hw-6 - 1 of accidents per airplane 0 1 2 3 4 a ii Random...

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1. # of accidents per airplane # of airplanes having the accidents 0 17 1 10 2 13 3 9 4 1 a. ii. Random Variable (X) Probability (x)
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1 0.34 2 0.20 3 0.26 4 0.18 5 0.02 Total 1.00 iii. For random variable 0, there are 17 accidents per airplane and total number of airplanes equals to 50. Therefore, 17/50= 0.34 b. The random variable of the frequency distribution is the number of accidents per airplane. c. CAP’s exposure unit is a property loss/damage to their airplanes. d. The expected frequency for the NE region is 1.34 for the year of 2013. 1 (.34)+ 1 (.20) + 2 (.26) + 3 (.18) + 4 (.02) = 1.34 losses per airplane e. $5,000 loss per accident X 1.34 E (f) = $6,700 EL per airplane. f. The expected loss in part e is nonrandom because CAP wants to use $5,000 as a nonrandom severity which means that the $ value of each loss will be the same when losses occur. g. The random variable in the Severity Distribution is the $ amount of losses per airplane. h. $ of Losses # of losses P[$ Severity] Total $ of losses 1,000 38 0.57 38000 5,000 17 0.25 85000 10,000 6 0.09 60000 20,000 5 0.07 100000 50,000 1 0.01 50000 Total 67 1.00 333000 i. The expected severity loss per exposure is $ 4,620; E(s) = 1,000 (.57) + 5,000 (.25) + 10,000 (.09) +20,000 (.07) + 50,000 (.01) = 570+1,250+900+1,400+500=
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