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Unformatted text preview: CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
Prellm #1 October 6, 1999 Use the text, notes, calculators, and your head, to answer the questions below.
The exam lasts 50 minutes. There are 50 points in total. 1) (6 pts) Consider three independent events denoted R1, R2, and R3 where
P[ R1] =0.4, P[R2] =0.3, P[R3] =0.2
(a) What is the probability of R1 and R2 and NOT R3?
(b) What is the probability of R1 or R2 (i.e. R1 U R2) occurring?
(c) If either R1 or R2 occurred (i.e. R1 U R2), what is the probability R2 occurred? Show work. 2) (4 pts) At an ASCE student chapter meeting, those present were unable to reach a decision so they
decided to refer the matter to a committee of 3; if the committee's membership is obtained by randomly
selecting from the 12 men and 6 women present at the meeting, what is the probability the committee
contains at least one woman? 3. (6 pts) A construction firm studied the differences between the firm's estimated costs E and the actual
costsA on small jobs. They found that the differences D were normally distributed with an average of
+6% with a standard deviation of 9%. [Here D = (EA)/ A.] What then is the probability that D < 15%. What is the lower quartile of the distribution of D? 4. (14 pts) Assume that large and destructive earthquakes occur worldwide at a rate of about 1 every 3.5
years. However, this year we had major and tragic earthquakes in both Turkey and Taiwan. (a) Brieﬂy, why might a Poisson Process be a good model of worldwide arrivals of major earthquakes?
(b) What is the probability of having exactly two such quakes in a single year? (c) What is the probability it will be more than 10 years before the next major quake? (d) What is the mean and variance of the time until the next 4 earthquakes occur? 5. (10 pts) The joint probability density distribution function for two random variables X and Y is:
fxy(x,y) = x/6 0sxs2,0sys3,
0 otherwise
(a) What is the mean value for X?
(b) What is fY(y)? (c) Are Y and X independent? justify your answer. 6. (10 pts) A camping club has equipment troubles. Often tents lack parts when they are returned. Tents
are checked by one of two quartermasters (Alice who checks 60% of tents and Jim who checks 40%) before
they are released for use. If Alice checks a bad tent there is an 80% chance she will notice and fix it; if
Jim checks a bad tent, the chances the problem is fixed are only 50%. Beth returned a bad tent last
weekend and later found that it was later released without being fixed. What is the probability a bad
tent is released without being fixed? What is the probability Jim checked Beth's tent? If 5 bad tents
were returned over 4 weeks, what is the probability exactly 2 of the 5 would not be fixed before release? CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
SOLUTIONS for Prelim #1 October 6, 1999 Axioms of Probability (Draw Venn diagram. Makes the problem easy. Many people missed key points.) 1a) Independent events so P(RlnRZnRB') = O.4*0.3 *0.8 = 0.096 2 pts
1b) P( R1 U R2) = P(Rl) + P(R1)— P[ R1 0 R2] = 0.4 + 0.3 — (0.4)(0.3) = 0.58 2 pts
10) P[ R2 IRIURZ] = P[R2r\(R1UR2)]/P[R1UR2] = P[R2]/P[R1UR2]= 0.3/0.58 = 0.517 2 pts Counting 2) Pr(At least 1 woman): 1 — Pr(no women) = l — (12 choose 3)/(18 choose 3) = 1 — (12*1 1*10)/( 18* 17*16) = 0.730 4 pts
Normal calculations 3a) Pr[D <—15% ] = <I>[ (15 — 6)/9 = 2.333 ] = 0.00982 3 pts
3b) p = 25%, use z0.25 = 0.6745. Thus d0_25 = p — 0.675 G = —0.07% or ~0% 3 pts Poisson process  Gamma and Poisson distributions (Fairly straightforward.) 4a) Poisson process is reasonable because earthquakes arrive separately, with p = lAt, we can
assume that worldwide arrival rate Ms constant, and arrivals in different periods are independent from one point on the globe to another. Answers should relate to earthquakes processes. 4 pts
4b) With 7! = 1/3.5 = 0.857; Pr(K = 2) = (Kt)2exp(M)/2! = 0.031 3 pts
4c) Pr(K = 0 Iv = M = 10/35 = 2.85) = exp(—v) = exp(2.85)= 0.0574 3 pts
[01". Pr[ T1 > 10] = 1  FT1[ t=10 l 7» 23.5] = exp(—7tt) = exp(2.85)= 0.0574; same answer! ]
4d) Arrival time to 4th is gamma with or = 4, B = ll?» :35. y = 043: 4(35) =14 years; 02 = 0432: 4/(3.5)2 = 49: (7)2 4 pts Pdfs, Moments and Independence 5a) Compute marginal pdf for x: fx(x) = x/2, 0 s x s 2 2 pts. I‘X = Integral from 0 to 2: IX fx(x) dx = x3/6 = 4/3 2 pts
5b) fy(y) = 1/3 for 0 s y s 3 => Uniform! 3 pts
Sc) Here fxy(x,y) = fx(x)fy(y) = (x/2)*(1/3), so they are independent! 3 pts
Bayes Theorem / —~—BT (0.20) / Alice (0.60)~~0—Okay (0.80)
————————— —Jim (0.40) ———°BT (0.50) \ —Okay (0.50)
6) ET is event that bad tent is released. P(BT) =P(BT!Alice)*P(Alice) + P(BTIJim)*P(Jim)
= 0.20(0.60) + 0.50(O.40) = 0.32 3 pts
Bayes Theorem: P[ Jim I BT ] = P(BTIJim)*P(Jim)/P(BT) = 0.625 3 pts Binomial dist. Pr[X=2  n25, p=0.32] = (5 choose 2) 0.322 0.683 = 0.322 4 pts. CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
Prelim #1 October 7, 1998 Revlsed Use the text, notes, calculators, and your head, to answer the questions below.
The exam lasts 50 minutes. There are 50 points in total. 1) (6 pts) Consider three events denoted Q, R, and W where
P[ Q] = 0.01, P[ R] = 0.02, P[ W] = 0.60,
P[QuRuW]=0.60 and P[ QnR] =0.005
(a) What is the probability of Q or R or both?
(b) Given that R occurred, what is the probability that Q also occurred?
(C) What is the probability of R u W? 2. (6 pts) A manual on wood engineering indicates that the strength of laminated wooden beams has a
10% chance of being greater than or equal to 140, and a 10% chance of being less than 80. Assuming
strengths are normally distributed: (a) What is the median strength of beams? (b) What is the
probability a beam’s strength exceeds 60? 3. (16 pts) In August a severe tornado killed two people at the State Fair in Syracuse.
On average 3 severe tomados occur every 20 years in Western New York.
(a) Brieﬂy, why might a Poisson Process be a good model of the arrivals of severe tomados?
(b) If a tornado occurred this year, what is the mean and variance of the time until 2 more tomados occur?
(c) What is the probability of 2 or more tornados in a 10 year period? 4. (12 pts) The time required to find a junction of buried pipe has probability density function: fT(t) = 3(1t)2 0 s t s 1 hr and 0 otherwise. Var[Tl = 0.0375
(a) What is the mean of T?
(b) Let T1, T2, and T3 correspond to the time required to find three different junctions in a local
neighborhood. Assume the pdf of all three is as given above. In addition Correlation[Ti, Tj] = 0.4 for i¢j.
What is the variance for the total time required to find the three junctions?
(c) Assume V, the volume of soil removed to find a junction, is 10 T14 where T is the time required to find the junction. Approximately what is the variance of the volume V of soil removed in any search? 5. (10 pts) Three librarians named a, B, and x file books. Because of their different duties, 0c, [3, and x
shelve respectively 60%, 30% and 10% of the books. The misshelving probability is 1 in 100 for both [3
and x. However, a is very careful and has misshelving probability of 1 in 200. (a) Of the mis—shelved
books, what fraction were handled by [3 ? (b) In the last 200 books shelved by [3, what is the probability
that B mis—shelves exactly 2 ? (c) <Think> If I start checking ALL books recently shelved, on average
how many do I need to check before I find one that was nusshelved ? CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
SOLUTIONS for Prelim #1 October 7, 1998 Axioms of Probability (Draw Venn diagram; then how can you miss 1c? Most peoople did.) 1a) P(QUR) = P(Q)+P(R)—P[ QmR] =0.01+0.02—0.005=2.5% 2pts
1b) PIQIR]=P[ QﬂR]/P[R]=0.005/0.02=0.25 2pts
1c)P[W]=O.60andP[QURUW]=0.60—>P[RUW]=P[W]=0.60 2pts Followsfrom0.60=P[W] s P[RUW]SP[QURUW]=0.60. {P[QnW]: ?} Normal calculations (Some people had trouble with this one. Need to know normal distribution!) 2a) Normal distribution is symmetric so the median = (140+80)/2 = 110. 2 pts
2b) 110 + 1.282 0' = 140 —> o = 30/1282 = 23.4; or (140—80) = 2*1.282*o —> o = 23.4 2 pts
Pr[S > 60] = 1—— Pr[ S < 60] = 1— d>[ (60 110)/23.4 = —2.14]=1—0.0162 = 0.984 2 pts Poisson process  Gamma and Poisson distributions (Fairly straightforward.) 3a) Poisson processes reasonable because storms arrive separately, with p = kAt, we can assume that average am'val rate it is constant, and arrivals in different periods are independent. 6 pts
3b) With it = 3/20 = 0.15, arrival time to 2nd is gamma with 0t = 2, B = 1/7t =20/3.
u = on = 2(20/3) =13.33 years; o2 = 0th = 29.2 = 2(20/3)2 = 89 = (9.4)2 5 pts
3b) Poisson dist.: P[K 2 2] = 1 — P[K = 1] — P[K = 0] = 1 —exp[M](1+7Lt); for Kt = 1.5
= 1 — 0.223 — 0.335 = 0.442 (Watch < versus S; Tables did not work for this one with M = 1.5.) 5 pts Pdfs, cdfs, moments and correlations/covariances of distributions
4a) uT=1ntegra1 fromO to 1: Itha) dt= 3[t2/2 2t3/3 +t4/4] =0.25 4pts To get variance integrate from 0 to 1: It2 fT(t) dt = [t3/3 2t4/4 + t5/5] = 1/10 = 0.1 Var[T] = E[T2]  E[T}2 = 0.0375 (Note: [1412 at Hz)
4b) Var[ T1 + T2 + T3 ] = 3 o2+ 2 Cov[T1, T2] + 2 Cov[T2, T3] + 2 Cov[T1, T3] 2 pts = 3 oT2 + 6 oTZ (0.4) = 0.2025 {Mean is 3 W = 0.75} 2 pts
4c) Use approximate variance formula:. Var[V] e (dV/dT)2 Var[T] = (14 uT04)2 oTZ 2 pts
= 2.42 = (1.56)2 {Means is about 10 uT14 = 1.43} 2 pts Bayes Theorem, and binomial and exponential distributions
5a) Bayes Theorem: P[ B I error] = P[ B and error ] / P[ error]
= 0.3 (0.01) / [0.6 (0.005) + 0.3 (0.01) + 0.1 (0.01) ] = 0.429 3 pts
5b) Binomial distribution Pr[ x = 2] = [200 choose 2] (0.01)2(0.99)198 = 0.2720. 3 pts
OR, use approximation X ~ Poisson[v = 200(0.01) = 2] > Pr[X=2] = vze'V/Z! = 0.2707
5c) Expected waiting time = Up = 1/ [0.6 (0.005) + 0.3 (0.01) + 0.1 (0.01) ] = 143 4 pts CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
Prelim #1 October 8, 1997 Use the text, notes, calculators, and your head, to answer the questions below.
The exam lasts 50 minutes. There are 50 points in total. 1) (10 pts) Consider two events denoted S and R where
P[S] =0.35, P[ SnR] =0.20and P[SUR]=0.70
(a) What is the probability of R?
(b) What would be the probability of S 0R R but NOT both ?
(c) What is the probability that R occurred if it is known that S occurred? 2. (10 pts) The distribution of the velocity V of artificial rain drops created in a lab experiment are thought to have a mean of 20 and a variance of 4. (a) What approximately are the mean and variance of the kinetic energy K where
K=3v2 (b) If the velocity V has a normal distribution, what is the probability K > 1500 ? 3. (15 pts) [This problem could be hard for some people. LOOK at graph on the board]
The joint probability density distribution function of two random variables X and Y is:
fXY(x,y) = 1/6 0 S x, 0 S y, and 6x + 2y 5 12
0 otherwise
a) What is the mean value for X?
b) What is the conditional cumulative distribution FYIX(yx) for Y given that X = x?
c) Are Y and X independent? Justify your answer. 4. (15 pts) The quality engineer for a large home construction firm collects records of the number
of defective appliances installed by the firm. Home furnaces made by Hot—Start seems to have an
abnormally large number of problems. Assume the arrival of reports of defective HotStart
furnaces is a Poisson process with a mean of 2 reports every 5 working days. (a) What is the mean and variance of the time until 4 defective Hot—Start furnace reports arrive? (b) What is the probability of at least one report arriving tomorrow? (0) The quality engineer will visit 3 homes that are under construction with HotStart furnaces.
If there are 20 such homes the engineer might visit, and exactly 2 have defective furnaces, what is
the probability she visits at least one home with a defective HotStart furnace? (d) The competing brand of furnace is Sure—Heat. The construction firm uses Sure—Heat furnaces
in 90% of its homes, and HotStart in the others. Sure—Heat furnaces have a failure rate in the first
year of 1in100, whereas 1—in20 Hot—Start furnaces fail in the first year. What fraction of the
firm’s first—year furnace failures are Hot—Start units? CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
SOLUTIONS for Prelim #1 October 8, 1997 Axioms of Probability 1a) P(R) + 0.35 — 0.20 = 0.70 > P(R) = 0.55 3 pts
1b) P(R or S but not both) = P[ S U R] —— P[ S n R] = 0.70 — 0.20 = 0.50. 4 pts
10) PIRIS]=P[SnR]/P[S]=0.2/0.35=0.571 3pts Derived Distributions and normal calculation 2a) E[K] z g(uV) = 1200 is worth 2 points, however the more accurate expression yields
instead E[K] z g(uV) + 0.5 g"(uv) Var[V] = 1200 + 0.5 (6) (4) = 1,212 3 pts
This is actually exact because 3V2 is a quadratic expression. Thus we can also use
Var[V] = E[V2] — E[V]2 —> E[K] = 3{ E[V]2 + Var[V] } = 1,212 exactly!
Var[K] e g'(uV)2 Var[V] = [6uV]2 6V2 = 57,600 3 pts
To estimate Var[K] as 32{Var[V] }2 makes no sense. Var[V2] ¢{Var[V] }2.
2b) Derived distribution: P[K > 1500 ] = P[ 3V2 > 1500] = P[ V > 22.36 ].
So: P[K > 1500 ] = P[ (V20)/2 > (22.3620)/2 = 1.18] = 11.9% 4 pts
K is NOT normally distributed; were EV=O, then K ~ gamma distribution with 0!. = 0.5 Pdfs, cdfs, and moments of distributions; independence
3a) To get fX integrate joint density over all y (fXY ¢ 0 for y = 0 to 63x) to find fX(x)=(1/6)[6—3x]=1—x/2for0SxS2. 3pts Then integrate from 0 to 2: Ix fX(x) dx = 4/2  8/6 = 2/3 = Hx 3 pts
3b) First get: fYX(ny) = fxy / fX = (1/6)/[1x/2] = 1/[63x] for 0 S y S 6—3x. 3 pts This result can also be seen from the graph of fxyz fXY is constant for 0 S y S 63x.
Conditional y distribution is uniform so: FYX(yX) = y/[63x] for O S y S 6—3x. 3 pts But, Pr[ Y S y  X = x ] = FYX(yx) at ny(x,y)/FX(x) = Pr[ Y S y  X S x ]
30) Not independent. Knowing value of one tells us the range of the other, so they
are very dependent. y=6 > x=0! Also: fXY ¢ fX 0 fY 3 pts Poisson process; Gamma and Poisson distribution; Counting; Bayes Theorem 4a) Poisson processes with )b = 0.4. Arrival time to 4th is gamma with on = 4, B = 1/7» = 2.5: u = OLB = 4/9» :10 working days; 0'2 = (182 = 40»2 = 25 (working days)2 4 pts
4b) Poisson dist.: P[K=0] = exp[—7tt] = exp[0.4] = 67% —> P[K 2 1] = 1 — P[K=0] = 33% 3 pts
4c) P[no defects] = (18 choose 3)/(20 choose 3) = (17*16)/(20*19) = 0.716 —> P[at least one] :1 — P[no defects] = 0.284. 4 pts OR, 2 defects aren’t in the 17 uninspected houses: 1 — (17 choose 2)/(20 choose 2) = 0.284
4d) Bayes Theorem: P[ HotStart I Failure] = P[ HotStart and Failure ] / P[ Failure] = 0.05 (0.1) / [0.01 (0.9) + 0.05 (0.1) ] = 0.357 4 pts CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
First Exam October 5, 1994 Use the text, notes, calculators, and your head. to answer the questions below.
The exam lasts 50 minutes. There are 50 points in total: one per minute. 1) Consider two events A and B where you know that
P(A) = 0.3; P(B) = 0.4; and P(A u B) = 0.5
(2 pts) Now, consider, what is the probability of A or B happening?
(2 pts) What is the probability of A and not B happening?
(2 pts) If B didn’t occur, what is the probability of A?
(2 pts) What is the probability of neither A or B happening? 2) (8 pts) Your division’s performance on construction jobs is represented by a score X which can be
thought of as a continuous random variable X, 0 s X s 1, where across jobs: P[XSs] =53 OSSSI. for s>1, P[X5s] = 1 What are the mean and variance of X? 3) (12 pts total) As part of the TQM effort, Prof. Stedinger decides to develop a weighted score 5
of his teaching using two different questions on a questionnaire:
S = 0.6 Q1 + 0.4 Q2 where Q1 = score question 1, Q2 = score question 2.
Q1 and Q2 are both random variables depending on who is in class and how students feel.
Assume that: EIQI] = 3.2, Standard deviation[Ql] = 0.2
E[Q2] = 3.6, Standard deviation[Q2] = 0.3; Correlation[Q1, Q2] = 0.5
(4 pts) If Q1 has a normal distribution, what is the probability a score on question Q1 exceeds 3.0 ? (3+5 pts) What are the mean and variance of S ? 4) Carolyn Stedinger plays girls soccer for Ithaca High School. Goals by either team seem to occur as a
Poisson process with an arrival rate of 1 per 15 minutes of play. a) (4 pts) What are the mean and variance of the number of goals scored in a 60 minute game? b) (4 pts) What is the probability of exactly 3 goals in the last 30 minutes of a game? c) (4 pts) When watching Carolyn's game Friday, her father wondered how long he had to wait to
see some action. What are the mean and. variance of the waiting time until two goals are scored? d) (4 pts) If the coach chooses the 11 players who start by randomly selecting from the 15 girls
available, what is the probability Carolyn gets to start? 5) (6 pts) On the Ithaca soccer team, Jennifer scores 70% of the goals and Amy the other 30%; Jennifer
scores 1 in 5 goals with her head  Amy scores 1 in 10 goals with her head. An Ithaca girl just scored a
goal with her head! What is the probability it was Jennifer? CEE 304  UNCEFITAINTY ANALYSIS IN ENGINEERING
~~~~~ First Exam Solution ~~~~~ October, 1994 1a) P[A or B] = P(A U B) = 0.5 [ A or B is just A U B; don’t make it hard!] (2 pts)
b) First compute P(A n B) = 0.2; then from P[A] = P(A n B’) + P(A n B). P[A and not B] = P(A n B’) = P(A) — P(A n B) 0.3  0.2 = 0.1 (2 pts)
C) P[A I B’] = P(A n B')/P[B’] = 0.1/0.6 = 0.167 (2 pts)
d) P[ (AUB)’] = P[A’ and B’] = 1  P[A or B] = 1  0.5 = 0.5 (2 pts)
1
2) pdfis 3s? foross $1. E[X] =J $352 ds =3/4 (2+2 pts)
0 1
EM] =J s2 3s2 ds = 3/5; Var[X] = E[x2] — E[x12 = (3/5)  (3/4)2 = 0.0375 (2+2 pts) 0
3) P[ Q1 > 3.0 ] = P] Z > 1 ] = 1  P[Z < 1] = 0.8413 = P[Z <1] (use tables) (4 pts)
E[ 0.6 Q1 + 0.4 Q2] = 0.6 E(Q1) + 0.4 E(Y) = 3.36 (3 pts) Var[ 0.6 Q1 + 0.4 Q2] = (0.6)2 Var(Q1) + (0.4)2 Var(Q2) + 2 (0.6) (0.4) Cov(Q1,Q2);
=(0.6)2 (0.2)2 + (0.4)2 (0.3)2 + 2 (0.6) (0.4) [(0.2) (0.3) (0.5)]
= 0.0144 + 0.0144 + 0.0144 = 0.0432 (all plus signs this year) (5 pts) 4) Arrival rate of Poisson process: A = 1 per 15 minutes = 1/ 15 per minute. a) Goals in 60 minutes is Poisson with v = 601 = 4; Mean = Variance = v = 4 (4 pts)
b) For 30 minutes, V: 307» = 2; P[X=3] = v3exp(—v) / 3! = 0.18 (4 pts)
c) E[Tz] = 2/ 7» = 30 minutes; Var[T2] = 2/ 12 = 450 min2 (4 pts)
d) P[Carolyn starts] =1 P[Carolyn doesn’t start] = 1  Comb(14,11)/Comb(15,11) = 1  Comb(14,3)/Comb(15,4) = 1—4/ 15 = 0.73 3213.; P[Carolyn starts] = Comb(14,10)/Comb(15,11) =
11/15 where Comb(14,10) = # teams with Carolyn (i.e. Carolyn plus 10 other girls from 14). QR; P[Carolyn starts] = 11/ 15 where Carolyn is assigned to one of 15 slots and the
first 11 slots are starters (imagine 11 black 8: 4 red balls in an urn; P[B] =11 / 15 ). (4 pts) 5) /  I ennifer, 0.7 /  Header, P[H [Jenny] = 0.2
Goal> —  l \ 
\———Amy, 0.3 /—Header, P[H  Amy] = 0.1
\ ........... P[Iennifer  Header] = (O.7)(0.2)/ [ (O.7)(0.2) + (0.3)(0.1)] = 0.14/0.17=0.824
Go Iennifer! (6pts) CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
First Exam October 6, 1993 Use the text, notes, calculators, and your head, to answer the questions below.
The exam lasts 50 minutes. There are 50 points in total, one per minute. If we do not have your secret code for 304, please write it on the inside cover of your exam book. 1) Consider three independent events D, E and G, where
P(D) = 0.1, P(E) = 0.2, and P(G) = 0.3
(3 pts) What is the probability that at least one of these events occurs?
(4 pts) What is the probability of DU(G’)? (where the prime denotes the complement of G)
(3 pts) What is the probability of D given that G occurred? 2) (6 pts) The 304 Total Quality Management team, consisting of 7 students, decides to form a committee
of 2 to write a survey. They will determine the committee membership randomly. Sam, Jill and lack really want to be selected: what is the probability that at least one of them is on the committee? 3) (10 pts) A quality—control engineer oversees application examinations. The number of correct answers
X and incorrect answers Y are recorded. She assigns a score 8 = X — 3Y to each exam, where the —3Y term
discourages guessing. If EX = 60, BY =5, Var[X] = 100, Var[Y] = 10, and Cov[X,Y] = 12, what is the
variance of S? If she intends to pass 90% of the students and scores are normally distributed, what should be the cutoff score for passing? 4) (8 pts) I looked up statistics on US murders using firearms. They were divided by age, sex, and race.
Let me use statistics for white 2024 year olds (”kids”), though the death of any citizen is tragic. For
such white ”kids”, the probabilities of being murdered in a year with a firearm are 1x10'4 and 2.3x10'5,
for males and females. The percentage of white ”kids” that are male and female are 51% and 49%,
respectively. What is the probability a 2024 year old white kid is murdered with a firearm during a
year. What is the probability the next two white ”kids” that are murdered are both male? 5) There was a very tragic earthquake in India last week. Worldwide the arrival rate of major
earthquakes could be described by a Poisson process. Furthermore, assume that 4 “major” earthquakes
occur per decade. a) (4 pts) What is the mean and variance of the number of earthquakes in a 30year period? b) (4 pts) What is the probability of exactly 2 earthquakes in a 5year period? c) (8 pts) If a scientific investigation making use of new worldwide instrumentation were
established, what is the mean and variance of the length of time they would have to wait to see 5 earthquakes? What is the probability they would have to wait more than 10 years? CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
~~~ First Exam Solution ~~~ October, 1993
1a) P(at least 1) = 1 — P(none) = 1 — (O.9)(0.8)(0.7) = 1 — 0.504 = 0.5 (3 pts)
b) P[DU(G’)] = P[D] + P[G']  P[(DhG’] = 0.1 + 0.7— (0.1)(0.7) = 0.73 (4 pts)
for independent D and (G').
OR: P[DU(G’)] = 1 — P[ {DU(G’)}' ] = 1 — P[ D’nG] = 1— (0.9)*(0.3) = 0.73
c) D and G are independent so P(D IO) = 0.1 (3 pts) 2) The probability that at least one is on the committee, is one minus the probability
that none are. The probability of none is (4 choose 2) divided by (7 choose 2) = 2/7. Thus the probability of at least one equals 1  2/ 7 = 5/7. (6 pts)
OR: consider two on the committe in (3 choose 2 ways) plus one on the committee in
(3 choose 1)*(4 choose 1) ways, divided by (7 choose 2), again yields 5/ 7. 3) E[X—3Y] =E(X)—3E(Y) = 6035 = 45 (ZptS)
Var[ X — 3Y ] = Var(X) + 9Var(Y)  2*3*COV(X,Y); Becareful of signs.
= 100 + 9*10 — 2*3*12 = 118 = (10.9)2; SD is square root of variance. (4 pts).
Cut off score = E[S] + SDIS] (—1.282) = 31 (4 pts)
4) /  —Male, 0.51 /Murdered, P[M I Male] = 1E4
Kid>  I \  Survive
\  Female, 0.49 /  Murdered, P[M I Female] = 2.3E5
\  Survive
P[Murder] = P[MIMale]P[Ma1e] + P[M I Female]P[Female]
= (1x10'4)(0.51) + (2.3x10'5)(0.49) = 5.1x10‘5 + 1.1x10'5 = 6.23x10’5 (3 pts) P[Male I Murder] = P[M I Male]P[Male]/P[Murder] = 5.1x10'5/6.23x10'5 = 0.819 (3 pts)
P[ next two murders are both males ] = P[Male I Murder]2 = (0.82)2 = 0.671 (2 pts) 5) First calculate arrival rate of Poisson process: 7L = 4/ 10 = 0.4 quakes/ year
a) Quakes in 30year period is Poisson with v = 30?» = 12; Mean = Variance = v = 12 (4 pts)
b) For 5 years, v = 5)» = 2; P[X=2] = vzexp(v)/ 2! = 0.271; Table A2 gives P[XsZ] (4 pts)
c) E[T5] = 5/7. = 12.5 years; Var[T10] = 5/7.2 = 31.3 years2 (4 pts) (subtracted 1 pt if mean and SD given in units of decades and units not speciﬁed)
P[ T5 2 10] = P[ XS 4 arrivals  v = 10(0.4) = 4] = 0.629; Poisson Table A2 with I. = 4.
Or using standard gamma described Devore Section 4.4 with Table A4 
P[T5 2 10] = 1 — F[ t/[3 ; on] = 1 — F[10/(1/0,4);51 = 1 — F[4;5] = 1 — 0.371 = 0.629 (4 pts). CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
First Exam October 12, 1990 Use the text, notes, and your head, to answer the questions below. The exam lasts 50 minutes.
There are 50 points in total, one per minute. 1. (10 points) Consider two independent events A and B where P[A] =0.30, and P[AnB]=0.06
What is the probability of B? What would be the probability that A or B (or both) occurred?
What is the probability that B occurred if it is know that A u B occurred? 2) (8 points) A platform is supported by 10 columns whose loadbearing capacities are normally
distributed with mean 50 tons and standard deviation 4.5 tons. Assume the loadbearing capacities
of the columns are independently distributed and the platform's capacity equals the sum of the
capacities of the ten columns. What design load can an engineer use and be 99% sure the platform
will support that value? 3) (8 points) AIDS is a national concern. Consider the division of individuals into "typical" and
"highrisk" groups representing 99.4% and 0.6% of the population, respectively. Let the incidence
rate in the "typical" group be 2 in 10,000 and the incidence rate in the "highris " group be 1 in 20.
What is the probability a randomly selected individual has the disease? What is the probability an
individual who is discovered to have the disease is from the "typical" group? 4) (8 points) A random variable X only takes on values between zero and one. In that range
P[ng]= x(2—x) 05x51
What is the mean of X? 5) Leak detection is a major concern for older water supply systems. In one city, leak detection
crews discover on average 3 leaks per 5day work week. Assume that the discovery of leaks
corresponds to a Poisson process. a) (5 points) What is the mean and variance of the number of work days until
10 leaks are discovered? b) (5 points) What is the probability that 1 or fewer leaks are discovered during a 4day week? c) (6 points) On any day, either one or more leaks are discovered, or none are discovered.
In that 4day week let K equal the number of days on which one or more leaks are discovered.
What is the mean and variance of K? What is formula for probability that K=k for any 0 g k g 4? 2) 3) 4) 5) b) CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
*** First Exam Solution *** October 12, 1990 Given that for independent events P[AnB] =P[A ] P[ B] = 0.06 and P[A]=0.3, one obtains P[B] =O.2; P[AuB] = P[A]+ P[B]—P[AnB]=0.30+0.20 — 0.06 = 0.44
P[BIAUB] = P[B]/P[AUB] = 0.2/0.44=0.45 B andA UB are not independent. X = strength of ten colums; these are independent random variables so means and variances add; furthermore, each strength is normal so sum of strengths is normal. Thus
x ~ N[ 10*50, 10*(4.5)2] = N[ 500, (14.2)2 ] 0.01 = Pr[ X g c ] = Pr[ (X — 500)/14.2 s (c — 500)/14.2 = —2.326] c = 500 — 2.33(14.2) = 467 tons < 500 tons! P[disease] = 0.994(0.0002) + 0.006(0.05) = 0.0002 + 0.0003 = 0.0005
P[ typical l disease ] = P[ typical and disease ] / P[disease] = 2/5 = 0.4 For QDF Fx(x)=x(2—x); thepﬂis fx(x) = 2(1—x) forO g x g 1 1 l
E[X] =1 xfx(x)dx =1 2x(1—x) dx=[x2—2x3]1 = 1/3
0 0 First calculate arrival rate of Poisson process: 7» = 3/5 = 0.6 events/working—day
Time to tenth arrival T10 has a distributed Gamma[ on = 10, B = 1M] E[Tlo] = 10/)» = 16.67 days; Var[TlO] = 10/)»2 = 27.78 days2 Number of leaks in 4 days is Poisson with v = 4*?» = 2.4 events
P[ 1 or fewer] = P[ K=0 ] + P[ K=1] = exp( —v)[ 1 + v/1!] = 0.31 K ~ Binomia1[n = 4, p]
where p = P[ X 21 Iv = 1 day * A] =1—p[0] =1—exp(—?t)=1—0.5488 = 0.45
E[K] = np = 1.8
Var[K] = np(l—p) = 4(.45)(.55) = 0.99
Formula: P[ K=k] =( E )pk(1—p)n'k = ( : )0.451<(0.55)4k 0 g k g 4
Note that expected number of detects in 4 days is 4(0.6) = 2.4.
However, the expected number of days with one or more detects is 1.8 days. 4 pts
3 pts
3 pts 4 pts 4 pts 4 pts
4 pts 4 pts 4 pts 2 pts
3 pts 1 pts
4 pts 3 pts
1 pts
1 pts
1 pts Hence average number of defects on day with at least one defect is 2.4/1.8 = 1.333. CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
First Exam October 14, 1989 Use the text, your notes, handouts, and what you have learned, to answer questions below.
The exam lasts 50 minutes allowing you one minute per point. 1. (10 points) Three events E1, E2 and E3 are subsets of the sample space S where
P[ E1] = 0.10; P[ E2] = 0.30; P[ E3] = 0.50
P[E1UE2]=0.35; P[E1 0E3] =0; P[E2nE3]=0
What is P[ E1 u E2 U E3 ] ? What is the probability of E2 given that E1 occurred? 2) (4 points) An engineer was told that the strength X of steel rods varies from rodtorod with a normal distribution whose mean is 5000 and whose variance is (200)2. What is the probability that
a randomly selected rod has a strength less than or equal to 4580? 3) (8 points) The concentration C of a pollutant in a river after dilution is a linear combination
C = 0.1 V + 0.6 W
of that V from a municipal waste water treatment plant and that W from a foodpacldng plant. If E[V] = 200, E[W] = 300, Var[V] = 1502, Var[W] = 3002 and the correlation pVW of V
and W is 0.40, what are E[C] and Var[C]? 4) (12 points) A random variable variable has the probability density function (pdf): 2— x Z 1
fX(x) = x3
0 otherwise What is it's CDF? What are the mean and variance of X? (The variance may be a surprise.) 5) (16 points) Prof. Sansalone noticed that among students taking CEE 371, those who took Eng.
116 seem to better than than those who didn't! Supppose that on a particular test problem 70% of
the students who took 116 answer the question correctly, and only 40% of the other students do.
Assume 60% of the class took 116 and 40% did not. (a) What fraction of the class answered the question correctly? (b) If a student answered the question correctly, what is the probability they took 116? (c) Among the 10 students in the ﬁrst two rows that took Eng. 116, what is the probability
that at least 8 got the problem right? (Be careful about what the questions asks.) (d) What are the mean and standard deviation of the number of students from among the 10
who took 116 who will answer the problem correctly? 2) 3) CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
*** First Exam Solution *** October 13, 1989 (Revised 10/94) Given that P[ E1 u E2] =0.35 P[E1UE2UE3] = P[E1UE2]+P[E3] —0 = 0.35+0.5 = 0.85 (because intersection P[ (E1 U E2) n E3] = P[ (E1 0 E3) u ( E2 0 E3 )] = 0)
P[ElnE2] = P[E1]+ P[E2]—P[E1UE2]=0.10+0.30 — 0.35 = 0.05;
thusP[ E2  E1] =P[E10E2]/P[E1]=0.05/0.10=0.5 Pr[ X g 4580 ] = Pr[ (X ~ 5000)/200 5 (4580 — 5000)/200 ]
= Pr[ Z s 2.12] = 0.0179 (See front cover of book.) E[C] = 0.1 E[V] + 0.6 E[W] = 200 Var[C] = 0.12 Var[V] + 0.62 Var[W] + 2(0.1)(0.6) Cov[V,W]
Cov[V,W] = pVW °V OW = 0.4*(150)(300) = 18,000
Thus one obtains Var[C] = 34,785 4) For the density function fX(x) = 1/(2x3) {which should be 2/x3}, one obtains 5) FX(x) =12? = [1x‘2] forleandOotherwise
l
E[X]=] 2.291 = 2
1 s Var[X] = E[X2] — E[X]2 E[XZ] =I 212% = I ;SL = 2[ln(s)]:° = 00; Hence Var[X] = 00
1 1
a) Pr[Correct] = 0.7(0.6) +0.4(0.4) =0.58
b) Pr[ 116 I Correct] = Pr[ 116 and Correct] /Pr[Correct] = 0.7(0.6)/ 0.58 = 0.72
c) The number of correct answers has a Binomial distribution.
Pr[ X _>_ 8 given p = 0.7] = 1  Pr[X _<_ 7 ] (now see Table A.1 in text), yielding
Pr[X_>_8]=10.617 = 0.383
d) EX = np = 10(0.7) = 7
Var[X] = np(lp) = 10(0.7)(10.7) = 2.1
Standard Deviation ofX = ME = 1.45 5 pts 5 pts 2 pts
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 Fall '08
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