Summer 2015/16- Main Exam STUDENT NUMBER: SURNAf4E: (FAMILY NAME) 'i~... OTHER ~~ES:

Transcription

1 Summer 2015/16- Main Exam ~UTS UNIVERSITY OF TECHNOLOGY, SYD,NEY STUDENT NUMBER: SURNAf4E: (FAMILY NAME) 'i~... OTHER ~~ES: This paper and all materials issued must be returned at the end of the examination. They are not to be removed from the exam centre. Examination Conditions: lt is your responsibility to fill out and complete your details in the space provided on all the examination material provided to you. Use the time before your examination to do so as you will not be allowed any extra time once the exam has ended. You are not permitted to ha'4! on yotir desk or on your person any unauthorised material. This includes but not limited to: Mobile phones Smart watches and bands Electronic devices Draft paper (unless provided) Textbooks (unless specified) Notes (unless specified) You are not permitted to obtain assistance by improper means or ask for help from or give help to any other person. You are not permitted to leave your seat (including to use the toilet): Until 90 mins has elapsed During the final15 mins During the examination you must first seek permission (by raising your hand) from a supervisor before: Leaving early (after 90 m ins) Using the toilet Accessing your bag Mathematical Modelling 2 Time Allowed: 3 hours and 10 mins Includes 10 minutes of reading time. Reading time is for reading only. You are not permitted to write, calculate or mark your paper in any way during reading time. This is a Closed Book exam Permitted materials for this exam: Calculators (non-programmable only) Drawing instruments i.e. Rulers, Set Squares and Compasses Materials provided for this exam: This examination paper Six (6) answer booklets (5-pages) Students please note: All questions are of equal value Answer all questions Answer each question in a separate booklet clearly indicating the q uestion number on the front cover Disciplinary action will be taken against you if you infringe university rules. po nbt open your exam paper until instructed ~ ~---~ ~~--. P~1~5

2 33230 Mathematical Modelling 2- M~in Exam Summer 2015/ ~ Rough work space Do not write your answers on this page ~--...,. ~---~ Page2of1p

3 Page 3 ******** ANSWER IN NSEPARA TE BOOKLET******** "'\-;' Question 1.( =20marks) (a) Find all ~lutions of the system oflinear equation~ax = b where: A=G or if there are no solutions, explain why this is so. (b) Given the transfolllultidn matrix T = ~ 2 (.;3 -~).;3" 2 (i) Find the eigenvalues of this matrix. (ii) Identify the angle of rotation in the xy-platl:e pro4uced by multiplication of a vector (x, y) by this matrix. (iii) Hence or otherwise describe the transformation produced by multiplication of a vector (x, y, z) by the matrix U = ; ;3 2 2 ( 11 (c) Let A = -;5..._4 6-4 (i) Calculate the trace of A. (ii) Given that ll 1 = 6 is an eigenvalue of the matrix A, find the eigenvectors u 1 associated with /t 1. (iii) Show that the vector u 2 = (1, -5, 3) is an eigenvector of A. State the corresponding eigenvalue /t 2. (iv) Hence determine the third eigenvalue /t 3.

4 Page4 ********ANSWER IN A SEPARATE BOOKLET******** Question 2. ( = 20 marks) (a) Let where w = z 2 cos(xy - z 2 )- xy + 2yz, x = 2u- 3v, y = 2u + 3v, z = 2u- 2v. aw Calculate au at the point u=5,v=2. (b) Find and classify the critical points of the function f(x, y) = 2x 3-2xy- y 2 (c) Use the method oflagrange multipliers to find the maximum value of the function on the circle f(x, y) = 2x 2 + 2xy + 2y 2

5 Page 5 ******** ANSWERI~ A'SEPARATE BOOKLET******** Question 3. ( = 20 marks) (a) Abus~walker is climbing a hill whose shape cah be described by the equation z = f(x, y) = ~100+x3+2yz. At what rate is the bushwalker ascending at the point P(S,O) if they are walking in the direction towards the point Q(0,3)? (b) Evaluate JL ex- l)yda over the region R bounded by the curves y = X and X = y 3. (c) Evaluate the integral JL 4xydx dy Over the region R bounded by the lines X + y = 0, X + y = 2,x - y = -1 and X - y = 1. It is recontmended that you transform the integral to be.in terms of new variables u and v, where U = X + y and v = X - y. (d) Use spherical coordinates to evaluate the mass of tl:le solid bounded below by the half cone and above by the spherical surface z ==.Jxz + yz The density of the solid is proportional to the distance from the origin at each point.

6 Page 6 ****:'**** ANSWER IN A SEPARATE BOOKLET ******** Question 4. ( = 20 marks) Part (a) The following data give the maximum daily rainfall (inches) over a forty year period for a town in northern New South Wales. Rainfall Minitab is used to obtain the following descriptive statistics. Descriptive Statistics: Rainfall Variable Rainfall N 40 Mean SE Mean StDev Minimum Q Median Q Maximum (i) The mean and median ofthis sample are not equal. Comment on possible reasons for the difference in their values. (ii) Which of the median and mean is the better measure of location for this sample? Give a reason for your answer....;over

7 Page 7 Part (b) L~t X denote the waiting time in seconds (rounded t~'~e tenth) for a large database update to be completed. The probability density function t;0i'x is as follows: X :P(X = x) ,, (i) W~at ili the probability that the waiting time for.<'~ database update is longer than 0.3 seconds? (ii) What is the mean waiting time for a database UJ?date? Part (c) A component in a remote sensing instrument in order to meet specifications needs to have a reliability of 99.9%. The maximum reliability which can be achieved for a single component is 72.5% and so several coponents are joined in parallel to achieve redundancy in a system and increase the reliability. How many such components need to be joined in parallel for the system to have a reliability of at least 99.9%?.../Over

8 Page 8 ******** ANSWER IN A SEPARATE BOOKLET ******** Question 5. ( = 20 marks) Part (a) A certain type of window is made by bonding together two different layers of "plastic" sheet. Each layer is produced by a different manufacturer but both manufacturers have cosmetic flaws occurring in their sheets. The number of flaws in the bottom sheet follows a Poisson distribution with an average of 4 flaws per 13quare metre and the number of flaws in the top sheet follows a Poisson distribution with an avel"'<ige of 6 flaws per square metre. (i) W~at is the average number of flaws in a two layer window of size 0.5 m 2? \'~ (ii) What is the probability that a two layer window has no flaws? (iii) What is the probability that a two layer window has five or more flaws? The cumulative probabilities below are obtained usij:lg.m:initab and should be used to answer the above questions rather than calculating the probabi:lities required using the appropriate probability density functions. Cumulative distribution probabilitie~ for Poisson distribution X P(X::; x) P(X::;x) P(X::; x) P(X::; x) P(X::; x) p,=2 p,=3 p,=4 p,=5 p,= ' jOvcr

9 Page 9 Part (b) To determine if children in families where at least one paretit smokes have a greater prevalence of a~thma, families are selected at random and for each iainily it is determined whether at least one parent in the family smokes and if any of th~. children have asthma. There are 120 children in families where at least one parent smokes and,pt these eighteen have asthma. Of the 100 children children in families where neither parent srnokcls, six have asthma. The following output is obtained using Mini tab. Test and Cl for Two Proportions Sample X N sample p <.p ''>'-, Difference = p (1) - p (2) Estimate for difference: ~Cl for difference: ( , ) Testfor difference= 0 (vs not= 0): Z = 2.23 P-Value Fisher's exact test: P-Value = Test and Cl for Two Proportions Sample X N Sample p Difference = p (1) - p (2) Estimate for di.ff~rence: % lower bound for difference: Test for di.ffe:tjence = 0 (vs> 0): Z = 2.23 P-Value = 0,,013 (i) Gi~e th~ null and alternative hypotheses for the test \vhich determines if the prevalence of asthma amongst children is greater for childre11 with at least one parent smoking. beatment is more effective than the placebo. (ii) What proportion of children in families with at least one parent smoking have asthma? (iii) What proportion of children in families with neither parent smoking have asthma? (iv) What is the difference between these proportions? (v) What is the p-value for the appropriatl;! test in this case? (vi) What is your conclusion concerning the prevalence of asthma on the basis of this p-value? (vii) Give an interpretation of a type Hel.TO+ in this situation..../over

10 Page 10 ******** ANSWER IN A SEPARATE BOOKLET ******** Question 6. ( = 20 marks) Part (a) The viscosity of two different brands of car oil, labelled A and B, is measured with the following results. Brand A Q.44 Brand B The following analysis is performed using Minitab. Two-Sample T-Test and CI: Viscosity A, Viscosity B Two-sample T for Viscosity A vs Viscosity B N Viscosity A 8 Viscosity B 9 Mean StDev SE Mean Difference = mu (Viscosity A) - mu (Viscosity B) Estimate for difference: % CI for differ~nce: ( , ) T-Test of difference = 0 (vs not =): T-Value P-Value 0.00:0 DF 15 Both use Pooled StDev = (i) What assumptions are being made about the distribution of the viscosity of the two brands of oil? (ii) What are the null hypothesis and the altcrn(lltive hypothesis for the test being performed? (iii) Is the alternative hypothesis in this case one-sided or two-sided? particular form for the alternative hypothesis be chosen? Why would this (iv) What conclusions would you reach on the basis of the test which was performed?.. -/Over

11 Page 11 Part (b) In order to determine the short-term effect of caffeine on heart, rate each of a group of forty volunteers has their heart rate measured. The group of forty i~""d\'vided at random into two groups of twenty in each, with one group of twenty being given six.(mnc!es"of normal coffee to drink and the other group of twenty being given six ounces of decaffeir1ated coffee to drink. Ten minutes after drinking the coffee each person's heart rate is measured again. It is thought that the amount of caffeine in n~rp:1al coffee will increase the heart rate wore thap: the amount of caffeine left in decaffeinated coffee and so the changes in heart rate, recorded as final heart rate minus initial heart rate are determined and are shown below. Normal coffee Decaffeinated coffee Four different analyses were performed using Minitab and.the results are shown on the following page. (i) Which analysis should be used in this particular situation. Explain why the analysis you choose should be used instead of any of the other three analyses. (ii) For the chosen analysis, what are the null and alternative hypotheses? (iii) What is your conclusion in this situation as a result of the chosen analysis?.../over

12 Page 12 Analysis 1 J'wo-sample J' for Normal vs Decaffeinated Normal Decaffeinated N Mean StDev SE Mean ~ Difference= mu (Normal)- mu 1 {Decaffeinated) Estimate for difference: % CI for difference: (-3.93, 7.43) T-Test of difference= 0 (vs not=): T-Value 0.62 P-Value DF 37 Analysis 2 Two-sample T for Normal vs Decaffeinated Normal Decaffeinated N Mean StDev SE Mean DiffErence = mu (Normal) - mu (Decaffeinated) Estimate for difference: % lower bound for difference: T-Test of difference= 0 (vs>): T-Value Analysis P-Value DF 37 Paired T for Normal - Decaffeinated N Mean StDev SE Mean Normal Decaffeinated Difference % CI for mean difference: (-4.80, 8.30) T-Test of mean difference 0 (vs not= 0): T-Value 0.56 P-Value Analysis 4 Paired T for Normal - Decaffeinated Normal N Mean StDev 9.37 SE Mean 2.10 Decaffeinated Difference % lower bound for mean difference: T-Test of mean difference= 0 (vs> 0): T-Value 0.56 P-Value 0.291

13 Page 13 FORMULA SHEET Table of Selecf:ed Derivatives and Integrals d - secx = secxtanx dx xn+1 J xn dx = --+ K. n 4-1 n+ 1 1 J sin x dx = - cos x + K J sinhxdx = coshx + K J 1 Ja2- x2. a 1 1. J. 2 dx ~ sin -I :?. + K sm x dx = 2" x - "4 sm 2x + K Formulae for multiple integrals J ~ dx = log lx I + K J cosxdx = sinx + K J coshxdx = sinhx + K J 2 J J cosn u du = -1 cosnn-1/ 1 u sin u + -- cosn- 2 u du n n. J sinn u du = -;, 1 sinn- n-1/ 1 u cos u + -n- 1 2 dx =. ~ tan - 1 :?. + K a+x Clt a cos 2 x dx = t x + ~ sin 2x + K sinn- 2 u du 1. Cylindrical coordinates: X = r cos e, y = r sine, z = z, dv = r dz dr db 2. Spherical coordinates: X = p sin cp cos e' y = p sin cp sine' z = p cos cp, dv = p 2 sin cp dp dcp db 3. Mass of a solid: m=!!! p(x, y, z) dv, where p(x, y, z) is the density 4. Centre of mass, (x, y, z), of a solid: x = ~ JJ J xp(x, y, z) dv, y = ~!!! yp(x, y, z) dv, z = ~!!! zp(x, y, z) dv 5. The Jacobian of the transformation given by x = x(u, v) and y = y(u, v) is J( ) = a(x, y) = u,v a(u,v) ox ox au av ay ay au av ox ay auav ox ay av au

14 Page 14 Statistical Formula! Basic Results 2 1 Ln -)2 s = -- (x; -x n-1 i=l Q;=i(n+l) 4 IQR = Q3- Ql x; -x Z-score = - s Probability Distributions - Discrete P(X ~ x) = L P(X = x;) E(X) = L x;p(x = x;) Xi x, Probability Distributions - Continuous P(X ~ x) = [xx j(t)dt E(X) =I: xj(x)dx E(X 2 ) =I: x 2 j(x)dx V(X) = E(X 2 )- E 2 (X)

15 Page 15 Statistical Formul~' Specific Distributions Binomial P(X=x)=(:)px(1-p)n-x :t:=0,1,2,...,n E(X)=np V(X)=np(1-p} Poisson e->...\x P(X = x) = ----;;y- x = 0, 1, 2,... E(X) =..\ V(X) =..\ Exponential f(x) =..\e->.x x > 0 E(X) = ~ 1 V(X) =..\ 2 Central Limit Theorem If a sample of n observations is drawn from,a population whose distribution has mean p and variance o-2, the sampling distribution of X has mean p and variance o- 2 /n. For sufficiently large n ( n at least 30) is approximately normal N(O, 1). Inference for a::single Mean (population variance known) X- Po z= -.-- o-/fo (} X± Zo:/2 fo _ (Znj20") 2 n-. d Inference for a Single Mean (population;;yariance unknown) x -po t=-- s/fo