Notes for tests 1

Notes for tests 1 - Lifetime 1 2 3 4 5 6 7 8 9 10...

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Unformatted text preview: Lifetime 1 2 3 4 5 6 7 8 9 10 probability .02 .04 .08 .10 .12 .18 .24 .15 .05 .02 Find the expected lifetime of the organism (lifetime) = sum of possibilities x corresponding probabilities =1(.02) + 2(.04) + 3(.08) + . + 10(.02) = 5.95 Now suppose the organism has lived to the age of 5.7 Find his/her expected residual life (expected additional life beyond 5.7) Soln: we need to find the conditional probabilities that the organism lives to age 6,7,8,9,10 given that it has lived to age 5.7 2 8 . 6 4 . 1 8 . 0 2 . 0 5 . 1 5 . 2 4 . 1 8 . 1 8 . ) 7 . 5 ( ) 6 ( ) 7 . 5 ( ) 7 . 5 6 ( ) 7 . 5 | 6 ( = = + + + + = = = = = = L P L P L P L a n d L P L L P Similarly, divide .24, .15, .05 & .02 by P(L > 5.7) to get the other conditional probabilities Similarly, divide ....
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This note was uploaded on 09/08/2008 for the course STAT 101 taught by Professor Russo during the Spring '08 term at University of Iowa.

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Notes for tests 1 - Lifetime 1 2 3 4 5 6 7 8 9 10...

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