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Unformatted text preview: CEE 304  Uncertainty Analysis in Engineering
First Examination
October 8, 2003 You may use text, your notes and calculators. There are 50 points in total, one per minute. 1. (8 pts) Consider three events denoted M and N, plus E which is independent of both M and N;
Pr[M] =0.6; Pr[N] =O.4; P[MnE] =02; P[MnN] =0.2 (a) What is the probability that E occurs? (b) What is the probability that neither M or N occurs? (c) If N occurs, what is the probability that M also occurs? Show work. (d) How many different committees of 2 can one form if 6 people are eligible to serve? 2. (20 points) Sam and Jacob are traveling by car with their parents; to pass the time they compete to spot Corvettes (a car type). They spot on average one Corvette every 8 minutes. a) What is the probability that 4 minutes passes without Sam or Jacob spotting a Corvette? b) What is the mean and variance of the number of Covettes they spot in 2 hours? c) Tony, their father, is getting tired of the game and says they will stop for lunch after the boys spot 5 more
Corvettes: what is the mean and variance for the time until lunch. d) Sam has better eyes than Jacob and is the first to spot a Corvette 75% of the time; so, what are the mean
and standard deviation of the number of cars they have seen when Jacob next sees a Corvette before Sam? e) What is the probability that in the next 5 cars, Sam is the ﬁrst to spot more Corvettes than Jacob? 3. (5 points) The annual load of nutrients from two streams, denoted SI and 82, are essentially normally
distributed with means 10 and 25; standard deviations 2 and 5, and covariance 8; however nutrients are not
equally destructive so that Carolyn, an ecologist, proposes the pollutant load index PLI = 1.0 $1 + 0.7 82.
Assuming PLI is normally distributed, what are the median and 99 percentiles of the PLI distribution? 4. (9 pts) A nursery grows trees for use in gardens. Unfortunately last season many of their trees died within
a year. Two thirds of the trees they planted were softwoods and one third hardwoods. Assume that 90% of
the softwoods and 60% of the hardwoods die. (a) What fraction of the trees died? (b) If the gardener sees one
of the trees that survived, what is the probability it is a softwood? (c) If a homeowner planted a dozen
softwoods, what is the mean and variance of the number that survive? 5. (8 pts) A random variable R has a triangular probability density function fR(r) = 1— 0.5 r for 0 s r s 2 and
zero otherwise. However, the interest in not really in R, but the weight of particles given by W = 7 R3.
(a) What is the mean and variance of R. Show work.
(b) What is the probability density function for W? Please answer all questions correctly to make the grading easier. CEE 304  Uncertainty Analysis in Engineering
Solutions First Examination
October , 2003 1. Axioms, independence, counting
1a) P[MnE] = P[M] P[E] = 0.6 P[E] = 0.2 => P[E] = 1 / 3 (use independence: easy!) lb) P[MUN] = P[M]+P[N] — P[MnN] = 0.6 + 0.4 — 0.2 = 0.8; so P[ (MUN)’ ] = 1—0.8 = 0.2 2 pts . 1c) P[M I N] = P[MnN]/ P[N] = 0.2/ 0.4 = 50% { Note: (MuN)’ = M’nN’ } 2 pts
1d) Combinations: 6 choose 2 is 6*5/1*2 = 15 2 pts 2. Poisson process—Exponential, Poisson, Gamma, Geometric, Binomial distributions
Setup: 7» = 1 / (8 minutes) = 0.125 per minute; Okay, lets go! a) Exponential: Pr[T1 > 4] = 1 — Fn(4 min) = exp(?\.*4 min) = 0.607 4 pts
Or solve using Poisson dist: Pr[K=0  v = 0.5] = exp(7t*4 min) = 0.607 «Same number!
b) Poisson dist; moments are: u = 62 = v = 7L*t = 120/ 8 = 15  4 pts c) Time to 5th Gamma: u = 5 / 7» = 40 mins; 0'2 = 5/ A2 = 320 minsz; a: 17.9 mins (not required) 4 pts d) Geometric distribution for time to ﬁrst success in Bernoulli trials. p = 0.25 a = 1/ p = 4; 0'2 = (lp)/ p2 = 12 ; O'= 3.46 (1 pt for standard deviation)
e) Binomial distribution n = 5, p = 0.25 (Jacob sees car first); from Devore tables,
probability of two or less (andthus Sam sees more) = 0.896. [Many missed this.]
3. Normal calculations, Moments of Sums
Given PLI = $1 + 0.7 $2, the mean is E{PLI} = E{Sl} + 0.7 E{32} = (10) + 0.7 (25) = 27.5
Var{ PLI } = Var{Sl] + 2 (0.7) Cov[Sl, $2] + 0.72 Var{SZ] = 22+ 2 (0.7) 8 + 0.72 52 = 27.45 = (5.23)2.
Normalmedian = mean = 27.5 ; PLI(,_99 = 99percentile = (a + 2.326 c) = 39.7
Bayes Theorem / —Died (0.90)
4. _ / ——— soft (0.667) ~—Live (0.10) DRAW THE PICTURE!
\
—hard (0.333) —°——Died (0.60)
\ ~— Live (0.40)
a) P(Died) = P(Died  soft) P(soft) + P(Died Ihard) P(hard) = 0.90 (0.667) + 0.60 (0.333) = 0.80
b) P[soft  Lived] = P(Lived  soft) P(soft)/ P(Lived) = 0.10 (0.667)/ 0.20 = 0.333
c) Binomial distr. (n=12,p=0.10): u = up = 12 (0.10) = 1.2 ,' 0'2: np(1p) = 12 (0.10)(0.90) = 1.08
5) Pdfs, Moment Computations, Derived Distributions
a) E{R} =11“ [10.5r] dr = [r2/ 2 — r3/ 6](for r=2) = 2/ 3; MW} = [13/ 3 — r‘/ 8](for r=2) = 2/ 3
Var{R} = Ele} — E{R}2 = 2/ 3 — (2/ 3)2 = 2/ 9 (Many students had trouble.) 1'
b) First get FR(r) =f [10.5s] ds = r—rz/4 for o 5 rs 2 2 pts 4 pts 4 pts 1 pt
2 pts
1+1 pt 3 pts
3 pts
3 pts 3 pts
1 pt Thean(w) = P{ W s w } = P{7 Rss w } = P{R s (w/7)“3 } = (w/7)"3 —(w/7)z/3/4 for 0 s ws 56
fw(w) = (1/ 21)(w/ 7)‘2’3  (1/ 42)(w/ 7)”3 for 0 s w s 56 (CDF transformation is slow but sure.) 4 pts Remember to give the range for which formula applies. Okay, now lets try direct transformation: 30 fw(W) = fR(r(W))  dr/ dw  = [(WI 7)“ — 05((W/ 7)“3)2] (1 / 21)(W/ 7)“.
= (1/21)(w/ 7)'2’3 —(1/42)(w/7)‘1’3for 0 s w s 56 (Good it worked.) CEE 304  Uncertainty Analysis in Engineering
First Examination
October , 2002 You may use text, your notes and calculators. There are 50 points in total, one per minute. 1. (8 pts) Consider three events denoted], R and S, where you know that
PU] =02, P[R] =0.6, P[S] =0.9 Assume the three events are independent. (a) What is the probability that J and R both occur? (b) What is the probability that either] or R occurs, but not both? (c) What is the probability that none of these events occur? Show work. 2. (18 pts) Stedinger’s research group is studying phosphorus loads in runoff from Catskill dairy farms.
Imagine that during the summer significant runoff and phosphorus producing events occur like a Poisson
processes with on average 4.5 events per 30 day month. (a) What is the mean and standard deviation of the number of events in the next two weeks (14 days)?
(b) A student plans to stay in the area until they have been able to observe the instrumentation
record 2 events. What is the mean and standard deviation of the time she needs to wait?
(c) What is the probability of observing at least 2 events in a 9 day period?
(d) What Microsoft Excel command containing the letters gamma generates the correct answer to part (c)?
(e) Only 10% of events are considered large. What is the probability that in the last 15 events there were
exactly two large events? 3. (9 pts) A simple model of the reliability of a structural system indicates that it can support a load of
L = 3 W + 1.8 Q wherein W and Q are normal such that W ~ N( 50, 1.32 ), Q ~ N ( 75, 1.52 ). Moreover,
Correlation(W,Q) = 0.5. What load can the engineer be 90% conﬁdent the system can support? 4. (5 pts) Security, Security — that is the word. A quality control specialist reviewed data for a baggage
checking system that uses sensors to ﬂag suitcases likely to contain dangerous materials. Trials indicate that ”safe” baggage results in a ﬂag only 3% of the time, while “unsafe” baggage generates a ﬂag 80% of the time. The specialist estimates that only 0.5% (or 1 in 200) bags carry ”unsafe” materials. If that is true,
what is the probability in practice that a suitcase ﬂagged by the machine is actually ”safe"? 5. Random variable Y has a probability density function (8 — y“)/ 12 for 0 s y s 2, and zero otherwise.
(a) (3 pts) What is the mean value of Y?
(b) (7 pts) It turns out that interest is not in Y, but in the variable K = 1 / Y. What is the pdf for K? CEE 304  Uncertainty Analysis in Engineering Solutions First Examination October , 2002
Axioms of probability and independence
1a) PUnR] = 0.12 (Easy start.) [For 1b: P[ ]U R] = PU] + P[R]— PUnR] = 0.68] 2 pts
1b) P( I or R but not both ) = P[ Iu R]  PUnR] = 0.68 — 0.12 = 0.56 3 pts or P( I or R but not both) = P[ I nR’ u I’ nR ] (disjoint sets) = 0.2(0.4) + 0.6(0.8) = 0.08 + 0.48 = 0.56
1c) P[ I’ n R’ n S’ ] = (10.2)(10.6)(1—0.9) = 0.8 (0.4) 0.10 = 0.032 (compliments independent) 3 pts 2. Poisson processPoisson, Gamma, Binomial distributions a) 7» = 4.5/30 days = 0.15 / day; number in 14 days is Poisson u = (I2 = v = 14}. = 2.1; o = 1.45 4 pts
b) Time to 2th Gamma with u = 2/ 7x = 13.33 days, 0'2 = 2/ 73 = 88.89 days2 ; o= 9.4 days 4 pts
c) Poisson dist: Pr[ K 2 2 I v = 97. = 1.35 ] = 1  P(K=0) — P(K=1) = [1/0!+(1.35)/ 1!]exp(1.35) = 0.390 4 pts
d) GAMMADIST( 9, 2, 6.6667, true) = 0.3908 == cdf of gamma distribution with «=2, [kl/71.. 2 pts
e) Binomial distribution n = 15, p = 0.10; probability of just 2 = (15 choose 2) pz(1p)13 = 0.267 4 pts 3. Normal calculations, Moments of Sums Given L = 3 W + 2 Q one obtains immediately E{L} = 3 (50) + 1.8 (75) = 285 1 pt Var(L} = 32 Var(W) + 2 [3*1.8] Cov{W, Q} + 1.82 Var{Q} where Cov(W,Q) = StDev(W} StDev(Q) Corr(W,Q) Thus Var{L} =32 13+ 2 (3*1.8) 0.5 (1.3*1.5) + 1.81 1.52 = 33.03 = (5.74)’. 5 pts Now 0.90 = P{ L > 10.10 } so can be 90% confident that L exceeds 285  1.282 (5.74) = 277.6 3 pts
/—Flag (0.03) 4. /  Safe (0.995) ..._,. ‘m DRAW THE PICTURE! Bayes \ Unsafe (0.005) won—«Flag (0.80)
\ « m (0.20)
P(Flag) = P(Flag lSafe)*P(Safe) + P(Flag lUnsafe)*P(Unsafe)
= 0.03(0.995) + 0.80(0.005) = 0.0299 + 0.0040 = 0.0339
P(Safe lFlag] = P(Flag lSafe)*P(Safe)/ P(Flag) = 0.0299/ 0.0339 = 0.88 or almost 90% ll! 5 pts 5) Pdfs, Moments, Derived Distributions, Integration of pdf to get a probability (Students had trouble.)
2 2
1 3 1 3’2 3’5
012 12 2 5 0
b) Let K = 1/Y. Note that the range ofK is 1/2 < K. So for 0.5 < k, one obtains: 2 4 2
FK(k)=P{Ksk}=P{1/Ysk}=P{l/ksY}= I113[8_y3]dy=112[8y—7] =1—[8/k—0.25/k‘]/12
l/k llk
One obtains the pdf by differentiating to obtain fK(k) = [8/1(2 — 1/k5]/12 for l_( > 1/2, and zero mouse. 2+2+2+1 pts becomes . Whow! dy/dk = l/k7‘. Direct transfonnation yields fK(k) = [8  y(k)]/ 12 ldy/dk] = [8/1‘:2  1/k’]/ 12, for k > 1/2, zero otherwise. CEE 304  UN CERTAINTY ANALYSIS IN ENGINEERING
PRELIM #1
October 12, 2001 Use the text, notes, calculators, and your knowledge, to answer the questions below.
The exam lasts 50 minutes. There are 50 points in total. 1. (6 pts) Consider three events denoted D, G, and], where D and G are independent.
P[D] =03, P[G] =05, P[I] =0.7, P[GU I] =O.8
(a) What is the probability of either D or G or both occurring?
(b) If G occurred, what is the probability that D will NOT occur?
(c) And if G occurred, what is the probability that I will occur? Show work. 2. (5 pts) Of 10 ASCE officers, 4 are randomly selected to choose refreshments for the next meeting. What is
the probability that Jim and Susan (2 of the officers), both serve on the committee of 4? 3. (12 pts) Annual pollutant loading into a lake is the sum of three sources. The load from source i is Xi for i
= 1,2,3. Assume that E[Xi] = 5000 and Var[Xi] = (500)2 for i = 1 and 2, and E[X3] = 15,000, and Var[X3] =
(700)2, and assume that the variations in loads are uncorrelated. (a) Then what is the mean and standard
deviation of the total annual load? (b) If all three loads are normally distributed, what annual pollutant loading has only a 10% probability of being exceeded? (c) If the loads were actually positively correlated, explain what effect that would have on the mean and variance. Justify answer. 4. (15 pts) At a large airport security does background checks on passengers. Every 20 minutes during the
afternoon the computer ﬂags a passenger who should be interviewed. (a) At this rate, what is the mean
and variance of the number of passengers ﬂagged for interviews during a 4 hour period? (b) What is the probability of just one interview in 30 minutes? (c) They have only 5 interview forms left:
what is the mean and variance of the time until they will need to conduct 5 more interviews? (d) Assume there is a probability p = 0.10 that a passenger interview results in the need for a further FBI
check  what then is the probability that the next 5 interviews result in the need for exactly one FBI check? (e) What is the mean and variance of the number of interviews until the next FBI check is required? 5. (8 pts) In a child’s game, a marker is thrown to determine the radius R of a circle, where
P[Rzr] = (1r)2 Osrsl, and P[Rzr] = 0,for 1<r, (a) What is the mean value of R? (b) Given that the area of a circle is nrz, determine the probability density function for the area of the circle defined by the generated radius. 6. (4 pts) Prof. Stedinger’s harddisk crashed this fall. It turns out that 20% of SpeedDrive harddrives used
as replacements fail in the 3year warranty period, and 5% of DriveFlash harddrives fail in the 3year warrenty period. However, because of a lower price, 70% of the replacements have been SpeedDrive units. Given that Stedinger’s drive failed, what is the probability it is a SpeedDrive? CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
SOLUTIONS for Prelim #1
October 12, 2001 Axioms of Probability (Draw Venn diagram. Makes the problem easy.)
1a) P[ D n G ] = 0.3*0.5 = 0.15 (independent) So: P(DUG) = P(D)+P(G)—P[ Dr) G] :03 +0.5—O.15= 0.65 2pts
1b) two are independent so :P[ D’ l G] = P[ D’ ] = 1  P[D] = 0.7 2 pts
1c) P[ J l G] = P] J n G] /P(G) = [Look atavenn diagram] > 0.4/0.5 = 0.8 2 pts Or,P(JUG) = P(J)+P(G)—P[JﬂG]=>O.7+O.5—P[JHG]=0.8=>P[JﬂG]=O.4 Counting: 2) 1*1*(8 choose 2)/(10 choose 4) = [8*7/2]/[10*9*8*7/4*3*2*1] = 4*3/[10*9] = 2/ 15
Must have both Jim AND Susan. Leaves 8 people from which to choose 2. (went poorly) 5 pts Normal calculations, Moments of Sums (#3 went well.) 3a) X = X1 + X2 + X2. E[X] = E[Xl] + E[Xz] + E[X3] 2 pts
2 5000 + 5000 + 15000 = 25,000. For independent variables: Var[X] = Varle] +Var[X2] + Var[X3] = 2(500)2 + (700)2 = (995)2 3 pts x090 = E[X] + 1.282 Stdev[X] = 25,000 + 1.282*995 = 26,280 3 pts If the Xi variables were positively correlated then the Var[X] would include the three terms
Cov(Xi, Xj) fori ¢j, and these would all be > 0, thus INCREASING variance of the sum X. 4 pts
Correlations would have NO effect on the mean: E[X] = E[Xl] + E[X2] + E[X3] Poisson processPoisson, Gamma, Binomial, Geometric distributions 4a) in = 3/hr. # in four hours is Poisson with p = v = 47» = 12. 02 = v = 4}» = 12 3 pts 4b) Poisson distribution: Pr[ K=1 I v = At = 1.5 ] = 1.5 e15/1! = 0.333 40) Time to 5th is Gamma with n = 50. = 5/3 = 1.667 hrs. 02 = 50.2 = 0.556 hrs2 3 pts
Or if A = 1/20 min1, then [1 = 100 min. 02 = 2000 min2 4d) Binomial Pr[ X=1 n = 5, p=0.01 ] = 0.328 3 pts 4e) Discrete waiting time  Geometric, p = 0.1; y = Up = 10; 02 = (lp)/p2 = 90. 3 pts A reasonable approximation is provided by the exponential distribution for continuous waiting time to the ﬁrst anrival. For it: 0.1 per interview, E(T)=1/}\.= 10 interviews; Var(T) = II)»2 = 100 interviewz.
Thus the exponential distribution gives about right answer and got full credit  THIS TIME. 5) Pdfs, Moments, Derived Distributions The complement of cdf was given! a) mm = 1—(1 — r)2 hence fR(r) = 2(1r); E[R] = [o r*2(1—r) = r2 — 213/31: 1/3 4 pts
b) FA(a) =Pr[Asa] = Pr[:tkls a] = Pr[ Rs Ja/tt ] = FRwa/tt 1 = 1—(1 — Ja/n )2 0 s a s n. PdfforA=fA(a)= adaFA(a)=2[1‘ Ja/n](0.5)[1/n/a/Ic]=(ilrr/a —1) 0 5a 51:. 4.5 pts / — Fail (0.20)
6) Bayes / —SpeedDrive (0.70) ————— Okay (0.80)
 DriveFlash (0.30) —°——Faii (0.05)
\ —Okay (0.95)
P[SpeedDrivelFail] = P(Fail ISpeedDrive)*P(SpeedDrive)/P(Fail) = 0.903 4 pts
where one needed: P(Fail) = P(Fail SpeedDrive)*P(SpeedDrive) + P(Fail IDiiveFlash)*P(DxiveFlash) = 0.155 ...
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