Homeowrk 2 Solution

Homeowrk 2 Solution - .92)=.092 A false positive happens with probability.9.06)=.054 Thus P(test positive)=.092.054=.146(b P(have disease given

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HW2 Solution 3.5 (a) In the population, 51% of the fish have 21 vertebrae. Thus P(Y=21)=.51. (b) In the population, the percentage of fish with 22 or fewer vertebrae is 3+51+40 = 94%. Thus, P(Y 22)=.94. (c) In the population, the percentage of fish with more than 21 vertebrae is 40+6=46%. Thus, P(Y>21)=.46. (d) In the population, the percentage of fish with no more than 21 vertebrae is 3+51=54%. Thus, P(Y 21)=.54. 3.6 (a) (.55)(.55)=.3025 (b) (.55)(.55)+(.55)(.45)+(.45)(.55)=.7975 Or 1-(.45)(.45)=1-.2025=.7925 3.8 .4+(.6)(.2)=.52 3.11 (a) There are two ways to test positive. A true positive happens with probability (.1)
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Unformatted text preview: (.92)=.092. A false positive happens with probability (.9)(.06)=.054. Thus, P(test positive)=.092+.054=.146. (b) P(have disease given test positive) = .092/.146=.63. 3.13 (a) 1016/6549=.1551 (b) 2480/6549=.3787 (c) (1016+2480-526)/6549 = 2970/6549=.4535 (d) 526/6549=.0803 3.16 (a) P(20<Y<30)=.41+.21=.62 (b) .41+.21+.03=.65 (c) .01+.34=.35 3.24 (0)(.15)+(1)(0.50)+(2)(.35)=1.2 3.25 2 2 2 ( ) (0 1.2) (.15) (1 1.2) (.50) (2 1.2) (.35) .46. VAR Y =-+-+-= Thus, the standard deviation is .46 .678. =...
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This note was uploaded on 03/28/2011 for the course AMS 110 taught by Professor N/a during the Spring '08 term at SUNY Stony Brook.

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Homeowrk 2 Solution - .92)=.092 A false positive happens with probability.9.06)=.054 Thus P(test positive)=.092.054=.146(b P(have disease given

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