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Unformatted text preview: III. Sumof(Days*prob)=3 Sumof(prob)=1.05 Expected value=3/1.05 =2 . 85714 IV . No. She needs a 166.449 (~166) to be in the 95 th percentile, her score puts her in the 93.1843% percentile. If her friend had a percentile score of 83.1843, her score comes to approximately 160. V . a) Sample mean=76 1.96(S x /sqrt(529))=2 S x = 23.4694 SE x =23.4694/sqrt(529) = 1.02041 b) Z 0.2/2 =1.28 76 +/ 1.28(23.4694/sqrt(529)) = [74.6939, 77.3061]...
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 Fall '08
 Lawrence
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