Answer:
Step-by-step explanation:
From the given information:
The uniform distribution can be represented by:
[tex]f_x(x) = \dfrac{1}{1500} ; o \le x \le \ 1500[/tex]
The function of the insurance is:
[tex]I(x) = \left \{ {{0, \ \ \ x \le 250} \atop {x -20 , \ \ \ \ \ 250 \le x \le 1500}} \right.[/tex]
Hence, the variance of the insurance can also be an account forum.
[tex]Var [I_{(x}) = E [I^2(x)] - [E(I(x)]^2[/tex]
here;
[tex]E[I(x)] = \int f_x(x) I (x) \ sx[/tex]
[tex]E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250) \ dx[/tex]
[tex]= \dfrac{1}{1500 } \dfrac{(x - 250)^2}{2} \Big |^{1500}_{250}[/tex]
[tex]\dfrac{5}{12} \times 1250[/tex]
Similarly;
[tex]E[I^2(x)] = \int f_x(x) I^2 (x) \ sx[/tex]
[tex]E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250)^2 \ dx[/tex]
[tex]= \dfrac{1}{1500 } \dfrac{(x - 250)^3}{3} \Big |^{1500}_{250}[/tex]
[tex]\dfrac{5}{18} \times 1250^2[/tex]
∴
[tex]Var {I(x)} = 1250^2 \Big [ \dfrac{5}{18} - \dfrac{25}{144}][/tex]
Finally, the standard deviation of the insurance payment is:
[tex]= \sqrt{Var(I(x))}[/tex]
[tex]= 1250 \sqrt{\dfrac{5}{48}}[/tex]
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