CEE_304_2nd_Exams2000_03

CEE_304_2nd_Exams2000_03 - CEE 304 - Uncertainty Analysis...

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Unformatted text preview: CEE 304 - Uncertainty Analysis in Engineering Second Examination November 14, 2003 (r) You may use text, your notes and calculators. There are 50 points in total, one per minute. 1. (7 pts) The sample variance S2 can be used as an estimator of the population variance 62. If one has normal data X,, what are the bias, E{ S2 - 62}, variance Var[Sz] and Mean Squared Error of the estimator $2? How can one reasonably compare different estimators to determine which is best? Explain. 2. (5 pts) Tests demonstrated that the strength of 1,000 foot cables has a Weibull distribution with mean 282.5 and standard deviation 39.5 (corresponding to parameters u = 300 and k = 8); what then would you estimated to be the 10 percentile of 4,090 foot cables made of the same material? 3. (9 pts. - harder) An environmental engineer is interested in the variation in the diameter of pathogenic bacteria found in drinking water. Assuming cells are roughly spherical, and that the cell volumes V have a lognormal distribution with mean E[V] = 125 and a coefficient of variation of 30%. What then are the mean and variance of the diameter of the cells? [Recall for a sphere that Volume = 11D3 / 6 for diameter D.] 4. (5 pts.) A charcoal filter can remove chlorinated hydrocarbons from drinking water. An environmental engineer collected 8 independent samples of water that passed through such a filter. The average of the 8 observed removal rates was i = 99.21% with sample standard deviation 5 = 0.27%. Assuming that the measurements are normally distributed about the true removal rate u, what is a 98% confidence interval for the true u? What is the probability the interval you just computed contains the true population mean? 5. (17 pts) A transportation engineer conducted a series of tests to demonstrate that a new cement mix for highway construction will withstand 25 years of service. She develops a wear index WI to assess the status of a sample after a demanding set of tests. A mix will be acceptable if its wear index has a mean value of less than 20 on a scale of 0-100. Fifteen sample values of the wear index values appeared to be normally distributed with a sample mean of 18.31 and sample stande deviation of 3.09. (a) Specify the null and alternative hypotheses, as well as the rejection region for a hypothesis test to demonstrate that mix does actually meet the target value of 20. Use a type I error of 5%. (b) What P-value do you obtain with this data set? What do you decide? (c) Suppose the actual wear index had a mean of 18.5 for a particular mix and test values have a standard deviation of 2.5; what is the probability future tests conclude that the mix does not meet the standard? (d) If one could vary the sample size, and again the true mean wear index is 18.5, how large would n need to be to ensure that the probability of incorrectly rejecting the mix is less than 7%? (e) With 11 = 15 and the 5% significance test above, for the mix to be accepted at least 97.5% of the time, how small would the actual mean wear index of the mix have to be? 6. (7 points) The distribution of the weight of packages in a given class is thought to have probability density function fw(w) (here 2 is the maximum weight of packages in this class), where: fw(w) = y (0.5)7 w"‘1 0 < w < 2 with E[W x"‘] = y 2‘“ / (7+m) (a) Suppose one had independent measurements, W1, wz . . ., w“, of the weights of different packages. What is the maximum likelihood estimator of the parameter Y? (b) Your boss is puzzled by the MLE estimators you are using to estimate 1. "Why does it make sense?” he asked, "Why not just use the sample average?” Please provide a SHORT 1-paragraph theoretical justification for the uSe of MLEs addressing this concern. CEE 304 - Uncertainty Analysis in Engineering Solutions Second Examination November 14, 2003 1. The bias, E{ S2 — 02} = 0. For normal data: variance Var[Sz] = 20‘/ (ml) where n is the sample size; so for zero bias, the Mean Squared Error of the estimator S2 equals its variance = 20‘/(n-1). 5 pts We would prefer to have zero bias and zero variance, and thus always be right. Zero variance is generally impossible and zero bias often impractical. So if we minimize the average squared error, incorporating bias and variance as errors, we end up with a minimum mean square error criteria. Some insist on zero bias, and then seek estimators with minimal variance (MVUE). 2 pts 2. The original parameters are u = 300, and k = 8. If the cable is made 4 times longer, the longer cables should have a Weibull distribution with a new u equal to 300/ 4“”) = 300/ 1.19 = 252.3 and same k = 8. 3 pts Solving F(x) = 1 — exp[ — (x/ u)" ] yields x = u [-ln(1-F)]”“". So for F = 0.10, k=8, and u = 252.3; one obtains for the longer cable: xmo = 190.4. 2 pts 3. Volume of cells V ~ Lognormal so In V = Y ~ N[ u, 02] where one obtains 0'2 = ln[ 1 + (0.3)2 ] = 0.0862; n = 111(125) - 0.5 0'2 = 4.785 3 pts Now Diameter D = (6V/ 10”” = [(6 / n) e" 10”” = [(6/ n) ] “333 em” = eW where W ~ Ni 11/3 + ln(6/Tt)/3, 0.3333262] = N[ 1.811, 0.00957] 4 pts So E[D] = exp[u/ 3+ln(6/n)/ 3 + 0.5(.33332 02)] = 6.144; Var[D] = E[D] 2 {exp(0.33332 0'2) —1 } = 0.363 2 pts Or compute mean and standard deviation of @333“, and then scale those results by [(6/ n) P3333. 4. (a) For n = 8 with i = 99.21% and s = 0.21%; a 98% Cl is i i 5 tom, HT! ; with tom, = 2.998 yields = 98.92 to 99.50% 4 pts (b) Interval either does or does not contain the true u: probability is one or zero. 1 pt 5. (a) Ho: u = 20; H,: u < 20 Thus must demonstrate clearly that mean is less than 20 ! 2 pts For n = 15, reject Ho if T < - tom“: - 1.761. 2 pts (b) Observe t = — 2.12, P-values is 0.026 ; reject 1-10 with this data set. 2+2 pts (c) d = (20—18.5)/ 2.5 = 0.6. For n = 15, a = 5%,‘d = 0.6: one-sided t test => B = 30%. 3 pts (d) Given on = 5% and want [5 = 7% when d = 0.6, so need d-f = 29 (about), 11 = 30 3 pts (e) Given n = 15, on = 5% and want 0 = 2.5% so must have d = 1 (about) so mix must have mean wear index 20 -— 1 (2.5) = 17.5 3 pts 5) a) We have the pdf fw(w) = y(0.5)Y w“ for 0 s w s 2 ; given a sample { wi } obtain the log-likelihood function: In L(w,, ...,w,,) = In I'[ fw(w) = n 1n 7 + 'y n ln(0.5) + ('y -1) 2i ln(wi ) 2 pts Maximizing: 0 = d(lnL)/ dy = n/ y + n ln(0.5) + E, ln(wi ) , so 2 pts 'Y =— 1/[ ln(0.5) + 2i1n(wi )/n] = -1/[ Ziln(wi/2)/n] > 0 1 pts b) Students need to mention likelihood principle that indicates the best way to understand the information in a sample is by considering the probability of observing the values that occurred. By maximizing the likelihood function we make the fitted distribution as consistent with the data as we can. Why should the simple average be the right statistic to use? It need not be best summary of the data. 2 pts CEE 304 - Uncertainty Analysis in Engineering Second Examination November 15, 2002 (revised) You may use text, your notes and calculators. There are 50 points in total, one per minute. 1. (6 pts) The distribution of the strength of structural members is thought to follow a Weibull distribution with mean 230 and shape parameter k = 16. What design value can an engineer be 99% certain the strength of a member will exceed? { numbers: I‘(1.0625) = 0.9676 ; I‘(1.125) = 0.9417; I‘(1.25) = 0.9064 } 2. (8 pts.) A high school teacher uses a tape to measure out both sides of a square play field. The tapes are cheaply made and as a result the length of a tape has a lognormal distribution with mean 25 m and standard deviation 1.2 111. What then are the mean and standard deviation the area of the square? 3. (9 pts.) Prof. Rachel Davidson is studying the effect of hurricane winds on electric utility distribution systems. Based upon a sample of 23 damaged utility poles in South Carolina, the average cost to replace a damaged pole was $882 with a standard deviation of $335. (a) Use this data to construct a 95% confidence for the true average cost to replace a damaged pole. (b) What is the probability the true average cost it to replace a damaged pole is in the particular interval you just constructed (assuming you did it correctly)? (c) Prof. Davidson plans to construct such intervals for 12 MORE states. What is the probability that all 12 of the intervals SHE WILL CONSTRUCT will contain the means u, for which they are estimators? 4. (17 pts) Friday Dr. Ed Calbrese gave a seminar on hormesis, which is a theory that while high doses of many compounds will harm a plant or animal, very low doses provide some benefit reflected in increased growth or immunity. To test that theory several professors applied small doses of a pesticide to different colonies of microorganisms and measured any increase in growth relative to the know growth rate of the controls. They found, based upon 10 samples, that the average response was 113% with a sample standard deviation of 18%. While the observed responses varied widely, they were well described by a normal distribution. (a) What is the rejection region for a hypothesis test to demonstrate that hormesis does in fact occur at these low doses (and thus response exceeds 100% )? Use a type I error of 0.01. What P—value do you obtain with this data set? What do you decide? (b) What hypotheses did you select for the test in part (a), and WHY? W (c) If in fact the true mean response really is 112%, wherein the true standard deviation is 15%, what would be the probability that one incorrectly concludes the pesticide provided no stimulation on average? (d) For the test in (a), if the true mean response is exactly 100%, with a standard deviation of 15%, what is the probability you conclude incorrectly that a stimulus occurred? (e) If the means under the null and alternative hypotheses are 100% and 112%, and the true standard deviation is 15%, how big a sample size is needed to keep the type II error below 0.10 for a = 0.01? 5. (10 pts) Rainfall exceedances R above a threshold are well described by the probability density function fR(r) =(4/ b‘)(r - b)3 for 0 s r s b (a) Given independent measurements, denoted r1 , r2 .. ., rn , of rainfall depths above the threshold, what nonlinear equation has as its solution the maximum likelihood estimator of b? [You don’t need to solve it] (b) Give an example of a distribution for which the MLEs and method—of-moments estimators of the parameter(s) Meme. What are the estimators? (c) Give an example of a Hamster distribution for which the MLEs and methodpf-moments estimators of the parameter(s) are Mm. What are the estimators? (Or what equation must be solved?) CEE 304 - Uncertainty Analysis in- Engineering Second Examination November 15, 2002 (revised) 1. Mean of Weibull distribution given by u = u I‘(1 + 1 / k) = 0.9676 11 => u = 237.21. 2 pts Need 1 percentile: xo‘01 = u [-ln(1-p)]“k = 178.31 (Note, 99 percentile equals 261.51) 4 pts 2. Let length of tape be X, where X ~ Lognormal, 1n X = U ~ N[ u, 02] 0’2 = ln[ 1 + (1.2 / 25)2 ] = 0.00230; 0 = 111(25) -- 0.5 02: 3.218 3 pts Now Area = X2 = [eU ]2 = e”, where 2U ~ N[ 211, 22 62] = N[ 6.435, 0.00921] 3 pts Thus E[A] = exp[ 2;; + 0.5 (402) ] = 626.4; StDev[A] = E[A] sqrt{exp(4oz) —1 } = 60.2 2 pts 3. (a) n = 23 with x = $882 a: s = $335; 95% CI = x a: s to_05,n_1/fi;t= 2.074 = $737.1 to $10269 5 pt (b) Interval either does or does not contain the true it: probability is one or zero. 2 pts (c ) That all 12 intervals that will be created contain u, p = (0.95)12 = 0.54 2 pts 4. (a) For n = 10, reject Ho if T > twig: 2.821. Observe t = 2.284, P—values is 0.024 ; accept Null hypothesis with this data set. 2+2+2 pts (b) H0: u = 100; Ha: u > 100 Thus must clearly prove response exceeds 100%. 2 pts (c) d = (112—100)/ 15 = 0.8; one-sided, on = 1% => B = 0.48 3 pts (d) This is just the type I error on = 1% 3 pts (e) d = (112—100)/ 15 = 0.8, B = 10% => need n = 30 (or a few less) using curves 3 pts 5) a) MLE: given pdf fR(r) =(4/b4)(r - b)3 for 0 s r s b and sample { ri}, obtain likelihood function: L(r) =IIER(r) ->lnL(r)=nln4~ 4n1nb +32ilnlri-b]. 3pts Maximizing: 0 = d(1nL)/dr = 4n/b + 3 2i 1/[ ri - b] , and that is as far as one can get, except perhaps to write b = 0.75 n/ 2i 1 /[ ri - b] 2 pts b) Answers vary. Poisson was used as an example in class. Exponential is another. The normal distribution is considered to be an example. Use of n-1 in denominator of S is not really important. 2 pts c) Answers vary. Lognormal was in lecture notes. 2-parameter Gamma was on homework. Gumbel and Weibull and 2-parameter exponential are examples: is that enough? 3 pts CEE 304 - Uncertainty Analysis in Engineering Second Examination November 16, 2001 You may use text, your notes and calculators. There are 50 points in total, one per minute. 1. (5 pts) Devore expresses a fondness for minimum variance unbiased estimators (MVUE). Why is the MVU E the best (or is it not the best) criterion for selecting point estimators? What would be an alternative and reasonable criterion? Provide an example that illustrates the relative advantage or disadvantage of the alternative criterion you mentioned relative to the MVUE criterion? 2. (10 pts) Prof. Stedinger and his students have been studying the concentration of cryptosporidium oocysts in natural waters used for drinking water. One source of these microbiological pathogens are wastewater treatment plants. Assume the NATURAL logarithm of the number of oocysts released per day from a water treatment plant is normally distributed with a mean and standard deviation (in natural logs) of 14.0 and 1.10. These are mixed with the natural flow of the river which is lognormally distributed with a mean of 1,000 cubic meters per day, and a coefficient of variation of 0.8 . The two are independent. What then in the mean and standard deviation of the concentration of oocysts in the water after the oocysts are released into and mixed in the stream? 3. (25 pts) On Nov. 1 under the Bush administration, EPA adopted new standards for Arsenic in drinking water equal to a maximum of 10 ppb. EPA estimates that 5% of water systems will need to install treatment to meet this standard. Susan is an environmental engineer with a county health department and has collected 7 water samples from a local drinking water utility (over the last two months); her job is to see if the data demonstrates that the mean concentration of Arsenic exceeds the 10 ppb standard. Assume that sample values are normally distributed about the true mean. 0 The data yield a sample average of 14.36 ppb with a sample standard deviation of 4.76 (n = 7). (a) Construct a 90% confidence interval for the true mean concentration of Arsenic in the water. What is the probability the true mean concentration in the interval you just computed? (b) What are the appropriate null and alternative hypotheses for Susan? (c) What is the rejection region for a test with a type I error of 1% ? (d) What is the t-value and the P-value for this data set? What do you conclude? (e) If the true concentration of Arsenic in the water is 14 ppb, and the true standard deviation of the measurements from sample—to—sample is 5 ppb, what is the probability the engineer will incorrectly conclude that the water meets the new EPA standard of 10 ppb? (f) If the true mean is 14 ppb and measurements have a standard deviation of 5 ppb, how many samples should be collected to ensure that the probability is 95% that one correctly concludes the water violates the EPA standard using the test in (c). 4. (10 pts) A structural engineer is considering the a new distribution for maximum crack lengths. The distribution has cumulative distribution function, 7x / (1 + 7x) for 0 g x , with probability density function y / [ 1 + 7x ]2. (a) Given independent observations of maximum crack lengths {xi | i=1,...,n }, what nonlinear equation has as its roots the maximum likelihood estimator of 7? [Don't solve it!] (b) The method of moments completely fails in this case. Can you see why? (c) Why is it that statisticians so often prefer MLEs to alternative estimators? CEE 304 - Uncertainty Analysis in Engineering Solutions Second Examination November 16, 2001 1. Many statisticians, including Devore, like the idea of considering only unbiased estimators — thus the estimator is centered on the true valule. But while s2 is an unbiased estimator of the variance 02, s is a biased estimator of the standard deviation 0 - unbiasness is not a consistent principle. One is also concerned with precision, or imprecision measured often by the variance of an estimator. Often there is a tradeoff between bias and variance. An alternative and reasonable criterion is the mean square error that considers the average squared error, whether it occurs due to bias or due to variance. Why distinguish between the two? As the first lecture example showed, (x+1)/ (n+2) is a biased estimator of p for a binomial distribution, but for most values of p it has smaller mean square error than does x/ n, which is always unbiased. Thus while MVUE criterion would not even consider (x+1)/ (n+2), the MSE criterion appropriately suggests that in most cases it is more attractive than the MVUE x/ 11. Many MLEs are unbiased though the bias is small and they asymptotically have the smallest possible variance. M SE and MVUE are criteria; MLE and MM are methods. 2. A variety of answers and methods showed up here. See lecture example on reliability. Let Oocysts concentration be 0, and natural logarithm ln 0 = W, where W ~ N( 14.0, 1.102 ). 2 pts Let streamflow be Q and its natural logarithm be T. Var(I') = ln[1+Var(Q)/ m; ]= 0.495 and M = ln[uQ] - 0.5* Var(T) = 6.66 3 pts Then Oocysts concentration in stream is O/ Q = exp(W-T) Here W — T = D ~ N[ uw — uT , Var(W) + Var(I') ] = N[ 7.34, 1.705 ]. [note signs] 1 pts E[O/ Q] = exp{ u(D) + 0.5*Var(D) } = 3612; Var(O/ q) = E[O/ q]2 [ exp(Var(D)) -— 1 ] = 5,870,000 3 pts So StDev[O/ q] = 7,660 { CV of concentration 0/ Q is 2.12! ) 1 pts 3. (a) For n = 7, the 90% CI is i i 5 tom,“ / fi = 10.86 - 17.86 where tom’s: 1.943 and n = 7. 5 pts. (b) H,,: u = 10; H“: u > 10 (Thus must prove concentration exceeds (>) 10 ppb.) 2 pts (c) One-sided t-test: reject Ho if «ft—1 (i - 10) / s > No.00“; = 3.143 3 pts (d) for n =7, )2 = 14.36, s = 4.76 -> t = 2.42 ; p is about 3% (more than 2.5%) 5 pts. (e) for u, = 14, o = 5, d = (14-10)/5 = 0.8 => [3 = 70% 5 pts (e) To get [3 = 5% with d = 0.8. need d-of-f = 29, n = 30. 5 pts 4) a) MLE: given pdf fx(x) = 7 / [ 1 + yx ]2 for x > 0 and sample {xi}, obtain the likelihood function: L(c) = H fx(xi) ———> In L(r) = n lny —— 2 2 ln[ 1 + 'yxi ]. 2+2 pts Maximizing: 0 = d(lnL)/dy = n/y— 2 2, xi/ (1 + yxi) , or y: 0.5 n / 2, xi/ (1 + yxi) 2 pts b) Method of moments: note E[X] = ~fyx/ (1+’yx)2 dx —> oo where the limits are from 0 to co. 1 pt This distribution has no finite mean! So how can one use the method of mOments? 1 pts (c) MLEs are based upon sound statistical principles, the are asymptotically as efficient (smallest variance of any other estimator in large sample), they are very flexible, they are invariant to the parameterization of the distribution, and they are widely accepted. I 2 pts CEE 304 - Uncertainty Analysis In Engineeran Second Exam November 10, 2000 You may use text, your notes and calculators. There are 50 points in total, one per minute. 1. (6 pts SHORT answers) a) Is the sample average it an unbiased estimator of u equal to E[X]? b) An engineer is conducting a hypothesis test to demonstrate that urban development increased the magnitude of flood peaks. They obtained a P-value of 4.3%. If they intended to use a test with a type I error of 1% (and a type I] error of 10%), should they with this data accept or reject of the null hypothesis of no change? c) A track student looked up the scores for the fastest times nationally to run 100 meters in each of the last 15 years. What is a particularly appropriate cdf to use to describe the distribution of these 15 values? Why? 2. (5 pts) Alexander noticed that 2x4 beams seem to be narrower than 2 inches. He measured the width of 14 beams and obtained an average width of 1.78 inches and a sample variance of (0.19 inches)2. Construct a 90% confidence interval for the true mean width of such beams from this supplier: assume the variability in widths are independent and normally distributed. What is the probability the true mean width is included in the interval you just computed. 3. (9 pts) Cynthia needed to compute the average travel times for delivery trucks to travel between the warehouse and the sales center in Newville. She took the sum of the times for the last 7 trips, and because she did not want to overestimate the time, she divided that total by 8 to obtain T = (X1+X2+X3+X4+X5+X6+X7)/8. Assume individual travel times Xi have a lognormal distribution with mean 3.6 hours and standard deviation of 0.6 hours. What are the mean, bias, variance and mean square error for T as an estimator of the true mean travel time? Does T have a smaller mean square error than the sample average of the 7 observations? 4. (6 pts) For problem 3, what is the probability the time Xi for a trip takes more than 4.5 hours? 5. (19 pts) Student hassled Prof. Jones because they claim his homeworks takes too much time. Jones collected data to demonstrate that students ON AVERAGE spend less than 7 hours a week outside of class on his course! He persuades 20 students to record how much time they spend studying during a week. The average study time turned out to be 6.41 hours with a sample standard deviation 1.39. (a) What are the appropriate null and alternative hypotheses (b) What is the rejection region of a test that has a type I error of 1%. (c) For the data Prof. Jones collected, what is the t value and the P-value for the test? (d) If the true mean study time is really 6.5 hrs with a standard deviation of 1.25 hrs, what is the probability of accepting the null hypothesis? (e) For the situation in (d), how large a sample would the professor need to have a probability of 70% of demonstrating that the mean was less than 7 hours? (t) What would the true mean study time have to be for the test to have only a 20% probability of incorrectly concluding the mean study time was seven hours using n = 20 observations (assume a = 1.25) ? 6. (5 pts) Consider the following triangular distribution: fx(x)= (2/a)[1-(x/a)l Osxsa Given observations {Xi l i=1,...,n }, our interest in estimation of a. What nonlinear equation has as its root the maximum likelihood estimator? CEE 304 - Uncertalnty Analysis In Engineering Second Exam November 10, 2000 SOLUTIONS (revised 12/01) 1) (a) YES: The sample mean is unbiased: Eli] = u. (b) Accept Ho. p—value = 4.3% > 1% (Ignore type II error given in problem). (c) Weibull: asymptotic distribution for the smallest (shortest) time in each year. Gave 1 pt. for people who said Gumbel, and thus were thinking about the right issue. 2 pts 2 pts 2 pts We want the person with the greatest speed (Gumbel), resulting in the smallest time (Weibull). 2. A 90% CI is i i s t0115M Nfi = 1.69 to 1.87 where tom,13 = 1.771 and n = 14. This interval either does or does not contain the true mean. We do not know which. 3. Cynthia uses as the estimator T where E[T] = 7 u / 8 = 3.15 The bias = {3.15 -— 3.60 } = -0.45 hours. [Estimator is biased] VarIT] = 7 02/ 82 = 0.039 = (0.20)2. MSE = Var + Bias2 = 0.039 + 0.452 = 0.242. If one used the sample average of 7 obs.: Bias=0; Var = MSE = 02/ 7 = 0.05 which is SMALLER than MSE of Cynthia's estimator due to the large bias Cynthia has. The fact that observations are lognormal has no effect on this answers. It enters in #4 below. 4) First get the variance of logarithms, which is b2 = ln [1+(oml/ and? ] = 0.0274 = (0.1655)2; Mean of logs a = lnlu] —- 0.5 b2 = 1.267. Pr[ X > 4.5 ] = 1 - <I>[(ln(4.5) - a)/b = 1.43] = 7.6%. {Note that use of a normal distribution yields: Pr[ W > 4.5 ] =1 — <I>[ (4.5 — 3.6)/ 0.6 = 1.5 ] = 6.7%.} 5) (a) Ho: u = 7; H,: u < 7 (He wants to prove times are less than 7.) (b) One-sided t-test: reject I-I0 if w’fi (5i — 7)/ s < — tom” = —2.539 (c) for n =19, i = 6.41, s = 1.39 => t = - 1.90 ; p-value is 3.6% (d) d = (7 — 6.5)]1.25 = 0.4 Use table to compute Pr[accepting H0 [Ha] = [3 = 60% (e) To get 6 = 30% need 11 = 50 or a few more (approximately) Problem specifies that the power = 1 — t3 = 70%. You need to know the difference. (0 For 0 = 20% with n = 20, one needs d = 0.8; hence u = 7.00 - d*1.25 =-- 6.00 6) MLE: for pdf fx(x) = (2 / a)(1 - x/ a). Then given sample {xi}, obtain likelihood function: L(a) = IIX fx(xi) —-> In L(r) = n in (2) —- n 1n(a) + 21110 - xi/ a) 0 = (d/ da) 1n L(a) = — n/ a + 2, (1 - xi/ a)'1(-xi)(~1/ a2) yields 11 = 2, [ xi/ (a - x9] ; no analytic solution exists. But a solution can be found within [ xmax , co] . Note: the summation in the answer that results from there being many observations xi. 4 pts. 1 pts 2 pts 1 pts 2 pts 2 pts. 2 pts 0 pts 2 pts 2+2 pts 3 pts 2 pts 2+1+1 pts 2 pts 4 pts 4pts 2 pts 2pts ...
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This note was uploaded on 02/02/2008 for the course CEE 3040 taught by Professor Stedinger during the Fall '08 term at Cornell.

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CEE_304_2nd_Exams2000_03 - CEE 304 - Uncertainty Analysis...

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