2008_test1sol - ACTL1001, S2, 2008. Test 1 Draft Solutions...

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ACTL1001, S2, 2008. Test 1 Draft Solutions 1. (a) Probability is s (45) s (15) = r 35 65 (b) 1 s (40) = 1 r 1 2 (c) s (45) s (65) s (20) = p 35 p 15 p 60 (d) 1 s (41) s (40) = 1 r 39 40 2. . (a) E [ X ] = 1 4 (1 + 2 + 3 + 4) = 2 : 5 (b) V ar [ X ] = 1 4 1 : 5 2 + 0 : 5 2 + 0 : 5 2 + 1 : 5 2 ± SD [ X ] = p V ar [ X ] = 1 : 118 (c) E ² X 3 ³ = 1 4 1 + 2 3 + 3 3 + 4 3 ± = 25 3. . (a) l 72 = l 70 d 70 d 71 = 2646 (b) 1 2 p 74 = 1 l 76 l 74 = 1 1376 1967 (c) 4 j 1 q 70 = ´ l 74 l 70 µ´ 1 l 75 l 74 µ = 308 3385 1
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(d) Now T 77 = 9838 L 76 = 1 2 (1376 + 1122) L 75 = 1 2 (1659 + 1376) hence e 75 = T 75 l 75 = T 77 + L 76 + L 75 l 75 = 12604 : 5 1655 = 7 : 5976 4. Smoker rates are now very cheap compared to real cost (and cost from com- petitors) - hence many smokers will buy. Reverse applies to non-smokers - not many non smokers will buy. This will lead to a situation where we are charging an averaged rate to a majority smoker group (i.e. rate too low for the risk). This will very likely lead to (potentially substantial) losses.
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This note was uploaded on 08/23/2010 for the course ACTL 1001 taught by Professor Bernartwong during the Three '10 term at University of New South Wales.

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2008_test1sol - ACTL1001, S2, 2008. Test 1 Draft Solutions...

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