Independent solution

How to solve this Uniform Distribution question

Setup

Setup

Let Z be the amount above the deductible before applying the selected percentage, and let X=pZ be the paid amount.

Z=(L−240)+,X=pZZ=(L-240)_+,\qquad X=pZ

Model

Model

Half of the uniform losses produce zero. On the positive range, integrate with the original uniform density 1/480.

E[Z]=∫240480(l−240)1480 dl=60E[Z]=\int_{240}^{480}(l-240)\frac{1}{480}\,dl=60
E[Z2]=∫240480(l−240)21480 dl=9600E[Z^2]=\int_{240}^{480}(l-240)^2\frac{1}{480}\,dl=9600

Compute

Compute

Compute the base variance, then use the square-law for a scaled random variable.

Var⁡(Z)=9600−602=6000\operatorname{Var}(Z)=9600-60^2=6000
2000=Var⁡(pZ)=p2(6000)2000=\operatorname{Var}(pZ)=p^2(6000)
p=20006000=0.5773502692…p=\sqrt{\frac{2000}{6000}}=0.5773502692\ldots

Answer

Answer

The selected percentage is approximately 57.7%.

57.7%(C)\boxed{57.7\%\quad\text{(C)}}