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Unformatted text preview: = 0.4167 ± 0 .2789 = (0.1378, 0.6956) 3. Given p = 0.75 (a) Since np = 75B 5, and n(1p) = 25B 5, by Central Limit Theorem, p is approximately Normally distributed. E( p )= p = 0.75 s.d. ( p )= 0.75 0.25 100 B = 0.0433 (b) ( 29 ( 29 0.65 0.75 0.85 0.75 0.65 0.85 0.0433 0.0433 2.31 2.31 2(0.4896) 0.9792 P p P Z P Z& ∏ = = = = (c) n = 12, p = 0.75, Binomial; X = # of applicants admitted, Y = # of applicants not admitted ( 10) (12 10) ( 2) 0.0317 .1267 .2323 0.3907 P X P Y P Y == = + + = (d) 2 2 0.05 2 2 ( ) (1 ) (1.645) (0.75)(0.25) 79.278 80 (0.08) Z p p n E= = =...
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 Spring '08
 Priya

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