Sample_Exam1_Key

# Sample_Exam1_Key - √ 10=261.4±80.455 =(180.95 to...

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MGMT 305 Spring 2007 Sample Exam 1 Solution Key Part I: C A C C B B D D C A C D D C C D A B A C A C B B Part II: 1. (a) z=1.28, minimum value is x=1050+1.28(200) = 1306 (b) P(X>1350) = P(Z>1.5) = 0.5 – 0.4332 = 0.0668 (c) , 0668 . 0 ) ( = p E 025 . 0 ) ( = p SD (d) 9082 . 0 4082 . 0 5 . 0 ) 328 . 1 ( ) 10 . 0 ( ) 10 discount on sold number total ( = + = = Z P p P P (e) X is Normal(1050, 200/sqrt(25)), that is, Normal(1050,40) (f) P(1000< X <1120) = P(-1.25<Z<1.75) = 0.3944 + 0.4599 = 0.8543

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2. (a) Sample Mean = 261.4 2 2 2 2 856,700 10(261.4) 19266.71 1 9 138.8 i x nx s n s - - = = = - = (b) t 0.05,9 =1.833 90% Confidence interval for µ: 261.4±(1.833)(138.8)/
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Unformatted text preview: √ 10=261.4±80.455 = (180.95 to 341.85) (c) p = 5/10 = 0.5 (d) (.5)(.5) ( ) 0.158 10 SD p = = 95% Confidence interval for p: 0.5 ± (1.96)(0.158)= (0.19 to 0.81) 3. (a) Point estimate: x = 568/10=56.8 (b) 6325 . 10 2 = = = n x σ (c) P(X<52)=P(Z< 2 8 . 56 52-)=p(Z<-2.4)=0.5-0.4918=0.0082 (d) P(Sum 36 < 2052) =P( X <57)=P(Z< 36 / 2 8 . 56 57-)=P(Z<0.6)=0.5+0.2257=0.7257...
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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.

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Sample_Exam1_Key - √ 10=261.4±80.455 =(180.95 to...

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