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Course Number: ESI 6321, Spring 2008

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-!; UF ESI 6321 Pre- Wark Assignment Due August 17, 2007 Luther Setzer 5 7 iz ) zJ / t/" .> ( / :r L f J 3( l---' (,./ Pre- J.'~ P( I rf JJ:-, J o-rl J,"<"- o.J{ '1' +. j) .~0 I~,". / '{'( (/ -- . UF-ESI-632I 7/11/2007 2-1. A four-sided die is engraved with the numbers 1 through 4 on its four different sides. Suppose that when rolled, each side (and hence each...

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ESI -!; UF 6321 Pre- Wark Assignment Due August 17, 2007 Luther Setzer 5 7 iz ) zJ / t/" .> ( / :r L f J 3( l---' (,./ Pre- J.'~ P( I rf JJ:-, J o-rl J,"<"- o.J{ '1' +. j) .~0 I~,". / '{'( (/ -- . UF-ESI-632I 7/11/2007 2-1. A four-sided die is engraved with the numbers 1 through 4 on its four different sides. Suppose that when rolled, each side (and hence each number) has an equal probability of being the bottom face when it lands. We roll two such dice. Let X be the sum of the numbers on the bottom face of the two dice. (a) What is the probability that X is at least five? Listing the possible combinations of the two dice: I 1 12 13 14 2I 22 23 24 3I 32 33 34 41 42 43 44 This yields 16 possiblities. Ten have the sum of five or greater. Thus, 10116 = 5/8 .,/* (b) How does your answer to (a) change if you are told that the bottom face of the first die has the number "3" on it? If the bottom number is three then the first die that means the only combinations you can count are the ones that have 3 as the first number. ... l I oJ' ( 3/16 1 I' J :, I I (c) How does your answer to (a) change if you are told that the bottom face ofone of the dice has the number "3" on it? I fl' This Would~be 3116 for the same reason as (b)~ ) Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 ?(A If Or ~) :- peA) + (X 1(6) - f(A~n~)JS) ~. ~vtl.) A ~ 0 g r (M< hM.\..f,..~ P( A .:- f (A) .-t P ( D) f CA a.l l)') : f CA) r (D IA) UF-ESI-632I 7/11/2007 page~ of 1~ 2-5. In Oblako County, any day can be either sunny or cloudy. If a day is sunny, the following day will be sunny with probability 0.60. If a day is cloudy, the following day will be cloudy with probability 0.70. Suppose it is cloudy on Monday. (a) What is the probability that it will be sunny on Wednesday? PTuesdaysunny= 0.30 _ PWednesday_sunny0.39 = ~ ... - (b) What is the probability that it will be sunny on both Tuesday and Wednesday? Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 p L\ r((~r (~~) ~... .., (,.,;J-"~-"'-' \..' ~. 7> &-< (=.:l )J ((~d a,,) ~~) -+ r (v",ii ~ 1 S<.,< c..,.. J. ::.- l( r~~) 'P (SAA~l.r~) -\_ / 7> ( , ) . r (yo.( I,...; 1) D.b' 0, \ .-- - UF-ESI-632I 7/11/2007 2-7. It is a relatively rare event that a new television show becomes a long-term success. A new television show that is introduced during the regular season has a 10% chance of becoming a long-term success. Pregular_success:= 10% A new television show that is introduced as a mid-season replacement has only a 5% chance of becoming a long-term success. Approximately 60% of all new television shows are introduced during the regular season. Pregular_new := 60% What is the probability that a randomly selected new television show will become a long-term success? Psuccess = 8% Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 (i;~J -7 J UF-ESI-632I 7/11/2007 pagS ofl~ 2-10. A hardware store has received two shipments of halogen lamps. The first shipment contains 100 lamps, 4% of which are defective. First:= 100 First defective = 4 The second shipment contains 50 lamps, 6% of which are defective. Second:= 50 Second_defective := p2Second Second defective = 3 Suppose that Emanuel picks a lamp (at random) offofthe discovers that the lamp he purchased is defective. shelfand purchases it, and he later Is the defective lamp more likely to come from the first shipment or from the second shipment? Defective:= First - defective + Second - defective Defective = 7 First defective := -----Defective = 57% V Fraction I Fraction I Second defective Fraction2 := ------Defective Fraction2 = 43% Emanuel has more likely chosen a defective lamp from the first shipment. Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632I 7/11/2007 2-15. An industrial training company that otTers week-long courses to corporations has three instructors on its permanent statT. Instructors := 3 The company receives request for its courses from its many corporate clients. The course fee charged by the company is $20,000 per course. RevenuelnstrUCIOrs 20000 := The company also has a pool of qualified instructors in the local area (predominantly retired business school faculty) that it can draw upon wherever demand for their courses exceeds their supply of permanent instructors. Pool:= 3 Under standardized arrangement, an instructor in the pool receives 55% of the course fee whenever he or she teaches a course. Commission := 55% Revenuepool:= 20000( 100% - Commission) Revenuepool = 9000 The weekly demand for courses obeys the probability distribution given in the table below. n:= 6 i:= 0 .. n x.:= I I Number of Courses x. = I Probability p i' .= o I 2 3 4 5 6 0.05 0.15 0.25 0.25 0.15 0.10 0.05 Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632I 7/11/2007 page10f14 The company will obviously utilize its own instructors to teach courses whenever possible. Pool instructors will be scheduled to teach only if the demand for courses exceeds the number of permanent statT. (a) VeritY that the data in the table corresponds to a probability distribution. All values ofp remain nonnegative and ~)=I and the weekly demand never exceeds n, so yes, it meets the definition of a probability distribution. (b) What is the probability that all of the permanent statTare idle in a particular week? Po = 0.05 (c) What is the probability that all of the permanent statT is busy in a particular week? 6 i=3 L p. = 0.55 I (d) One of the permanent statT is a star teacher. The company schedules her to teach on every possible occasion to maximize the chances of repeat customers. What is the probability that she will be busy in a given week? 6 i= I L p. = 0.95 I Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 o -;f(VO =-J-)~ ":. I L , (5 ,'}....) ._ 2$vO ..~ lJ;) (7 '5" 'I"; ~ , ) ~ riTO f1 ~ , .l v-o a-f I _ - ~ 71'0-0 ...} . J""1l :- S 7 / )1s" i"" ,I C' 5 7 )~ ... ,,~:> :; 7, \"6"0 )'O ':I: '17)--U-O ...~. 0 C S~<,0\) .(1 ) J~(~ P(-r<) ~ +CJ~"f:>C -::75;~ Z~ -rrJ,~ ~ ~ ~ --- --------..... I~' ~ ~ -:: _ )-~o () ,0"5 ,t("' o - ~'O -5f'7o '" ~ ~(1"If' 1501 UF-ESI-632I 7/11/2007 (e) What is the mean and standard deviation of the weekly revenue after deducting payments to pool instructors? Number of Courses Probability Revenue X. I = p i' '= I Revenue. 0 20000 40000 60000 69000 78000 87000 = I o 2 3 4 5 6 0.05 0.15 0.25 0.25 0.15 0.10 0.05 ~:= L i=O 50500 n (p(ReVenuei) ~ = i=O (jx = 22600 t/ Luther Setzer 1 NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ES[-632 [ 7/[ 1/2007 pagejof1~ (f) Adding more instructors to the staff involves an incremental cost of$2,500 per instructor per week. How many instructors, ifany, should the company add to its permanent staff in order to maximize expected profit? Compare three new means against old one given the fixed cost of each new instructor. ~ = 50500 Revenueone_new:= Revenue Revenueone new .- Revenueone new + Revenuelnstructors 4 3 ~_one_new := L i=O n (p(Revenueone_neWj) - (12500) Revenut1wo_new:= Revenueone new Revenut1wo new := Revenut1wo new + Revenuelnstructors 5 4 ~_two_new:= L (p(Revenut1wo_neWj) i=O n - (22500) Revenut1hree_new := Revenut1wo new RevenUt1hree new .- Revenut1hree new_ + Revenuelnstructors 6 ) ~ three new:= L (p(ReVenut1hree_neWj) i=O n - (32500) The company should not add any new permanent instructors to its staff. , I I Luther Setzer [ NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 ~.' 4~ (L- (~'/) --------_.---_. Vu (4 X +L}- ~\ i ~ Ig ~J q,,,yt- 'f v \ (Xi ~v t-~~(~) ----...J -::5) ~~ ~" , 0:" l7'J.. (/1 ~lf) UF-ESI-632I 7/11/2007 page~of74 2-19. A package delivery company experiences high variability in daily customer demand, which in turn results in high variability in the daily workload at the central sorting facility. The company relies on its sorting facility employees working overtime to provide on-time delivery when the workload demand is very high. A sorting facility employee receives a salary of$12 per hour for a 40 hour week, and the employee receives $18 per hour for every hour worked overtime, that is for every hour worked over 40 hours in a given week. Week:= 40 Base:= 12 18 Overtime:= The number of overtime hours that an employee works in any given week is a random variable, with a mean of 15 hours and a standard deviation of 4 hours. Mean:= 15 Deviation := 4 What are the mean, the standard deviation, and the variance of an employee's total weekly salary? Basic-pay:= Base Week Basic-pay = 480 Deviation basic:= 0 Mean overtime:= Overtime Mean Mean overtime = 270 Deviation overtime:= Overtime Deviation Deviation overtime = 72 Mean total = 750 Luther Setzer 1 NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632I 7/11/2007 page[of1Y Deviation_total := Deviation_basic + Deviation overtime Deviation total = 72 . V anance:= D" eVl3tlon tota ,2 Variance = 5184 Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UFF-ESI-6314 2007-05-22 2-22. A large retail company has stores at two locations in a city: a large department store and a discount outlet store. Weekly sales of umbrellas at the department store have a mean of 147.8 and a standard deviation of 51.0. MeanLarge:= 147.8 S~arge:= 51.0 Weekly sales of umbrellas at the discount outlet store have a mean of 63.2 and a standard deviation of37.0. MeanOutlet:= 63.2 SDOutlet:= 37.0 Sales of umbrellas in the two stores have a correlation of 0.7. Correlation := 0.7 Umbrellas cost$17 each at the department store and $9 each at the outlet store. Cost Large:= 17 CostOullet:= 9 Compute the mean, variance, and standard deviation of the total (combined) umbrella sales revenue from the two stores. Mean_Sales:= MeanLargeCostLarge+ MeanOutle,Costoutlet Mean Sales = 3081.40 Covariance = ------ Correlation SDLarge DOutlet S Covariance:= Correlation S~arge' SDOutle, Covariance = 1320.9 I /I ? (J' x := " Covariance I (J'" = 36.344 /J5s,(l.M( iN~1~ ~~ rr Jccr~ ::. ,:>.J 1~ (~SN'A(,) ['f-~ ~~ -fr -' ,,{, 5 ( (1- ) :: (,) 7. g L(.;1-' _y, ,I d .1>5 tt.~ I V' -< -" L 7.') .:1)~(p.) \ln~ ( ! ~~ 0 ~ 11,\ "\ ). ~rc ~ -= d J.::l~ ,~ UF-ESI-632I 7/11/2007 page~Of~~ 2-33. The FNO Market Index Fund is a mutual fund that trades on the stock exchange, and whose stock price varies from day to day. In any given month, the probability that its stock price will increase is 0.65. Pincrease 0.65 := Pdecrease:= I - Pincrease Pdecrease= 0.35 On average, on months that its stock price increases, the expected increases in its stock price is 5%. Increase:= 5% On months that its stock price decreases, the expected decrease in the stock price is 4%. Decrease := 4% (a) What is the probability that the FNO Market Index Fund will increase in exactly seven of the next twelve months? 12! 7 5 Pincreaseyear:= 7! .5! 'Pincrease 'Pdecrease Pincreaseyear= 0.204 (b) What is the expected change in the FNO Market Index Fund for next year? For the purposes of this question you may assume that the increase in the stock price is exactly 5% in months when the stock price increases, and the decrease in the stock price is exactly 4% in months when the stock price decreases. Change := [I + (Pincreaselncrease - Pdecrease' Decrease)J 12 Change = 1.246 Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 ,j ft,. _~_'(/~ J~ ( )' ( /-,'l) 5 (6- I).' UF-ESI-6321 7/11/2007 2-36. The owner of a charter fishing boat has found that 12% of his passengers become seasick during a half-day fishing trip. p:= 12% He has only two beds below deck to accommodate those who become ill. Beds:= 2 About to embark on a typical half-day trip, he has 6 passengers on board. n:= 6 What is the probability that there will be enough room below deck to accommodate those who become ill? P(p,n,x):= n! ---.p x!(n - x)! x n x (1 - p) -. The acceptable range of sick passengers becomes x:= 0 .. 2 P(p,n,x) 0.464 = 0.380 0.130 Summing these probabilities together gives the probability of having enough beds. p':= L P(p,n,x) x P' = 0.974 Luther Setzer 1 NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 ,1.-- P (b bC F 6r IJ) -/I) ~ P (D) + P( Ci) + :P (A ') _ P(b_1 ~1 - pCD"~ A) - P (f~l e---- + N D ~l r "....J _ c-~\)(,::<) - ,,) C.,)(.s') _T-~)(.'\) + (.}.)).S-( ) (I 5 r) ( UF-ESI-6321 7/11/2007 page 15of34 3-2. The men's 100 meter sprint at the 1996 Olympic Games in Atlanta was a hotly contested event between Donovan Bailey of Canada, Frankie Fredericks of Namibia, and Ato Boldon of Trinidad. Assume that the probability distribution of the time to run the race is the same for all three runners, and that this time obeys a (continuous) uniform distribution between 9.75 seconds and 9.95 seconds. a:= 9.75 b:= 9.95 I h:=-- b-a h=5 f(t) := if a ~ t ~ b b-a o otherwise t := 9.5,9.55 .. 10.1 f(t) 3 2 o 9.6 9.7 9.8 9.9 10 10.1 (a) What is the probability that Donovan Bailey's time will beat the previous of9.86 record seconds? t:= 9.86 Breaking this rectangle into two smaller ones at t yields \ AI := h(b - t) . 1 -' Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632\ 7/11/2007 (b) What is the probability that the winning time will beat the previous record of9.86 seconds? Begin with the probababilty of one runner not beating the previous record. A2 := h(t - a) A, = 0.55 Use binomial distribution to calculate probability of all runners failing to beat previous record. n:= 3 r:= 3 Pall fail:= P (1 -p) 0.166 r n-r PaliJaiJ= This means that if at least one runner beats the record, then I (c) In answering part (b) with the data you were given, you had to assume that the running times for the three athletes were independent of each other. Give one reason why this may not be valid. Among many reasons, one that comes to mind involves air drag effects of one upon the other. Should they run on independent tracks, psychological effects can still come into play. Ie, 1'1 ~A ... R>it- (V\aR-C Loo /C...,'w '- ~ ~ /l-()"'''''''~'S rl~tL ~~H'rVh. ro w- '(( c...A-reH u~ <';) ! ., ~ "'-tk"'A CkfU: "T}f-l=.-f fl. T')f-e p",'C;""t) OC"j)CN~""W'T orJ <.. ~~1e ~ .. E Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 -----IZ,,; ~ ~ X ::-:z.~"1- \ -< r- o )\.e:. _ '.t]~X "55" 4/5 ~ 2:;)'1> :::- 1,37 (~ ~ ~ P(X ~<1737) =- p( z-L.. 'I ~I -).$5-0 ) -- 't /5 - ~ P(2"--I:'b) UF-ESI-6321 7/11/2007 pagellf ,y 3-3. The weekly sales of a brand-name kitchen cleanser at a supermarket is bel ieved to be Normally distributed with a mean 2,550 bottles and a standard deviation 415 bottles. ~:= 2550 0":= 415 The store manager places an order at the beginning of each week for the cleanser. She would like to carry enough bottles of the cleanser so that the probability of stocking out (i.e., not having enough bottles of cleanser) is only 2.5%. p:= 2.5% How many bottles should she order each week? Begin by determining the desired percentage probability of stocking enough bottles. Penough:= I - P Penough= 0.975 Now refer to Table A.I to locate the corresponding value of Z. Z:= 1.96 Now calculate the value of X that will yield the needed number of bottles. Z=-- X-~ 0" X := ceil(ZO" + ~) X = 3364 Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UFF-ESI-6314 2007-05-22 3-8. MITs pension plan has two funds, the fixed income fund and the variable income fund. Let X denote the annual return of the fixed income fund. Let Y denote the annual return of the variable income fund. We assume that X obeys a Normal distribution with mean 7% and standard deviation Y obeys a Normal distribution with mean 13% and standard deviation 8%. 2%, and that ~x:= 7% <Tx:= ~y:= 2% 13% <Ty:= 8% = -0.4. We also assume that CORR(X,Y) CORRXY := -0.4 A particular professor has invested 30% of his pension money in the fixed income fund and 70% of his pension money in the variable income fund. a:= b:= 30% 70% annual return of the Professor's pension money? (a) What is the expected ~u:= Rx + Ry ~u = 11.2% (b) What is the standard deviation 2 of the annual return of the Professor's pension money? <Tu:= J2a '<Tx 2 + b '<Ty 2 + 2ab<Tx<TyCORRXY <Tu= 5.39% V""" (c) What is the distribution of the annual return of the Professor's the distribution pension money? of the annual return of the Because the weightings a and b were constants, Professor's pension money is Normal. L/~ UFF-ESI-6314 2007-05-22 page\1of~Y (d) What is the probability that the annual return of the Professor's pension money is between 10% and 15%? 10% -I-Lu llo:= ---------- llO = -0.22 15% -I-Lu Zl5 := ---------- Per Table A.I : PIO:= 0.4129 PIS:= p:= 0.7611 PIS - PIO P = 35% v UF-ESI-6321 7/11/2007 page100f 3Y 3-15. A software company sells a shrink-wrapped product at a price of$500 per package under a 30-day return policy. C:= 500 This policy permits a customer to return the product within 30 days of purchase for a refund of the full purchase price if the customer is not satisfied for any reason. (a) In January of the current year, the firm's retail sales division sold 2,500 packages. n := 2500 Historically, 10% of the products sold by the division are returned for a full refund. p:= 10% What is the mean and standard deviation of the number of packages that will be returned from January's sales? I..L:=np I..L 250 = (J":= ~n.p.(1 (J"=15 - p) V (b) What is the mean and standard deviation of revenue from January sales, after allowing for returns? This solution relies on number of units sold. Ps:= I - P Ps = 90% I..Ls= nps : I..Ls 2250 = (J"s = 15 I..LR:=Cl..Ls I..LR=1125000 (J"R= 7500 " Luther Setzer 1 NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-6321 7/11/2007 (c) What is the probability that January revenue (after allowing for returns) will exceed the division's target of$1.3 million? 6 X:= 1.310 X - Z:=--- ~R Z = 23.333 This result means that the probability approaches zero. Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632I 7/11/2007 page1:&f JY 3-19. Investor's Market Times is a magazine that rates stocks and mutual funds and publishes predictions of stock performance. Suppose that Investor's Market Times has predicted that stock A will have an expected annual return of 10% with a a standard deviation of 4%, and stock B will have an expected annual return of 20% with a standard deviation of 10%. IJ.A:= 10% IJ.B:= 20% Investor's Market Times has also estimated that the correlation between the return of these stocks is -0.20. CORR:= -0.20 (a) What fraction of your portfolio should you invest in each of stocks A and B so that the expected annual return is 13%? lJ.u:= 13% a:= 30% Given guess value a := Find(a) a = 70% b:= I - a b = 30,% .,' ./ Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632I 7/11/2007 (b) What is the standard deviation of the return of the portfolio that has an expected annual return of 13% (part (a? CTu = 3.67% V (c) The magazine recommends 50 small international stocks as great buying opportunities for the next year. They claim that each of these 50 stocks has an expected annual return of20% and a standard deviation of20%. They also claim that these returns are independent. For diversification reasons, they recommend that investors invest 2% of their money in each stock and hold the portfolio for a year. What is the expected return and standard deviation of the annual return of this Dortfolio? n:= 50 i;= I ..n ~:= 20% CT:= 20% c:= 2% E:= L (c~) i= I n E = 20% CORR:= 0 := ~ 2 2 CTE 50CCT CTE = 2.83% Luther Setzer 1 NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632I 7/11/2007 (d) What is the probability that the portfolio of 50 stocks will have an annual return between 18% and 24%? XI:= 18% X2:= 24% XI -I-l ZI:=-(TE ZI = -0.707 PI := 0.2389 X2 -I-l Z2:=-(TE I ,<..-1 ;, P2:= 0.9207 p:= P2 - PI p=0.682 V Luther Setzer 1 NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632I 7/11/2007 4-3. According to annual precipitation data kept for the past 75 years, the average precipitation in New York City is 41.76 inches per year and the observed sample standard deviation is 5.74 inches per year. n:= 75 S:= 5.74 (a) Construct a 99% confidence interval for the mean of the distribution of yearly precipitation in New York City. Z99:= 2.58 ~upper:= Xbar + Z99~pper = 43.47 S -Ft v S ~Iower:= Xbar - Z99-Ft ~Iower= 40.05 - . {. (~ . I ). (b) Determine the required sample size in order to estimate the mean of the distribution of yearly precipitation in New York City to within 1.2 inches at the 99% confidence level. Z99- S -Ft = 1.2 n=153 1"- Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-6321 7/11/2007 (c) During the same 75-year period, the average precipitation in Tokyo was 34.37 inches per year and the observed sample standard deviation was 4.98 inches per year. Construct a 95% confidence interval for the difference between the yearly precipitation in New York and Tokyo. StoJ...yo 4.98 := Z95:= 1.96 ""_10." ,; ",", ~d ","'_<ok yo - '''.( -+--n ? S- Stokyo n 2) lower = 6.186 -+--n ? S- Stokyo n 2) I / q Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632I 7/11/2007 4-8. A supermarket manager is trying to decide how many checkout registers to keep open during the early-morning hours. To aid her analysis, she recorded the number of customers who came to the supermarket during the early-morning hours for the past 10 days. The data collected is presented in the table below: n:= 10 i:= I ..n x I '= 167 172 150 178 160 164 167 140 190 150 Xbar := mean( X) Xbar = 163.8 ./ , S:= n-I S = 14.673 Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ES[-632\ 7/[ 1/2007 (a) Construct a 95% confidence interval for the main number of early-morning customers in the supermarket. k:= n - I k=9 (3:= 95% t:= 2.262 ~upper := cei{ Xbar + t ~) ~pper = 175 / ~Iower:= flOO{ Xbar - t ~) ~Iower = 153 -../ (b) What assumptions must you make in order to compute your confidence interval? I. ~he u~erlying data approximatel,y fits a Normal curve . 2. The data flowed from random sampIing. 3. e a a mcluded no outliers such as speci~[ sales that would skew data. -#.;>f ~t'<) ",,"az~ '" 1+N1 0. N iY\C>~l"I'J'- .- ('..H:>epc:~()erv"" of- ~ O""N"Ctl- ; N G. . Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (32\) 544-7435 UF-ESI-6321 7/11/2007 4-12. A subscription literary magazine would like to "profile" their subscribers by mailing a questionnaire to a random sample of their subscribers. In particular, they would like to determine an estimate of the average income of their subscribers to the nearest $1,000 with 99% confidence. The standard deviation of the distribution of their subscribers' incomes is believed to be at most$5,000. Determine a conservative estimate of the required sample size. Z:= 2.58 S:= 5000 z- S Ft = 1000 n = 167 -/ Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632I 7/11/2007 4-15. A public opinion group is trying to estimate the percentage of registered voters in the state who are registered as "Independent." It is desirable to estimate the true percentage to within 3% at the 99% confidence level. Determine an estimate of the required sample size that is needed. Z:= 2.58 lJ.p_upper = Pbar + z = Pbar - ~ - lower z ----- n J Pbar'( I - Pbar) z -----=3% 2 4 Z .n =(3%) 2 n := cell Z .--(3%)2 n = .24 I] [ 1849 Luther Setzer 1 NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632I 7/11/2007 4-17. At two different branches ofa department store, pollsters randomly sampled 100 customers at Store I and 80 customers at Store 2, all on the same day. At Store I, the average amount purchased was $41.25 per customer with a sample standard deviation of$24.25. At Store 2, the average amount purchased was $45.75 with a sample standard deviation of$34.76. n,:= 100 SI := 24.25 S2:= 34.76 (a) Construct a 95% confidence interval for the mean amount purchased per customer in each of the two stores. Z:= 1.96 Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 UF-ESI-632I 7/11/2007 (b) Construct a 95% confidence interval for the difference between the means of purchases per customer of the two stores. ~d lower = -13.48 / Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 Of UF-ESI-632I 7/11/2007 4-19. A book club is planning to conduct a telephone survey to estimate the proportion of members of the club who would be inclined to enroll in a new service specializing in cassette-tape recording of books. The book club would like to estimate this proportion to within plus or minus 5 percentage points at the 95% confidence level. Z:= 1.96 (a) What is the size of the sample that would need to be surveyed? (b) Suppose that the book club has decided to call 500 randomly selected members to perform the telephone survey. Suppose that there is a 10% chance that any randomly selected member will not answer the telephone. Let X be the number of randomly selected members among the 500 who will answer the telephone call. What is the distribution of X? n := 500 Pno answer:= 10% The distribution of X will be binomial. (c) What is the expected value of X? p:= I - Pno_answer p = 90% j.L:= np j.L=450 V' (d) What is the standard deviation of X? 0":= ..In.p.(1 - p) 0" = 6.708 /' Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435 , . 4- UF-ESI-632I 7/11/2007 J (e) What is the approximate probability that X is larger than 400? X:= 400 To solve this problem requires use of the Normal approximation to the binomial. X-IJ. Z:=-(]' Z=-7.454 This result means that the probability approaches 100% or I. Luther Setzer I NASA Pkwy EStop NEM3, Kennedy Space Center, FL 32899 (321) 544-7435

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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 &amp; 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
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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 &quot;Group B&quot; 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 &lt; 60) = P Z &lt; = P ( Z &lt; -2.20 ) = 0.0139 4.1 (d)y = 123.0 + ( - 2.16) 40 = 36.6 40 - 36.6 30 - 36.6 P( 30 &lt; Y &lt; 40 ) = P &lt;Z&lt; = P( - 1.61 &lt; 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, &quot;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
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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&lt;0&gt; .1 1 1 1 1 x := 1 1 1 1 1 10.500000000
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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
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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