Answer Key to Sample Midterm Exam
Section I:
11: N
12: I
13: R
14: R
15: N
16: O
17: R
18: O
19: R
110: O
21: Y
22: Y
23: N
24: Y
25: N
26: N
27: N
28: Y
29: N
210: Y
211: Y
212: Y
31: Y
32: Y
33: Y
34: Y
35: Y
36: Y
37: Y
38: Y
39: Y
310: N
311: Y
312: Y
41: P(GS) = 72/150 = 0.48
42: P(GSFT) = P(GS and FT)/P(FT)
= (48/150)/(64/150)
= 48/64
= 0.75
43: Not independent since P(GS) is not equal to P(GSFT)
44:
The two probabilities are not the same.
P(GSFT): This is the probability that given an appliance repair person is factorytrained,
s/he will provide good service, or the probability that a factorytrained repair person will
provide good service.
P(FTGS): This is the probability that given an appliance repair person has provided good
service, s/he is factorytrained, or the probability that a goodservice providing repair
person is factorytrained.
51: (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 ‘relativefrequence’ interpretation.
52: (Bayesian Statistics) The “raintomorrow” event needs not to be repeatable, and instead, it
can be an onetime 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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 Fall '04
 MYUNG
 Psychology, Normal Distribution, Standard Deviation, Mean, Buzzy

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