For an insurance:
I. Loss amounts are uniformly distributed on the interval [0,12]
II. There is an ordinary deductible of 3 per loss.
Determine the 90th percentile of the insurance payment.
a) 7.80
b) 7.30
c) 0
d) 6.80
e) 8.00
The answer is a.
Here is my approach. I am getting an answer close to 7.8 but I don't know if what I am doing is actually correct or if I am just getting lucky. It seems a little bit too off for it to be a correct approach, if not please explain why I can't do it this way.
f(x)=1/12
.90= integral from 3 to x of (x-3)/12 dx
after solving for x I get an answer of 7.647.
I. Loss amounts are uniformly distributed on the interval [0,12]
II. There is an ordinary deductible of 3 per loss.
Determine the 90th percentile of the insurance payment.
a) 7.80
b) 7.30
c) 0
d) 6.80
e) 8.00
The answer is a.
Here is my approach. I am getting an answer close to 7.8 but I don't know if what I am doing is actually correct or if I am just getting lucky. It seems a little bit too off for it to be a correct approach, if not please explain why I can't do it this way.
f(x)=1/12
.90= integral from 3 to x of (x-3)/12 dx
after solving for x I get an answer of 7.647.
Question on percentiles and deductibles
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