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**Unformatted text preview: **CEE 304 - UNCERTAINTY ANALYSIS IN ENGINEERING
PRELIM #1 October 4, 2000 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 two events denoted W and H where
P[W] =0.6, P[H] =02, P[HU W] =r-0.7
(a) What is the probability of both H and W occurring? (b) What is the probability of W not occurring at the same time that H does occur?
(c) If one has observed that H occurred, what is the probability W occurs? Show work. 2) (8 pts) A student is trying to assemble a ﬂashlight becauSe of a power failure. Their younger brother
took the three matching ﬂashlights apart. There are now 3 bulbs, 3 ﬂashlight bodies, and 5 batteries.
Two of the three bulbs work, and 4 of the 5 batteries work. The bodies all work. If one bulb, one body, and
2 batteries are selected at random, what is the probability the ﬂashlight when assembled will fail?
How many different (unique) pairs of batteries can the student select? Show work. 3. (8 pts) Professor Stedinger is scoutmaster of Boy Scout troop 2 in Ithaca. The troop is planning its
annual fall fund raising drive: selling popcorn. Suppose 2 patrols will in expectation sell $600 and $800
worth of popcorn, with standard deviations of $150 and $200, respectively. The correlations among the
amounts each patrol sells is 0.60. What total value of popcorn (the sum of that sold by both patrols) can
Scoutmaster Stedinger be 95% certain the patrols will sell. (Use a normal distribution.) 4. (13 pts) In the Western United States in August, major forest ﬁres in a dry year start somewhere at the
average rate of two every 5 days. (a) Brieﬂy, why might a Poisson Process be a good model of the occurrence of such ﬁres?
(b) What is the probability a whole week (7 days) goes by without a fire starting? (c) What is the probability of having 2 or more fires begin between August 12 & August 18 ? (7 days)
(d) Starting on August 2nd, what is the mean and standard deviation of the time until 8 fires start? 5. (8 pts) The probability density distribution function a random variables Q is:
fQ(q) = =(3-q)/4 Osqu,
0 otherwise
Let E = 00-5. What are the probability density function and the cdf for E? 6. (7 pts) Sally is in charge of checking the tires on the corporate ﬂeet of SUVs. Careful inspection
reveals that 60% of the FireBridge and 5% of the GoodMonth tires have problems, the two sources for
tires. If one third of the tires are made by GoodMonth, then overall what fraction of the tires have
problems? Across the entire ﬂeet, what fraction of the tires with problems were made by FireBridge? CEE 304 - UNCERTAINTY ANALYSIS IN ENGINEERING
SOLUTIONS for Prelim #1 October 4, 2000 Axioms of Probability (Draw Venn diagram. Makes the problem easy. Many people missed key points.) 1a)P(WUH)=P(W)+P(H)-P[WﬂH] => P[WﬂH] =0.10 2pts 1b) P[W' 0H] =0.1 [Lookatdiagram] 2pts
01‘ P(H) = P(WﬂH) + P(W'ﬂH) => 0.2 = 0.1 + P(W'ﬂH) => P(W'ﬂH) = 0.1 1c) P[ W l H] = P(WﬂH)/P(H) = 0.1/0.2 = 0.5 2 pts Counting (trouble here) 2a) Compute ﬁrst probability of non—failure:
Pr(Works) = (2/3) (1/1) (4/5) (3/4) = 2/5 = 0.4 [Drawing without replacement] 5 pts
Hence Pr(Failues) = 1— Pr(Works) = 0.6 1 pts
To work one must have one of 3 good lights, and must have selected
ﬁrst one of 4 good batteries from 5, and then one of 3 good batteries from remaning 4.
Equivalently: Pr(Works) = C(2,1)C(3,1)C (4,2)/{C(3,1)C(3,1)C(5,2)} = 2/5 = 0.4
Note: (4/5) (3/4) = C(4,2)/C(5,2) = Perm(4,2)/Perm(5,2) = 3/5 2b) Combinations(5,2) = C(5,2) = 5*4/2 = 10. 2 pts
Normal calculations 3a) S = P1 + P2. Sum of RVs. E[S] = 600 + 800 = 1,400 2 pts
Var[S] = 1502 + 2002 + 2(0.60)(150)(200) = 98,500; StDev[S] = 313.9 3 pts
5005 = E[S] —— 1.645 Stdev[S] = 1400 — 1.645*313.9 = $884 3 pts Poisson process -- Gamma and Poisson distributions (Straightforward)
4a) Poisson process is reasonable because ﬁres arrive separately, with p = hAt, we can assume
that arrival rate 7» is constant across August, and beginning of ﬁres in distant places is likely to be
independent of one another, once we have conditioned upon it being a dry year in the West. 3 pts
4b) With )t = 2/5 ; Pr(K = 0 for t :7) = exp(-)tt) = 0.0608 3 pts
40) Pr(K 2 2) = 1—Pr(K = 0 or 1) = 1 - exp(-)tt) - (M) exp(-}tt)/1! = 1—0.06—0.17 = 0.770 3 pts
4d) Arrival time to 8th is gamma with a = 8, B = ll)» :25. y = 043 = 20 days; 02 = 0:02 = 50 ; o 2 St, [291 = LQZ days [Remember units] 4 pts Pdfs, Moments and Independence 5) Compute FQ(q) = (6q — q2)/8 for o s q s 2 3 pts.
cdf:FE(e)=Pr{Ese}=Pr{Q0-5se}=Pr{Qse2}=(6e2-e4)/8, Oses 1.414 3pts
pdf: fE(e) = (12c - 4e3)/8 = (3e - e3)/2 o s e s 1.414 [Range is important] 2 pts 0R: fE(e) = fQ( q[e] ) | dq/de | for l—to—l-monotone funcs => {(3 - e2)/4}*(2e) = (3e - e3)/2 Bayes Theorem / ---—Bad (0.60)
/ —-—Firebridge (2/3)~----0-----Okay (0.40)
——————— —-°Goodmonth (1/3) —-—0———-—Bad (0.05) \ —---Okay (0.95)
6) Fraction bad tires: P(Bad) =P(BTlFireBdge)*P(HreBdge) + P(BadIGoodmth) *P(Goodmth)
= 0.60(2/3) + 0.05(1/3) = 0.4 + 0.01667 = 0.41667 ‘ 4 pts Bayes Theorem: PI Firebdge l Bad 1 = P(Bad l Firebdge)*P(Firebdge)/P(Bad) = 0.96 3 pts WS/ 2590 :0) Agﬁvm H m we .5 w mo wig ma H md o macros—E “£980 5:33—ch No
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- Fall '08
- Stedinger
- Normal Distribution, Probability distribution, Probability theory, pts, probability density function