Question

The owner of an automobile insures it against damage by purchasing an insurance policy with a deductible of 250. In the event that the automobile is damaged, repair costs can be modeled by a uniform random variable on the interval (0, 1500). Determine the standard deviation of the insurance payment in the event that the automobile is damaged.

Answer 1

Explanation:

From the given information:

The uniform distribution can be represented by:

`f_x(x) = (1)/(1500) ; o \le x \le \ 1500`

The function of the insurance is:

`I(x) = \left \{ {{0, \ \ \ x \le 250} \atop {x -20 , \ \ \ \ \ 250 \le x \le 1500}} \right.`

Hence, the variance of the insurance can also be an account forum.

`Var [I_((x)) = E [I^2(x)] - [E(I(x)]^2`

here;

`E[I(x)] = \int f_x(x) I (x) \ sx`

`E[I(x)] = (1)/(1500) \int ^(1500)_{250{ (x- 250) \ dx`

`= (1)/(1500 ) ((x - 250)^2)/(2) \Big |^(1500)_(250)`

`(5)/(12) * 1250`

Similarly;

`E[I^2(x)] = \int f_x(x) I^2 (x) \ sx`

`E[I(x)] = (1)/(1500) \int ^(1500)_{250{ (x- 250)^2 \ dx`

`= (1)/(1500 ) ((x - 250)^3)/(3) \Big |^(1500)_(250)`

`(5)/(18) * 1250^2`

∴

`Var {I(x)} = 1250^2 \Big [ (5)/(18) - (25)/(144)]`

Finally, the standard deviation of the insurance payment is:

`= √(Var(I(x)))`

`= 1250 \sqrt{(5)/(48)}`

≅ 404