# exam1_solution - STAT200 Exam1 Solution Min 55 Q1:75...

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STAT200 Exam1 Solution Min: 55 Q1:75 Median:85 Q3:93 Max:100 IQR = Q3-Q1=18 Q1 - 1.5 × IQR = 48 Q3 + 1.5 × IQR = 120 No outliers Smallest z-score : ggG±².³² ´².µ³ = −2.198 Largest z-score : ´µµG±².³² ´².µ³ = 1.245

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STAT200 Exam1 Solution Admitted : 900/2200 Not admitted : 1300/2200 Men : 600/1300 Women : 300/900 Department Men Women A 440/740=0.595 50/70=0.714 B 150/350=0.429 200/450=0.444 C 10/210=0.048 50/380=0.132 b This is a typical example of Simpson’s paradox. By ignoring a lurking variable, original conclusion in (b) has been misleaded.
STAT200 Exam1 Solution X 50 100 Probability 0.5 0.5 μ g = 50 × G ± + 100 × G ± = 75 σ g = ² G ± (50 − 75) ± + G ± (100 − 75) ± = 25 μ ³ = 2μ g + 70 = 220 σ ³ = 2σ g = 50

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STAT200 Exam1 Solution μ g G = 75 σ g G = 25/√50 = 3.536 X G ~N(75,3.536) P(X G < 70) = P(Z < −1.414) = 0.0787
D1 : The first card is a diamond, D2: The second card is a diamond P(D1) = 13/52 , P(D1 g ) = 39/52

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## This note was uploaded on 09/11/2011 for the course STAT 200 taught by Professor Agniel during the Spring '09 term at University of Illinois at Urbana–Champaign.

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exam1_solution - STAT200 Exam1 Solution Min 55 Q1:75...

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