# 1819SEM1-CE2407.pdf - NUS CONFIDENTIAL CE2407 NATIONAL...

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NUS CONFIDENTIALCE2407NATIONAL UNIVERSITY OF SINGAPORE______________________________________________________________________INSTRUCTIONS TO CANDIDATES1.Please write your student number only. Do not write your name.2.This assessment paper containsFOUR(4)questions and comprisesNINETEEN(19)printed pages.3.Answer ALL questions. All questions carry equal marks.4.Please start each question on a new page.5.This is an “CLOSED BOOK” assessment.6.Programmable or Graphical calculator is NOT allowed for this examination.(eg.Electronic calculator is permitted for this exam).CE2407 – ENGINEERING & UNCERTAINTY ANALYSES(Semester 1: AY2018/2019)Time Allowed: 2.5 Hours
NUS CONFIDENTIAL- 2 -CE2407Useful General Formulae for CE2407
NUS CONFIDENTIAL- 3 -CE2407߲ݑ߲ݔݑ௜ାଵ,௝െ ݑ௜,௝ሺ∆ݔሻ߲ݑ߲ݕݑ௜,௝ାଵെ ݑ௜,௝ሺ∆ݕሻForward-difference approximation߲ݑ߲ݔݑ௜ାଶ,௝െ 2ݑ௜ାଵ,௝൅ ݑ௜,௝ሺ∆ݔሻ߲ݑ߲ݕݑ௜,௝ାଶെ 2ݑ௜,௝ାଵ൅ ݑ௜,௝ሺ∆ݕሻ߲ݑ߲ݔݑ௜,௝െ ݑ௜ିଵ,௝ሺ∆ݔሻ߲ݑ߲ݕݑ௜,௝െ ݑ௜,௝ିଵሺ∆ݕሻBackward-difference approximation߲ݑ߲ݔݑ௜,௝െ 2ݑ௜ିଵ,௝൅ ݑ௜ିଶ,௝ሺ∆ݔሻ߲ݑ߲ݕݑ௜,௝െ 2ݑ௜,௝ିଵ൅ ݑ௜,௝ିଶሺ∆ݕሻ߲ݑ߲ݔݑ௜ାଵ,௝െ ݑ௜ିଵ,௝2ሺ∆ݔሻ߲ݑ߲ݕݑ௜,௝ାଵെ ݑ௜,௝ିଵ2ሺ∆ݕሻCentred-difference approximation߲ݑ߲ݔݑ௜ାଵ,௝െ 2ݑ௜,௝൅ ݑ௜ିଵ,௝ሺ∆ݔሻ߲ݑ߲ݕݑ௜,௝ାଵെ 2ݑ௜,௝൅ ݑ௜,௝ିଵሺ∆ݕሻ
NUS CONFIDENTIAL- 4 -CE2407Note: Before checking this Table, round-off the variablezinto only two decimals, e.g.,0.548~0.55, 0.234~0.23, 0.765~0.77).
NUS CONFIDENTIAL- 5 -CE2407
NUS CONFIDENTIAL- 6 -CE2407
NUS CONFIDENTIAL- 7 -CE2407Lecture 2 Probability (1)Mutually exclusive2 events)P(E)P(E)EP(E21213 events123123P(EEE )P(E )P(E )P(E )Not mutually exclusive121212P(EE )P(E )P(E )P(EE )123123122313123P(EEE )P(E )P(E )P(E )P(E E )P(E E )P(E E )P(E E E )Operational rulesCommutative AB = BA,AB = BAAssociative(AB)C = A(BC)(A B)C = A(BC)Distributive(AB) C = ACBC,(AB)C = (AC) (BC)De Morgan’s rule12n12nEEEEEEKK,12n12nEEEEEEKKLecture 3 Probability (2)P(E1|E2) = P(E1E2)/P(E2)1212P(EE)1P(EE),but1212P(E E )1P(E E ) P(E1E2E3) = P(E1|E2E3)P(E2|E3)P(E3)Statistically independentP(E1|E2) = P(E1),P(E1E2) = P(E1)P(E2),P(E1E2...En) = P(E1) P(E2)...P(En)Theorem of total probabilityP(A) = P(A|E1)P(E1) + P(A|E2)P(E2) + … + P(A|En)P(En)Bayes theoremP(E1|A) = P(A|E1)P(E1)/P(A)1122(|) ()(|)(|) ()(|)()...(|)()iiinnP A E P EP EAP A E P EP A EP EP A EP ELecture 4 Random variablesContinuous( )()for allXFxP XxxDiscrete()()()iiXiiXall xxFxP Xxpx( )( )XXdFxfxdx,( )()xXXFxfxdx()( )( )XXP aXbFbFa1[]()or( )NXiXiXiE Xxpxx fx dx
NUS CONFIDENTIAL- 8 -CE24071[ ()]()()( )( )NiXiXiE g Xg xpxorg x fx dx1[]()or( )NnnniXiXiE Xxpxx fx dx22221[() ]()()or()( )NXXiXXiXXiEXxpxxfx dx222[]XXE X,Coefficient of variationXXX,Skewness3333[() ]1()( )XXXXXEXxfx dxLecture 5 Distribution (continuous)Normal distribution211( )exp(,)22XxfxNx    21(0,1)( )exp22xNx( )( )()sSsFsx dx()1()1ppssp  Whenp< 0.5,1(1)psp Lognormal distribution-4-3-2-1012340.00.10.20.30.40.5N(0,1)area = 1-pprobability = pspsfS(s)211ln( )exp(,)022XxfxLNxx   Parameters:= E(lnX),2= var(lnX)2[]exp(0.5)XE X,2ln0.5X22222ln 1ln 1(for0.3)XXXXXln( )XaFa 

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