2013-Quiz 1 Solutions - Name Statistical Analysis Quiz 1(15 points Instructions This is an open book quiz You have 15 minutes to complete the quiz To

2013-Quiz 1 Solutions - Name Statistical Analysis Quiz 1(15...

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Name: _______________________________ Statistical Analysis Quiz 1 (15 points) Instructions This is an open book quiz. You have 15 minutes to complete the quiz. To get full credit for each question, please provide your reasoning and justification, NOT only the result. SECTION F4 1. 60% of all students at a given university are female. 30% of all students eat in the university cafeteria. 40% of the male students eat in the cafeteria. a. What percent of students are female and don’t eat in the cafeteria ? (2 points) Eat at cafeteria Do not eat at cafeteria Total Male 16 24 40 Female 14 46 60 Total 30 70 100 P ( Female∩Doesnot eat cafeteria ) = 46 0.46 b. What is the probability that one randomly selected student will be either a male or eats in the cafeteria ? (2 points). P ( Male Eats cafeteria ) = P ( Male ) + P ( Eats Cafeteria ) P ( Male∩ Eats Cafeteria ) = 40 + 30 c. Are the events “female ” and “eating in the cafeteria” independent? Please justify your answer (2 points). If independent, P ( Eats Cafeteria | Female ¿ = P ( Eats Cafeteria ) Here, P ( Eats Cafeteria | Female ¿ = P ( Female∩ Eats cafeteria ) P ( Female ) = 0.14 0.6 = 0.23 and P ( Eats Cafeteria ) = 0.3
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Since P ( Eats Cafeteria | Female ¿ ≠P ( Eats Cafeteria ) , the events are not independent. 2. Four small companies that are managed by one large company were being evaluated for their performance. As a part of quality improvement initiative, employee satisfaction was measured. a. If your friend works for any of these companies, what is the probability that he is satisfied working there? (3 points) Define 4 mutually exclusive and exhaustive events: P ( A ) = 80 200 = 0.4 ; P ( B ) = 50 200 = 0.25 ;P ( C ) = 30 200 = 0.15 ; P ( D ) = 40 200 = 0.2 Other Information: P ( Dissatified | A ) = 0.05 ; P ( Dissatisfied | B ) = 0.1 ;P ( Dissatisfied | C ) = 0.06 ; P ( Dissatisfied | D ) = 0.04 ( 0.4 ) ( 0.05 ) + ( 0.25 ) ( 0.1 ) + ( 0.15 ) ( 0.06 ) + ( 0.2 ) ( 0.04 ) = 1 [ 0.02 + 0.025 + 0.009 + 0.008 ] = 1 0.062 = 0.938 P ( Dissatified | A ) P ( A ) + P (
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