0.0010 | 10,000.00 | 98,674,422,250.00 |
| Expected value( μ) | 66,500.00 | 179,322,750,000.00 | |
| variance | 179,322,750,000.00 | ||
| standard deviation=(variance)^(1/2) | 423,465.17 |
x refers to the costs
P(x)=probability of costs
x *P(x)=cost*prob. of costs
μ=Expected value
b.
expected gain/(loss)=expected cost-cost of purchasing a network interruption insurance
cost of purchasing a network interruption insurance=$100,000
expected gain/(loss)=$66,500-$100,000=-$33,500
c.
expected gain/(loss)=expected cost-cost of purchasing a network interruption insurance
cost of purchasing a network interruption insurance=$50,000
expected gain/(loss)=$66,500-$50,000-$16,500
