89 0187 256 p z 089 0187 256 5 61 x time

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Unformatted text preview: ed. a) E(D) = 1/8 1 V(D) = ( 1 ) 2 + ( 16 ) 2 = 8 5 256 5-37 σ T = 5 256 = 0.1398 b) P ( D > 1 ) = P( Z > 4 c) P ( D < 0) = P ( Z < 11 − 48 5 0− 1 8 5 ) = P( Z > 0.89) = 0.187 256 ) = P( Z < −0.89) = 0.187 256 5-61 X : time of ACL reconstruction surgery for high-volume hospitals. X ~ N(129,196). E(X1+X2+…+X10) = 10* 129 =1290 V(X1+X2+…+X10) = 100*196 =19600 5-62 a) Let X denote the average fill-volume of 100 cans. σ X = 0.5 2 100 = 0.05 . 12 − 12.1 ⎞ ⎛ ⎟ = P( Z < −2) = 0.023 0.05 ⎠ ⎝ 12 − μ ⎞ ⎛ c) P( X < 12) = 0.005 implies that P⎜ Z < ⎟ = 0.005. 0.05 ⎠ ⎝ 12 − μ Then 0.05 = -2.58 and μ = 12.129 . b) E( X ) = 12.1 and P ( X < 12) = P⎜ Z < ⎛ d) P( X < 12) = 0.005 implies that P⎜...
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This document was uploaded on 02/09/2014.

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