CEE_110_PS2_08_Solutions

CEE_110_PS2_08_Solutions - | sand)P(sand) + P(2 |...

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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 2008 M-W 2-4 PM 3400 Boelter Hall Prof. K. D. Stolzenbach 5732J Boelter Hall, 206-7624 [email protected] Problem Set 2 Solutions ___________________________________________________________________________ 1. 2.
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3. 4. 5.
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6. (a) P(1) = P(1 | sand)P(sand) +P(1 | clay)P(clay) + P(1 | rock)P(rock)= (14/100)(100/250) + (6/100)(100/250) + (4/50)(50/250) = 24/250 = 0.096 (b) P(sand | 2) = P(2 | sand)P(sand)/P(2) = P(2 | sand)P(sand)/[P(2 | sand)P(sand) + P(2 | clay)P(clay)+ P(2 | rock)P(rock)] = (10/100)(100/250)/[(10/100)(100/250) + (2/100)(100/250) + (2/50)(50/250)] = 10/14 = 0.714 (c) P(rock AND 0) = P(0 | rock)P(rock) = (44/50)(50/250) = 44/250 = 0.176 (d) P(clay OR 2) = P(clay) + P(2) – P(clay AND 2) = P(clay) + [P(2
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Unformatted text preview: | sand)P(sand) + P(2 | clay)P(clay)+ P(2 | rock)P(rock)] – P(2 | clay)P(clay) = (100/250) + [(10/100)(100/250) + (2/100)(100/250) + (2/50)(50/250)] – (2/100)(100/250) = 112/250 = 0.448 (e) Dependent , because, for example, P(rock AND 0) = 0.176 [see (c)] ≠ P(rock)P(0) = (50/250)(212/250) = 0.170 7. (a) P(steep AND 3) = 2/80 = 1/40 (b) P(mild OR 2) = P(mild) + P(2) – P(mild and 2) = 30/80 + 10/80 – 4/80 = 36/80 = 9/20 (c) P(mild | 1) = P(mild and 1)/P(1) = (10/80)/(21/80) = 10/21 (d) If independent, P (flat and 2) = P(flat) x P(2) = (40/80) x (10/80) = 400/6400 = 1/16 Therefore, the number of hills with flat and 2 is 80 x 1/16 = 5 8. 9. Variability in horsepower is clearly greater in the US and least in Sweden and Italy....
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This note was uploaded on 05/19/2011 for the course CEE 110 taught by Professor Stolzenbach during the Spring '08 term at UCLA.

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CEE_110_PS2_08_Solutions - | sand)P(sand) + P(2 |...

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