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key-to-mid

# key-to-mid - Answer Key to Sample Midterm Exam Section I...

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Answer Key to Sample Midterm Exam Section I: 1-1: N 1-2: I 1-3: R 1-4: R 1-5: N 1-6: O 1-7: R 1-8: O 1-9: R 1-10: O 2-1: Y 2-2: Y 2-3: N 2-4: Y 2-5: N 2-6: N 2-7: N 2-8: Y 2-9: N 2-10: Y 2-11: Y 2-12: Y 3-1: Y 3-2: Y 3-3: Y 3-4: Y 3-5: Y 3-6: Y 3-7: Y 3-8: Y 3-9: Y 3-10: N 3-11: Y 3-12: Y 4-1: P(GS) = 72/150 = 0.48 4-2: P(GS|FT) = P(GS and FT)/P(FT) = (48/150)/(64/150) = 48/64 = 0.75 4-3: Not independent since P(GS) is not equal to P(GS|FT) 4-4: The two probabilities are not the same. P(GS|FT): This is the probability that given an appliance repair person is factory-trained, s/he will provide good service, or the probability that a factory-trained repair person will provide good service. P(FT|GS): This is the probability that given an appliance repair person has provided good service, s/he is factory-trained, or the probability that a good-service providing repair person is factory-trained. 5-1: (Standard Statistics) First, we must assume that “rain tomorrow” is a repeatable event. The probability statement means that out of all cases in which the weather person makes this statement, it will actually rain the next day 70% of the time, but no rain 30% of the time. This is what is called the ‘relative-frequence’ interpretation. 5-2: (Bayesian Statistics) The “rain-tomorrow” event needs not to be repeatable, and instead, it can be an one-time only event. The statement means that the weather person’s own subjective belief (degree of confidence) that it will rain tomorrow is 0.70 on a scale of 0 to 1. It is worth

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key-to-mid - Answer Key to Sample Midterm Exam Section I...

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