Test_2_Review_KEY

Test_2_Review_KEY - NORMDIST(30,45,12,TRUE 7b 0.6915...

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Test 2 Extra Problem Answers 3b 0.039 BINOMDIST(65,100,0.59,FALSE) 3c 0.043 BINOMDIST(50,100,0.59,TRUE) 3d 0.462 1-BINOMDIST(59,100,0.59,TRUE) 4 x P(x) 0 0.312 BINOMDIST(A8,10,0.11,FALSE) 1 0.385 BINOMDIST(A9,10,0.11,FALSE) 2 0.214 BINOMDIST(A10,10,0.11,FALSE) 3 0.071 BINOMDIST(A11,10,0.11,FALSE) 4 0.015 BINOMDIST(A12,10,0.11,FALSE) 5 0.002 BINOMDIST(A13,10,0.11,FALSE) 6 0.000 BINOMDIST(A14,10,0.11,FALSE) 7 0.000 BINOMDIST(A15,10,0.11,FALSE) 8 0.000 BINOMDIST(A16,10,0.11,FALSE) 9 0.000 BINOMDIST(A17,10,0.11,FALSE) 10 0.000 BINOMDIST(A18,10,0.11,FALSE) 5 200.2 2600*0.077 13.6 SQRT(2600*0.077*0.923) 6b 0.9562 NORMSDIST(2.12)-NORMSDIST(-1.93) 6c -1.48 NORMSINV(0.07) 1.48 NORMSINV(1-0.07) 7a 0.1056
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Unformatted text preview: NORMDIST(30,45,12,TRUE) 7b 0.6915 1-NORMDIST(39,45,12,TRUE) 7c 0.3944 NORMDIST(60,45,12,TRUE)-NORMDIST(45,45,12,TRUE) 7d 0.0312 NORMDIST(40,45,12/SQRT(20),TRUE) 7e 57.4 NORMINV(0.85,45,12) 6 7a 7b 0.9562 7c 1.48 8a 8b 0.0055 8c 0.0064 8d 0.6778 9a 0.779 9b 0.118 9c 0.031 And .031 < .05 so the probability of 7 or more is unusual. See page 207 in text. μ= np σ = sqrt(npq) µ = 0 σ = 1 Yes because it meets all 4 criteria: 1. There are a fixed number of trials (10) 2. Each trial is independent. 3. Each trial has only 2 possible outcomes 4. p is constant for all 10 trials....
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This note was uploaded on 10/22/2011 for the course ACCT 3551 taught by Professor Brown during the Spring '11 term at UNC.

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