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Homeowrk 3 Solution

# Homeowrk 3 Solution - H W3 Solution 3.27(a.42 4(b 4 =.1132...

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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 1-p=.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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Homeowrk 3 Solution - H W3 Solution 3.27(a.42 4(b 4 =.1132...

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