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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+2480526)/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.
 Spring '08
 N/A

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