excans:B5fe42cdd5: Difference between revisions
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(Created page with "'''Key: B''' For Product X: aggregate losses have mean 10*20 = 200 and variance 10*25 + 400*9 = 3850. For Product Y: aggregate losses have mean 2*50 = 100 and variance 2*100...") |
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P( S > 400) = P \left( \frac{S-300}{\sqrt{6550}} > \frac{400-300}{\sqrt{6550}} \right)= P(Z > 1.24 ) = 0.11 | \operatorname{P}( S > 400) = \operatorname{P} \left( \frac{S-300}{\sqrt{6550}} > \frac{400-300}{\sqrt{6550}} \right)= \operatorname{P}(Z > 1.24 ) = 0.11 | ||
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{{soacopyright | 2023}} | {{soacopyright | 2023}} | ||
Latest revision as of 13:50, 14 May 2023
Key: B
For Product X: aggregate losses have mean 10*20 = 200 and variance 10*25 + 400*9 = 3850.
For Product Y: aggregate losses have mean 2*50 = 100 and variance 2*100 + 2500 = 2700.
Because Product X and Product Y are independent, total aggregate losses, S, have mean 300 and variance 6550.
Using the normal approximation, we have:
[[math]]
\operatorname{P}( S \gt 400) = \operatorname{P} \left( \frac{S-300}{\sqrt{6550}} \gt \frac{400-300}{\sqrt{6550}} \right)= \operatorname{P}(Z \gt 1.24 ) = 0.11
[[/math]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.