HW3 Solution
3.27
(a)
4
(.42)
.1132
=
(b)
3
1
4
1
(.42) (.58)
.3278
C
=
(c)
2
2
4
2
(.42) (.58)
.3560
C
=
(d) .1132+.3278+.3560=.7970
(e) .3278+.3560=.6838
3.28
(a)
20
(.9)
.1216
=
(b)
19
1
20 (.9) (.1)
.2702
⋅
=
(c)
18
2
20
2
(.9) (.1)
.2852
C
=
(d)
18
2
190 (.9) (.1)
.2852
⋅
=
3.30
(a)
(10)(.6)
6
np
=
=
(b)
(1
)
(10)(.6)(.4)
1.55
np
p

=
=
3.42
(a)
3
2
(10)(.5) (.5)
.3125
=
(b)
3
2
4
1
5
(10)(.5) (.5)
(5)(.5) (.5)
(1)(.5)
.5
+
+
=
(c)
5
5
(.5)
(.5)
.0625
+
=
3.45
The probability that an innocent subject will have a higher electrodermal response on any
critical word (compared to the three controls for that critical word) is 25%. Thus, P(failure on a
critical word)=.25.
Let us identify “success” as “success on a critical word.” Thus, p=.75 and 1p=.25. We then use
the binomial formula with n=6 and p=.75.
To be labeled “guilty” a subject must fail 4 or more critical words out of 6. We find the
probability of 4,5, or 6 failures by setting j=2,1, or 0, as follows:
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 Spring '08
 N/A
 25%, µ, 144.8mg

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