Final ADM2303 Exam 2005 Fall Semester Q1. Because cheating has been on the increase, the Rector has set up a special investigation. As part of this, a private investigation company has been engaged, and they arrange for one of their agents to "register" for a particular statistics course. The agent finds out that a group of 10 students met at a bar and bought a copy of the mid-term quiz and its solution in advance. Unfortunately for them, the professor who wrote the exam made a rather unusual mistake in the solution to question 2 of the quiz, so it is pretty easy to spot who was cheating if their answer matches this mistake. There are 65 students registered for the course. 4 a) [ 4 ] If the marker checks 5 papers as a sample, what is the chance none of them belongs to one of the cheaters? Be sure to name any methods or models you use and show how you get your answer. b) In a separate investigation, professors belonging to the activist group Cascade Rounding Angers Professors want to correct the improper practice of "Cascade Rounding" where 1.23456 is rounded to 1.24 by first rounding to 1.2346, then 1.235, and finally 1.24. (THIS IS WRONG, for those who do it The correct answer is 1.23). First, they want to know how many students do this, so set up a calculation where the mean of seventeen numbers is $1.23456, then ask for it to be rounded to the nearest cent. The believe fully 12% of students are CR users. 4 1b1) [ 2 ] If they sample 7 papers from a very large collection of exams, what is the probability no student cascade rounds? Be sure to name any methods or models you use and show how you get your answer. 3 1b2) [ 3 ] If they sample 7 papers from a very large collection of exams, what is the probability at least 2 students of the seven cascade rounds? 5 1c) [ 5 ] A nationwide study is conducted on cascade rounding and there are many thousands of high school graduates asked to round 1.23456 to 2 decimals. What is the probability that a random sample of 198 such graduates returns less than 12 "cascade rounders"? Bonus Question: 2 1d) [ 2 ] Suppose you had conducted the study in ( c ) only for the U of Ottawa where 546 students were in the population. What happens to your answer in ( c )? Specify your answer with the new value of the resulting probability. Q2. BuzOff is a cosmetics maker that gets its revenue from specialty products that combine cosmetics with insecticides and insect repellents. In particular, their two key products are a lice-killing shampoo (Niet!) and a mosquito-repellent skin cream. (Zeroz). Due to the specialty nature of their business, Buzoff gets mixed prices for their products. However, they have fairly good evidence that the average profit per case of Niet is $23.45 with standard deviation $1.29, and for Zeroz is $31.23 with standard deviation $2.45. Because Niet tends to sell better during schooltime when children take too literally their parents' advice to share (at least where lice are concerned), while Zeroz gets sold mainly in the summertime when mosquitos are active, the correlation between profits of these two products is -0.4. mu(N)= 23.45 sigma(N) = 1.29 mu(Z)= 31.23 sigma(Z)= 2.45 4 2a) [ 4 ] If Buzoff sells 1225 cases of Niet and 3543 cases of Zeros, what is the coefficient of variation of the resulting profit?
2 2b) [ 2 ] Suppose the profit on a case of Zeroz is Gaussian (Normally) distributed. What is the probability a case earns a profit of more than $35? 2 2c) [ 2 ] Suppose the profit on a case of Niet is exponentially distributed with mean $23.45. What is the probability the profit on a case is less than $20? 2 2d) [ 2 ] Why does the information about Niet in part ( c ) contradict information in the introduction to this question?(Hint: Compare basic parameters.) Q3. Blotz-a-Lotz makes ink cartridges for printers. The cartridges have to satisfy many specifications, but the label says "50 ml". Blotz decides that it is unlikely consumers can measure to better than 1ml, so specifies that the cartridges must contain between 49 and 51 ml.of ink. Note that the specification refers to measurements on individual cartridges rather than their means. 2 3a) [ 2 ] What is a reasonable value to use for the standard deviation of "fill" (i.e., the number of ml. of ink in the cartridge) in order to meet the "specification"? Explain briefly. 2 3b) [ 2 ] Blotz looks at the histograms of ink content for each batch of 475 from 2 machines. Figures 3-2 present the histograms and Figures 3-3 are the corresponding normal probability plots for these two machines labelled AA and DD.. Can you presume that ink fill is Gaussian distributed? Justify your answer BRIEFLY. 3c) Two Xbar/s charts are given for each of these two machines using samples of 5 cartridges at a time. The commands that drew them are given as well. 2 i)State which charts (by name of chart) you would use and why. 2 ii)Then explain whether Blotz can use these machines for filling cartridges AA has too much variability -- uncontrollable DD is OK on variability -- controllable, but level too low (out of control) 1 + 1 MUST mention variability or controllability properly and if so, 1 iii) how they need to proceed to get into production. Use DD but adjust fill level. 5 3d) For a particular machine, Blotz is convinced that ink fill is Gaussian distributed. However, the colour and consistency of the ink affects the variability of fill. If it is assumed that the mean fill over the whole production is 49.6 ml., what is the probability, possibly approximate, of observing the following sample or one with a mean as or more different from the given production value. 49.2, 49.9, 49.7, 50.1, 50.2, 49.8, 49.6 (ml.)
2 3e) If the standard deviation of fill in (d) is reliably known to be .35 ml over production in general, work out the probability requested in part (e). Q 4. Electonic communication errors are considered to happen at a rate per time. Combined with the transmission capacity of the line, this can be translated into an error rate per volume of data. On your Internet connection, it therefore turns out that data-transmission errors happen at the rate of 1 error per 350,000 kBytes of data transmitted. (If you know about such things, please ignore the fact that we
usually measure in bits rather than bytes.) You want to transfer large three files. File: AA: BB: CC: Size in kBytes 600,000 600000 60,000 60000 120,000 120000
2 4a) [ 2 ] What model should you use for computing probabilities of errors in transmission. Justify briefly. 4 4b) [ 4 ] What are the probabilities of transfering each of the three files separately without error? 4 4c) [ 4 ] What is the probability exactly one of the three files has an error (i.e., at least one error) when transmitted separately and in the order given? 2 4d) [ 2 ] What is the probability the transfer of file AA occurs with two or more errors? Q 5. Bankruptcy is a serious social and economic problem. The following table shows the numbers of personal bankruptcies in Ontario by different categories. (Assume the table is complete for all ages and degrees of debt on bankruptcy.) Note that 0 Debt is simply the end of the range. Age 20-29 30-39 40-49 50-59 60-69 TOTAL Debt midpt midpt * TOTAL (midpt^2)*TOTAL Total Debt Outstanding ($1000s) 0-100 100-200 100 345 123 345 321 335 23 122 12 99 579 1246 50 28950 1447500 150 186900 28035000 200-300 554 765 876 324 231 2750 sum = 4575
250 687500 sum = 903350 171875000 sum = 201357500
3 5a) [ 3 ] Are the variables "Total Debt Outstanding" and "Age" statistically independent? You must justify your answer. 2 5b) [ 2 ] What is the probability that the person is in his/her 40's (40-49) if he/she has declared bankruptcy with 200-300K debt? For the following part of the question, it is possibly you may NOT have seen formulas for the calculations and may have to "work them out". Because the numbers of bankrupts are quite large, you need not worry about differences between samples and populations and may use "population" formulas as a starting point. The purpose of these questions is to determine which students understand and can apply the concepts. Some calculations that may help you have been done in the table. 3 5c) [ 3 ] Estimate the standard deviation of debt outstanding. Explain your work BRIEFLY.
Solutions 1(a) Hypergeometric N,S,n,K : 65 10 5 0 P(K=0| N, S, n) = C(10, 0) C(55, 5) / C(65, 5)= 1*3478761 / 8259888 = 0.4212 1 Hypergeom, 1 params, 1 setup, 1 answer. Also possible by 55/65 * 54/64 * 53/63 * 52/62 * 51/61= (b1) Binomial (1) n=7, p=.12 (1) P(K=0 | n, p) = (1-p)^n = 0.40867559636992 0.421163216740953
1+1
marks
(b2) 1 - P(0) - P(1) = 1 - 0.40867559636992 - 0.39009943289856 = 0.20122497073152 1 setup, 1 parts, 1 answer (c) P(K<12 | n=198, p=.12) ~= P(z < (11.5 - np) /sqrt(npq) = P(z < -2.68117965919122 ) but use z = -2.68 to get 0.00368110800917498 or 0.37 % probability 1 setup, 1 CC, 1 z, 1 lookup, 1 answer (d) Must reduce std deviation of "K" by using FPC, so z gets bigger. FPC = sqrt((N-n)/(N-1)) = 0.799082042153209 1 mark z --> -3.35532463220735 use -3.55 giving Prob .00019 1 mark 2. (a) W = 1225*N+3543*Z E(W)= 1225*23.45+3543*31.23= 139374.14 $ (1 mark) V(W)=(1225^2)*(1.29^2)+(3543^2)*(2.45^2)+2*1225*3543*1.29*2.45*(-0.4)= (1 mark) 66871967.715 $^2 ==> SD(W)= 8177.52821548174 $ (1 mark) CV(W) = sd/mean= 0.0586732102202155 (1 mark) (b) P(Zeroz > 35) = P(z > (35-31.23)/2.45) = P(z> 1.53877551020408 use 1.54 = 0.0617801767118119 1 for setup, 1 answer (c) For exponential, E(N)=23.45 = 1/R where R is exponential parameter. P(N < 20) = 1 - exp(-20/23.45) = 0.573813597907662 (Probability of < is same as <= for continuous RV.) 1 1
(d) For exponential, SD(N) = 1/R also, so SD would be 23.45. Does NOT match what we are given. Argument of this sort accepted, but only full 2 if answer uses the numbers somehow. 3. (a) Since range is 49 to 51 ml, use empirical rule and suggest sigma(fill) = 2 ml / 6 = 1/3 ml. 1 for empirical rule or equivalent, 1 for answer. (b) Histograms are "mound shaped" consistent with Gaussian 1 NPPs are straight (note Anderson Darling p value is nearly .5 too) also consistent with Gaussian. (c) Use 3-4 top and 3-5 top because they use the specifications. The other graphs use means and variabilities from within data. 1 answer + 1 reason (d) n= 7 1
X_bar= 49.7857142857143 1 s= 0.333809184158512 1 Sigma unknown, use t with 6 d of f t = (Xbar - 49.6)/(s/sqrt(n))= 1.47196014438796 P(Xbar >= 49.785) = P(t > 1.47) 1 use t=1.47 Tables: P(t6>1.4398) = .1, P(t6>1.9432) = .05 1 Thus almost 10% chance of observing a sample like this or more extreme. 1 (e) Now have sigma, so xbar is distributed as Gaussian (EXACT) (Should mention something about Gaussian or lose .5) P(Xbar > 49.785) = P(z> (49.785-49.6)/(.35/sqrt(7)) = P( z > 1.40386804260568 ) Use 1.40 1 mark = 0.0807566592337711 1 mark 4. (a) Poisson 1 mark We have "hits" over time and are told that the chance is uniform of a hit.
1 mark
(b) P(0 | mu) = exp(-mu) But we need a mu for each, given by size/350 kBytes 1 mark File: Size in kBytes mu P(0 hits) AA: 600000 1.7143 0.18009 1 mark each BB: 60000 0.17143 0.84246 CC: 120000 0.34286 0.70974 (c) P(exactly 1 has error) = P(AAbad, BBok, CCok) + P(AAok, BBbad, CCok) + P(AAok, BBok, CCbad) = 1 mark = (0.819907687852048 * 0.842460441616771 * 0.709739595689125 + 0.180092312147952 * 0.157539558383229 * 0.709739595689125 + 0.180092312147952 * 0.842460441616771 * 0.290260404310875) = 0.49025 + 0.02014 + 0.04404 2 marks = 0.55442 1
(d) P(AA 2 or more errors) = 1 - P(0) - P(1) = 1 - exp(-1.714)*(1 + 1.714) = 0.511178009884129 1 mark setup 1 answer 5 (a) To have VARIABLES independent, must have all cross pairs of events independent. 1 mark Events indep if P(A and B) = P(A) * P(B) P(60s) = (12+99+231)/4575= 0.0747540983606557 P(debt 0-100K)=579/4575= 0.12655737704918 1 mark Product = 0.00946068261220102 BUT P(60s AND 0-100K)= 12/4575= 0.00262295081967213 Therefore NOT independent. 1 mark (b) (c ) P(40s) = (876)/2750= 0.31855 1 method, 1 answer
Mean approx = 197.45355 = 903350 / 4575 Need V(Debt)=E(Debt^2) - (E(Debt)^2) = 44012.5683060109 38987.9051628893 = 5024.66314312162 SD(Debt)= 70.8848583487448 $ This is approximate because we use midpoint as estimate of mean of class. Alternative:
number 579 1246 2750 total 4575 4575 4575 fraction 0.12656 0.27235 0.60109 midpt - mean -147.45 -47.45 52.55 square(midpt-mean) 21742.5499716325 2251.83958911882 2761.12920660515 * fraction 2751.68009 613.28790 1659.69515 sum=app var = 5024.66314 Thus SD = sqrt(Var)= 70.8848583487448