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Svet_ta [14]
3 years ago
15

The owner of an automobile insures it against damage by purchasing an insurance policy with a deductible of 250. In the event th

at 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.
Mathematics
1 answer:
choli [55]3 years ago
4 0

Answer:

Step-by-step explanation:

From the given information:

The uniform distribution can be represented by:

f_x(x) = \dfrac{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)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250) \ dx

= \dfrac{1}{1500 } \dfrac{(x - 250)^2}{2} \Big |^{1500}_{250}

\dfrac{5}{12} \times 1250

Similarly;

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

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

= \dfrac{1}{1500 } \dfrac{(x - 250)^3}{3} \Big |^{1500}_{250}

\dfrac{5}{18} \times 1250^2

∴

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

Finally, the standard deviation  of the insurance payment is:

= \sqrt{Var(I(x))}

= 1250 \sqrt{\dfrac{5}{48}}

≅ 404

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