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University of Florida - ESI - 6321
2.10) For a more technical approach, see the solutions.3.14.a) incorrect SD (-2)3.14.b) right approach but incorrect answer because of part a (-2)
University of Florida - ESI - 6321
ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 1 Spring 2008Reading material: Chapters 2, 3 and 4 from Bertsimas & Freund (Prework coursepack).Problems: 2.2, 2.6, 2.10, 2.17, 2.32, 3.6, 3.14, 4.4, 4.10, and 4.13.Due date: December 23, 2007
University of Florida - ESI - 6321
100CHAPTER 2Fundamentals of Discrete ProbabilityDana Meseroll thought things over. She felt very uneasy about the possibility of installing the G-LAN, having it not perform up to specifications, and suffering the embarrassment of having to rebuild it,
University of Florida - ESI - 6321
3.9Exercises141pate in class. However, before she goes to bed, she must start and then complete a homework assignment which is due tomorrow morning. According to her experience, the time it takes her to complete a homework assignment for this class is
University of Florida - ESI - 6321
190CHAPTER 4Statistical Sampling(d) K11allenge) The company's primary concern is to avoid underestimating the average amount claimed by its policyholders. Determine a value b so that the insurance company is 95% confident that the average amount claime
University of Florida - ESI - 6321
ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 1 Spring 2008 SolutionsExercise 2.2(a) A total of 8 outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT. (b) P(HHT) = 1/8. (c) P(First two tosses are heads) = 2/8. (d) P(Two heads in a row) =
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-02-02.xmcdpage 1 of 12.2 We toss a coin three times. Let the outcome of this experiment be the sequence of heads (H) and tails (T) resulting from the three tosses. (a) Enumerate all of the possible outcomes of this experiment. Li
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-02-06.xmcdpage 1 of 22.6 An athletic footwear company is attempting to estimate the sales that will result from a television advertisement campaign of its new athletic shoe. The contribution to earnings from each pair of shoes so
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-02-10.xmcdpage 1 of 12-10. A hardware store has received two shipments of halogen lamps. The first shipment contains 100 lamps, 4% of which are defective. n1 := 100 p1 := 4% D1 := p 1 n 1 D1 = 4 The second shipment contains 50 la
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-02-17.xmcdpage 1 of 32.17 In a particular town there are two automobile rental agencies that offer different prices for a weekend out-of-state automobile rental. An automobile rental at Express Car Rentals (ECR) costs $195 and in
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-02-32.xmcdpage 1 of 22.32 It is estimated that f := 80 % of all customers at Speedy Pizza prefer thin crust pizza to thick crust pizza. The remaining g := 20 % prefer thick crust pizza to thin crust pizza. Of the many orders rece
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-03-06.xmcdpage 1 of 23.6 Winter lasts from December 21 through March 21. The average winter temperature in Boston is Normally distributed with meanB:= 32.5degrees Fand standard deviationB := 1.59degrees FIn New York City,
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-03-14.xmcdpage 1 of 13.14 Wenjun Chen and Tim Schwartz have analyzed the sport of five-pin bowling, which is the leading participant sport in Canada. After studying scores from an Ontario league, they concluded that the logarithm
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-04-04.xmcdpage 1 of 24.4 A company manufacturing stereo equipment claims that their personal CD player can be used for approximately t := 8 hr of continuous play when used with alkaline batteries. To provide this estimate, the co
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-04-10.xmcdpage 1 of 14.10 During a local election between two candidates, exit polls based on a sample of n := 400 voters indicated that pbar := 54 % of the voters supported the incumbent candidate. (a) Construct a 98% confidence
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-04-13.xmcdpage 1 of 14.13 A potting company is trying to estimate the average age of a Boston resident. the company woutd tike to be 95% confident that their estimate witt be within L := 3 yearsof the true mean of the poputation
University of Florida - ESI - 6321
Time Slot Morning Afternoon Prime Time Late EveningCost of Advertisement in Dollars per Minute $120,000.00 $200,000.00 $400,000.00 $150,000.00Estimated Number of Viewers 1000000 1300000 3200000 800000
University of Florida - ESI - 6321
0.36 Wins Point 0.60 Loses Point 0.40In Play 0.60Not in Play 0.40 Hard Serve0.45 Wins Point 0.50 Soft Serve In Play 0.90 Not in Play 0.10 Loses Point 0.50
University of Florida - ESI - 6321
ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 2 Spring 2008Reading material: Winston, Chapter 13, Sections 13.4 and 13.5 (Coursepack II), as well as course slides.Problems: 13.2-10, 13.4-2, 13.4-10 (read text below "Group B" heading), 13.4
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-632 1-13-02-1 O.xmcdpage I of 213.2-10 My current income isI := 40000dollarsI believe that I owe T:= 8000 in taxes. For C:= 500 dollars dollars92= (currentincome) - (taxes) - (payment toI can hire a CPA to review my tax return;
University of Florida - ESI - 6321
ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 2 Spring 2008 SolutionsExercise 13.2-10Hiring a CPA has an expected utility of .2u(40,000-4,000-500) + .8u(40,000-8,000-500) = .2(35,500)1/2 + .8(31,500)1/2 = 179.7 while not hiring a CPA has a
University of Florida - ESI - 6321
$3000 $3500 Purchase Machine 1 $2000 + 10($150) $4000 $3000 + 10($100) 0.4 $3000 + 10($0) 0.2 $3000 + 10($100) 0.4 $3000 Repairer Does Not Evaluate $3000 + 10($0) 0.4 $3000 + 10($100) 0.4 $3000 + 10($200) 0.2 $4000 $3000 + 10($200) 0.4 $5000 $4000 0.2($30
University of Florida - ESI - 6321
12.6.4 A data set has n x y xy x2 y2 = = = = = = 25.000 1,356.250 -6,225.000 -738,100.000 97,025.000 10,414,600.000Compute the analysis of variance table and calculate the coefficient of determination R2. 1 0 SSE SST SSR MSR 2 MSE F critical value p R2 =
University of Florida - ESI - 6321
UF-ESI-6321 UF-ESI-6321-12-01-02.xmcd page 1 of 212.1.2 Suppose that the purity of a chemical solution y is related to the amount of a catalyst x through a linear regression model with0 := 123.0 1 := - 2.16 and with an error standard deviation := 4.1 :=
University of Florida - ESI - 6321
12.1.2 (a) (b) (c)y = 123.0 + ( - 2.16) 20 = 79.8 - 2.16 10 = -21.6 y = 123.0 + ( - 2.16) 25 = 69 60 - 69 P( Y < 60) = P Z < = P ( Z < -2.20 ) = 0.0139 4.1 (d)y = 123.0 + ( - 2.16) 40 = 36.6 40 - 36.6 30 - 36.6 P( 30 < Y < 40 ) = P <Z< = P( - 1.61 < Z
University of Florida - ESI - 6321
12.1.2 Suppose that the purity of a chemical solution y is related to the amount of a catalyst x through a linear regression model withand with an error standard deviation(a) What is the expected value of the purity when the catalyst level is(b) How mu
University of Florida - ESI - 6321
12.6.4) incorrect table because of computational mistakes (-5)
University of Florida - ESI - 6321
ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 3 Spring 2008Reading material: Hayter, Chapter 12, Sections 12.1-12.7, 12.10 (Coursepack II), as well as course slides 1-80. Problems: 12.1.2, 12.2.2, 12.3.1, 12.4.2, 12.5.1, 12.6.4, 12.11.12(c,
University of Florida - ESI - 6321
ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 3 Spring 2008 SolutionsExercise 12.1.2(a) Since x=20, we have E (Y ; x ) = 0 + 1x = 123.0 + ( -2.16 20) = 79.8 (b) y = 1x = -2.16 10 = -21.6 (c) Y1 ~ N 123.0 + ( -2.16 25), 4.12 = N 69, 4.12(
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-01-02.xmcdpage 1 of 212.1.2 Suppose that the purity of a chemical solution y is related to the amount of a catalyst x through a linear regression model with0 := 123.0 1 := -2.16and with an error standard deviation := 4.1 :=
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-02-02.xmcdpage 1 of 112.2.2 A data set has n := 20x := 8.552 y := 398.2 xy := 216.6 x_squared := 5.196 y_squared := 9356Calculate1 :=n xy - x y n x_squared - ( x )2= 30.1010 :=yn- 1 xn= 7.039_squared :=y_squared
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-03-01.xmcdpage 1 of 212.3.1 In a simple linear regression analysis with n := 18 data points, an estimate1 := 0.522is obtained with SE_1 := 0.142 (a) Calculate a two-sided := 1 - 99% = 1 %confidence interval for the slope p
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-04-02.xmcdpage 1 of 112.4.2 In a simple linear regression analysis with n := 17 data points, the estimates0 := 12.08 1 := 34.60 _squared := 17.65are obtained. If Sxx := 1096 xbar := 53.2 construct a two-sided := 1 - 95% = 5
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-05-01.xmcdpage 1 of 112.5.1 In a simple linear regression analysis with n := 17 data points, the estimates0 := 12.08 1 := 34.60 _squared := 17.65are obtained. If Sxx := 1096 xbar := 53.2 construct a two-sided 95% prediction
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-06-04.xmcdpage 1 of 112.6.4 A data set has n := 25x := 1356.25 y := -6225 xy := -783100 x_squared := 97025 y_squared := 10414600Compute the analysis of variance table and calculate the coefficient of determination R2.1 :=n
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-11-12.xmcdpage 1 of 112.11.12 Are the following statements true or false? (c) The standard fitted regression line is obtained from a least squares fit. TRUE. As state on page 542, "The fitted line minimizes the sum of squared
University of Florida - ESI - 6321
Depth (feet) x 5000 5200 6000 6538 7109 7556 8005 8207 8210 8600 9026 9197 9926 10813 13800 14311 137498Cost ($thousands) y 2596.80004882812 3328.00000000000 3181.10009765625 3198.39990234375 4779.89990234375 5905.60009765625 5769.20019531250 8089.500000
University of Florida - ESI - 6321
ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 4 Spring 2008Reading material: Hayter, Chapter 12, Sections 12.1-12.7, 12.10 (Coursepack II), as well as course slides 1-80. Problems: 12.2.4, 12.3.3, 12.4.4, 12.5.3, 12.6.5, 12.7.1, 12.11.11. R
University of Florida - ESI - 6321
ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 4 Spring 2008 SolutionsExercise 12.2.4Cost is the dependent variable, y, and Depth is the explanatory variable, x. We need y , x , i xi , i y i , i xi y i , i xi2 , i y i2X Depth (feet) 5000 5
University of Florida - ESI - 6321
dt( -2.145 , 14) = 0.047 dt( 2.145 , 14) = 0.047 pt( -2.145 , 14) = 0.025pt( 2.145 , 14) = 0.975
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-02-04.xmcdpage 1 of 312.2.4 Oil Well Drilling Costs Estimating the costs of drilling oil wells is an important consideration for the oil industry. DS 12.2.1 contains the total costs and the depths of 16 offshore oil wells loca
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-03-03.xmcdpage 1 of 312.3.3 Oil Well Drilling Costs Consider the data set of oil well drilling costs given in DS 12.2.1. dollar := amp for unit handling 5000 5200 6000 6538 7109 7556 8005 8207 ft x := 8210 8600 9026 9197 9926
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-04-04.xmcdpage 1 of 212.4.4 Oil Well Drilling Costs Consider the data set of oil well costs given in DS 12.2.1. dollar := amp for unit handling 5000 5200 6000 6538 7109 7556 8005 8207 ft x := 8210 8600 9026 9197 9926 10813 13
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-05-03.xmcdpage 1 of 312.5.3 Oil Well Drilling Costs Consider the data set of oil well costs given in DS 12.2.1. dollar := amp for unit handling 5000 5200 6000 6538 7109 7556 8005 8207 ft x := 8210 8600 9026 9197 9926 10813 13
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-06-05.xmcdpage 1 of 312.6.5 Oil Well Drilling Costs Consider the data set of oil well costs given in DS 12.2.1. dollar := amp for unit handling 5000 5200 6000 6538 7109 7556 8005 8207 ft x := 8210 8600 9026 9197 9926 10813 13
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-07-01.xmcdpage 1 of 312.7.1 Oil Well Drilling Costs Consider the data set of oil well costs given in DS 12.2.1. dollar := amp for unit handling 5000 5200 6000 6538 7109 7556 8005 8207 ft x := 8210 8600 9026 9197 9926 10813 13
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-11-11.xmcdpage 1 of 512.11.11 Gold Extraction from Unprocessed Ore Consider the data set in DS 12.11.9 from the mining industry that shows the amount of ore processed in various locations together with the resulting amount of
University of Florida - ESI - 6321
University of Florida - ESI - 6321
Depth (feet) 5000 5200 6000 6538 7109 7556 8005 8207 8210 8600 9026 9197 9926 10813 13800 14311Cost ($1000) 2596.8 3328 3181.1 3198.4 4779.9 5905.6 5769.2 8089.5 4813.1 5618.7 7736 6788.3 7840.8 8882.5 10489.5 12506.6
University of Florida - ESI - 6321
Ore Processed (kilotons) 7.3 9.1 10.2 11.5 13.2 16.1 18.4Gold Obtained (oz) 8.9 11.3 10.6 11.6 12.2 15.7 17.6
University of Florida - ESI - 6321
x [Depth (feet)] 5000 5200 6000 6538 7109 7556 8005 8207 8210 8600 9026 9197 9926 10813 13800 14311 137498 n = 12.2.4 (a)y [Cost ($1000)] 2596.80004882812 3328.00000000000 3181.10009765625 3198.39990234375 4779.89990234375 5905.60009765625 5769.200195312
University of Florida - ESI - 6321
ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 5 Spring 2008Reading material: Hayter, Chapter 12, Section 12.8; Chapter 13, Sections 13.1-13.4 (Coursepack II), as well as course slides 81-112. Problems: 12.8.5, 13.1.1, 13.2.3, 13.4.2a,b, 13.
University of Florida - ESI - 6321
ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Homework 5 Spring 2008 SolutionsExercise 12.8.5To obtain the linear model, the logarithmic transformation is used in the following way:ln( yield ) = ln 0e 1 (temprature ) = ln ( 0 ) + ln e 1 (tempratur
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-12-08-05.xmcdpage 1 of 212.8.5 An experimenter has data on the yield and the temperature of a chemical process and wishes to fit the model yield = 0 e1 temperatureA linear regression model is fitted to the data y = ln( yield )
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-13-01-01.xmcdpage 1 of 313.1.1 The multiple linear regression model y = 0 + 1 x1 + 2 x2 + 3 x3 is fitted to a data set of n := 30 observations. The total sum of squares is SST := 108.9 and the error sum of squares is SSE := 12.4
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-13-02-03.xmcdpage 1 of 713.2.3 Oil Well Drilling Costs Consider again the problem of estimating the costs of drilling oil wells, which was originally discussed in Problem 12.2.4. The data set given in DS 13.2.3 contains the varia
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-13-04-02.xmcdpage 1 of 413.4.2 Oil Well Drilling Costs Consider the modeling of oil well drilling costs described in Problem 13.2.3 and the data set in DS 13.2.3. Suppose that a model is used with cost (y) as the dependent variab
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-13-06-01.xmcdpage 1 of 613.6.1 Consider the data set in DS 13.6.1. This worksheet will use Mathcad and the matrix methods introduced later in the chapter by inserting a value of 1 for x<0> .1 1 1 1 1 x := 1 1 1 1 1 10.500000000
University of Florida - ESI - 6321
UF-ESI-6321UF-ESI-6321-13-06-08.xmcdpage 1 of 213.6.8 The multiple linear regression model y = 0 + 1 x1 + 2 x2 is fitted to a data set of n := 30 observations. (a) Suppose that1 := -45.2se_1 := 39.52 := 3.55se_2 := 5.92Is it clear that both x1 and
University of Florida - ESI - 6321
Depth (feet) 5000 5200 6000 6538 7109 7556 8005 8207 8210 8600 9026 9197 9926 10813 13800 14311Geology 319 241 268 345 549 552 417 568 618 767 675 504 639 641 953 1165Downtime (hours) 0 160 121 48 120 608 391 471 80 140 51 137 0 158 0 56Rig-index 0.62
University of Florida - ESI - 6321
x 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5y 5.9 8.9 7.1 10.1 15.9 14.8 23.3 28.2 31.3 44.8 49.1