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posledela
3 years ago
6

Suppose there is an ordinary deductible of $40 and the maximum payment by the insurer is $130. Determine the standard deviation

of annual losses incurred by the automobile owner before the insurance reimbursement.
Business
1 answer:
Sedaia [141]3 years ago
5 0

An automobile owner has 40% chance of having exactly one accident in a year and 60% chance of having no accidents in a year. There is no chance that the automobile owner incurs more than one accident in a year.

If there is an accident, the loss amount has the following distribution:

Loss amount & Prob

30 0.3

60 0.15

100 0.45

180 0.1

Suppose there is an ordinary deductible of $40 and the maximum payment by the insurer is $130.

Determine the standard deviation of the payment made by the insurer to the automobile owner.

Answer:

The answer is 32.437

Explanation:

Let X be the loss and let Y be the payment by the insurer.

Hence, we have Y = o if x<= 40,

x-40 if 40 < x < 130

130 if x > = 170

Therefore, Var (y) = E (y^2) - [E(y)]^2

E(y) = 0.4[130 (0.1)+60 (0.45) +20 (0.15)] = 17.2

E(^2) = 0.4 [130^2(0.1) + 60^2(0.45) +20^2(0.15)] = 1348

Finally, Var(y) = 1348 -17.2^2 = 1052.16

Sqrt(1052.16) = 32.437

Therefore, the standard deviation of the payment made by the insurer to the automobile owner is 32.437

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Answer:

See the explanation below.

Explanation:

a. The number of widgets that will be sold

Let y represent the number of widgets that will be sold, and and we already have x as price of widget, we therefore have:

y - 1,000 = (-600/10) * (x - 389)

y - 1,000 = -60 * (x - 389)

y = 1,000 - [60 * (x - 389) ]

y = 1,000 - 60x + 23,340

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b. The revenue generated by the sale of widgets

Let R represent Revenue, therefore we have:

R = xy

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C = 8,000 + 2,367,065  - 5,835x

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P = 24,340x - 60x²  - (2,375,065  - 5,835x)

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I found this online :

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