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CEE 110 PS3 11 Solutions

# CEE 110 PS3 11 Solutions - UNIVERSITY OF CALIFORNIA LOS...

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UNIVERSITY OF CALIFORNIA, LOS ANGELES Civil and Environmental Engineering Department CEE 110 Introduction to Probability and Statistics for Engineers Spring Quarter 2011 MW 12-2 PM Franz 1178 Prof. K. D. Stolzenbach 5732J Boelter Hall, 206-7624 Problem Set 3 Solutions ___________________________________________________________________________ 1. a) i) P(X 1)=P(X=1)= 4/7 ii) P(X>1)= 1-P(X=1)=1-4/7=3/7 iii) P(2<X<6)=P(X=3)= 1/7 iv) P(X 1 or X>1)= P(X=1)+ P(X=2)+P(X=3)=1 b) i) P( X < 1.5) = 4/7 ii) P( X 3) = 1 iii) P( X > 2) = 1 – P( X 2) = 1 – 6/7 = 1/7 iv) P(1 < X 2) = P( X 2) – P( X 1) = 6/7 – 4/7 = 2/7 c) μ = E ( X ) = 1 f (1) + 2 f (2) + 3 f (3) = 1(4/7) + 2(2/7) + 3(1/7) = 11/ 7 V ( X ) = 1 2 f (1) + 2 2 f (2) + 3 2 f (3) + μ 2 V ( C ) = 1(4 / 7) + 4(2 / 7) + 9(1/ 7) (11/ 7) 2 V ( C ) = 3- (121/ 49) = 0.5306 2. a) i) P ( X < 2) = 3 x 4 1 2 dx = ( 1 x 3 ) 1 2 = ( 1 8 ) ( 1) = 7 / 8 = 0.875 ii) P ( X > 5) = 2 x 4 5 dx = ( 1 x 3 ) 5 = 0 ( 1 125 ) = 0.008 iii) P (4 < X < 8) = 2 x 4 4 8 dx = ( 1 x 3 ) 4 8 = ( 1 512 ) ( 1 64 ) = 0.014 iv) . From part (iii), = 0.014. Therefore, P ( X < 4 orX > 8) = 1 – 0.014 = 0.986

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b) Now, f ( x ) = 3 x 4 for x > 1 and F X ( x ) = 2 u 4 1 x du = ( 1

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CEE 110 PS3 11 Solutions - UNIVERSITY OF CALIFORNIA LOS...

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