FTDipAcct20 Quantitative Analysis Class Name Module Quantitative Analysis Assignment Instructions 1. 2. 3. 4. 5. Complete all the THREE questions Your assignment may be hand-written or typed on A4 papers (with or without lines) Do not need to copy the questions. Include the 'Assignment Coversheet cum Lecturer's Comment Form' at the front of your assignment. On the 'Given Name' section in the Coversheet, write done your FULL NAME as indicated in the class register 6. Submit your assignment to your tutor on the due date. No email submission is allowed. 7. This assignment does not require Turn-it-in submission and no submission through e-learn portal is available. 8. Please note the penalty for plagiarism and late submission stated in the Study Guide Assignment Questions Question 1 a) The following data represent the number of years patients survived after being diagnosed with terminal cancer: 0.4, 0.5, 0.6, 0.6, 0.6, 0.8, 0.8, 0.9, 0.9, 0.9, 1.2, 1.2, 1.3, 1.4, 2.1, 2.4, 2.5, 4.0, 4.5, 4.6 (i) Construct a stem-and-leaf display (6 marks) (ii) Supposedly you are inserting the above stem-and-leaf display in a report to be submitted to management, write a short comment on the diagram. (4 marks) b) The following data shows the weight (in kg) of 13 crabs found in a restaurant on a particular evening: 3.4 1.2 1.7 2.4 2.4 1.1 0.9 0.8 1.2 1.6 0.7 1.2 1.3 (i) Compute the mean and median. (3 marks) (ii) Determine the shape of the distribution based on the sample data. Explain your conclusion. (2 marks) Question 2 (a) It is noted that 8% of Kaplan students are left handed. If 20 (TWENTY) students are randomly selected, calculate the i. probability that none of them are left-handed, (2 marks) ii. probability that at most 2 are left-handed, (3 marks) iii. standard deviation for the number of left-handed students (2 marks) (b) If 50 (FIFTY) classes of 20 (TWENTY) students are randomly selected, what is the probability that 10 (TEN) classes have no left-handed students? (3 marks) Question 3 (a) Superior Construction Pte Ltd is a successful company dealing with many major projects in Singapore. Recently, it has submitted its biddings for two major Government projects. Project A worth about $120 million and the company believes it has 40% chance of securing the project. Project B worth $1.8 billion and there is 30% chance the company can win the project. Both projects are independent of each other. What is the probability that the company: i. will secure Project A or B but not both (3 marks) ii. will not secure Project A or will not secure Project B (3 marks) (b) Do you agree that \"if two events are mutually exclusive then these two events will be independent\"? Why? (5 marks) (c) Provide one business-related example each, with explanation, for mutually exclusive and independent events. (4 marks) -The End- In question 3, each project is independent of the other. This is to mean that the selection of one project by the company is independent of the other project. Total probability is always 1, therefore since project A has 40% chance of being secured means that it has 60% chance of being NOT secured. These percentages are equivalent to 2/5 and 3/5 respectively. Likewise, project B has 30% chance of being secured implying that it has 70% chance of NOT being secured. These percentages are equivalent to 3/10 and 7/10 respectively. i. Probability that the company will secure project A or B but not both is the probability that project A is secured or B is secured but not both projects. So, the company can only secure one project and NOT the two projects. The two projects are independent. Probability of securing either project A or project B = probability of securing project A + probability of securing project B. this is equal to 2/5 + 3/10 = 0.7 ii. Probability that the company will NOT secure project A or will not secure project B implies that none of the two projects will be secured. Therefore this probability is equivalent to1 - probability of securing project A + probability of securing project B. = 1-(2/5 + 3/10) = 0.3 b. YES, two mutually exclusive events are also independent events. Mutually exclusive events mean that the events cannot occur together. Only one event can occur first then followed by the other event. On the other hand, independent events mean that the occurrence of one event does not affect or influence the occurrence of the other event. Therefore, mutually exclusive events are also independent events. a. 8% is equivalent to 0.08. This is the probability of getting a left hand student implying that the probability of NOT getting a left handed student is 1-0.08=0.92. We have a sample size of 20, which is our n and X as the number of left handed students in chosen group. The distribution of this data is binomial distribution and therefore p which is the probability of success(getting a left handed student) = 0.08, q which is the probability of failure (NOT getting a left handed student) = 1 - 0.08= 0.92, n, which is the sample size = 20 students and X which is the number of left handed student in the chosen sample size. By binomial distribution, P(X) = (nCx)*pxqn-x. i. Probability that none of them are left handed implies that X=0, that is, getting 0 left handed students in the sample size. Following binomial distribution P(X=0) = (20C0)*0.080*0.9220=0.188693329. this is the probability of NOT getting a left handed student in the sample size of 20 students. ii. Probability that at most 2 are left handed means that in the sample size of 20 students we get 0 left handed students, or 1 left handed student or 2 left handed student. Note that we are using \"or\" thus we add the probabilities. Therefore, P(X2) = P(X=0) + P(X=1) + P(X=2). We already calculated P(X=0) = 0.188693329 in i above. Now we calculate P(X=1) and P(X=2). P(X=1) = (nCx)*pxqn-x = (20C1)*0.081*0.9219 = 0.328162311 and P(X=2) = (nCx)*pxqn-x = (20C2)*0.082*0.9218 = 0.271090605. we now have P(X2) = P(X=0) + P(X=1) + P(X=2) = 0.188693329 + 0.328162311 + 0.271090605 = 0.787946245 b. We have the sample size 50 classes each with 20 students and X is 10 classes. Since we are interested in finding the probability of students who are not left handed, we need to get probability of students who are right handed. Remember that probability of getting a left handed in a class of 20 students is 0.08 therefore probability of getting a right handed student tin a class of 20 students is 0.92. We have 50 classes each with 20 students. This implies that probability that a class has right handed student is 0.9220 = 0.188693329. This question also follows binomial distribution with n = 50, X = 10, p = 0.188693329 and q = 1 - 0.188693329 = 0.811306671. By binomial P(X) = (nCx)*pxqn-x, this means that P(X=10) = (50C10)*0.18869332910*0.81130667140 = 0.136972731. thus the probability of 10 classes having right handed student is 0.136972731 In question 3, each project is independent of the other. This is to mean that the selection of one project by the company is independent of the other project. Total probability is always 1, therefore since project A has 40% chance of being secured means that it has 60% chance of being NOT secured. These percentages are equivalent to 2/5 and 3/5 respectively. Likewise, project B has 30% chance of being secured implying that it has 70% chance of NOT being secured. These percentages are equivalent to 3/10 and 7/10 respectively. i. Probability that the company will secure project A or B but not both is the probability that project A is secured or B is secured but not both projects. So, the company can only secure one project and NOT the two projects. The two projects are independent. Probability of securing either project A or project B = probability of securing project A + probability of securing project B. this is equal to 2/5 + 3/10 = 0.7 ii. Probability that the company will NOT secure project A or will not secure project B implies that none of the two projects will be secured. Therefore this probability is equivalent to1 - probability of securing project A + probability of securing project B. = 1-(2/5 + 3/10) = 0.3 b. YES, two mutually exclusive events are also independent events. Mutually exclusive events mean that the events cannot occur together. Only one event can occur first then followed by the other event. On the other hand, independent events mean that the occurrence of one event does not affect or influence the occurrence of the other event. Therefore, mutually exclusive events are also independent events. a. 8% is equivalent to 0.08. This is the probability of getting a left hand student implying that the probability of NOT getting a left handed student is 1-0.08=0.92. We have a sample size of 20, which is our n and X as the number of left handed students in chosen group. The distribution of this data is binomial distribution and therefore p which is the probability of success(getting a left handed student) = 0.08, q which is the probability of failure (NOT getting a left handed student) = 1 - 0.08= 0.92, n, which is the sample size = 20 students and X which is the number of left handed student in the chosen sample size. By binomial distribution, P(X) = (nCx)*pxqn-x. i. Probability that none of them are left handed implies that X=0, that is, getting 0 left handed students in the sample size. Following binomial distribution P(X=0) = (20C0)*0.080*0.9220=0.188693329. this is the probability of NOT getting a left handed student in the sample size of 20 students. ii. Probability that at most 2 are left handed means that in the sample size of 20 students we get 0 left handed students, or 1 left handed student or 2 left handed student. Note that we are using \"or\" thus we add the probabilities. Therefore, P(X2) = P(X=0) + P(X=1) + P(X=2). We already calculated P(X=0) = 0.188693329 in i above. Now we calculate P(X=1) and P(X=2). P(X=1) = (nCx)*pxqn-x = (20C1)*0.081*0.9219 = 0.328162311 and P(X=2) = (nCx)*pxqn-x = (20C2)*0.082*0.9218 = 0.271090605. we now have P(X2) = P(X=0) + P(X=1) + P(X=2) = 0.188693329 + 0.328162311 + 0.271090605 = 0.787946245 b. We have the sample size 50 classes each with 20 students and X is 10 classes. Since we are interested in finding the probability of students who are not left handed, we need to get probability of students who are right handed. Remember that probability of getting a left handed in a class of 20 students is 0.08 therefore probability of getting a right handed student tin a class of 20 students is 0.92. We have 50 classes each with 20 students. This implies that probability that a class has right handed student is 0.9220 = 0.188693329. This question also follows binomial distribution with n = 50, X = 10, p = 0.188693329 and q = 1 - 0.188693329 = 0.811306671. By binomial P(X) = (nCx)*pxqn-x, this means that P(X=10) = (50C10)*0.18869332910*0.81130667140 = 0.136972731. thus the probability of 10 classes having right handed student is 0.136972731