Exam1-Spr07_key - -= 0.4167 0 .2789 = (0.1378, 0.6956) 3....

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MGMT 305 Spring 2007 Exam 1 Solution Key Part I: D B C C A C B A D A D B D C D B A C B A Part II: 1. (a) z = 1.645, Minimum Value is x = 1800+1.645(250) = 2211.25 (b) P(X>2010) = 1800 2010 1800 250 250 X P - - = P(Z>0.84) = 0.5 – 0.2995 = 0.2005, i.e., 0.20 (c) P(X>3) = P(XB4) = 1 – P(XB3) = 1 – (0.0115+.0576+.1369+.2054) = 1 – 0.4114 = 0.5886 (d) X is Normal(1800, 250/sqrt(25)), that is, Normal(1800, 50) P(1750< X <1900) = 1750 1800 1900 1800 50 50 P Z - - < < = P(-1<Z<2) = 0.4772 + 0.3413 = 0.8185
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2. (a) 1998 166.5 12 x = = (b) 2 2 336128 12*(166.5) 3461 314.6364 12 1 11 s - = = = - s = 17.738 (c) / 2, 1 17.738 166.5 1.796 166.5 9.196 (157.304,175.696) 12 n s x t n α - = = = (d) p = 5/12 = 0.4167 n p = 5, n(1- p ) = 7 n p p z p ) 1 ( 2 / - ± =0.4167 ± 1.96 0.4167(1 0.4167) 12
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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(1-p) = 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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This note was uploaded on 06/29/2008 for the course MGMT 305 taught by Professor Priya during the Spring '08 term at Purdue University-West Lafayette.

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Exam1-Spr07_key - -= 0.4167 0 .2789 = (0.1378, 0.6956) 3....

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