Research Methodology and Statistical Techniques Operations Research

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) 2 (a) It is noted that 8% of Kaplan students are left handed. If 20 (TWENTY) students are randomly selected, calculate the probability that none of them are left-handed, (2 marks) 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) 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: will secure Project A or B but not both (3 marks) 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) Answer: 1 (i) Based on the given data, the stem and leaf plot is as drawn below (Hillier, 2006). Stem Leaf 0 4,5,6,6,6,8,8,9,9,9 1 2,2,3,4 2 1,4,5 4 0,5,6 Key: 0l6 = 0.6 (ii) The above plot suggests that 50% of the patients could not even survive for a year. Besides, the maximum life for a cancer patient post diagnosis based on the given sample is 4.6 years. Only three patients i.e. 15% of the sample size survived for a period in excess of four years. Thus, from the available sample data, it may be concluded that post diagnosis also, the terminal cancer patients do not survive for long which implies that most of them are detected at the last stage. The patient data above is skewed towards the left and hence is negative skewed which implies that the above distribution is non-normal. (Gupta, 2008). (b) (i) Mean = Sum of data/Count of data values = 19.9/13 = 1.531 kg The data values given count to 13 and hence median value is captured by [(13+1)/2]th value which comes out to be the 7th value when the given data is organised in ascending or descending order. The 7th value when the data is arranged in ascending order is 1.2 kg which is the median value. From the above calculation, it is apparent that the given distribution is not normal as the median value does not coincide with that of the mean. Besides, as the datas mean is higher than the corresponding median, thus the data would be skewed towards the right and the presence of this skew further indicates the non-normal nature of the data (Jackson, 2012). 2. Probability of left land usage by the usage is 0.08 since only 8% of the students use their left hands. The given question considers a binomial distribution with probability of success as 0.08 and total number of trials as 20. As per the binomial distribution, the formula for probability of x successes is shown below. P(X=x) = nCx px (1-p)n-x i) Using the formula given above, we get Probability of zero success i.e. P (X=0) = 20C0 (0.08)0(0.92)20 = 0.1887 ii) Using the formula given above, we get Probability of at most 2 successes i.e. P (X2) = P(0) + P(1) + P(2) Hence, P(X2) = 20C0 (0.08)0(0.92)20 + 20C1 (0.08)1(0.92)19 + 20C2 (0.08)2(0.92)18 = 0.7879 (ii) Standard deviation = (np(1-p)) = 20*0.08*0.92 = 1.213 Probability associated with student not being left handed = 1-0.08 = 0.92 Requisite probability that given 20 students are all not left handed = 0.9220 = 0.1887 After performing the step shown above, the problem can be represented in the form of a binomial distribution where from 50 trials, the number of successes desired are 10. Requisite probability or P(X=10) = 50C10*0.188710(1-0.1887)40 = 0.137 3. (i) In the given case, we would consider the following two cases. Case 1: Secures project A but does not secure project B Requisite probability in this case = P(A) *P(B-bar) = 0.4*0.7 = 0.28 Case 2: Secures project B but does not secure project A Requisite probability in this case = P(A-bar) *P(B) = 0.6*0.3 = 0.18 Hence, total probability = 0.28 + 0.18 = 0.46 (ii) Probability of not winning project A i.e. P(Not A) = 1-P(A) = 1-0.4= 0.6 Probability of not winning project B i.e. P(Not B) = 1-P(B) = 1-0.3= 0.7 Again, using the addition theory, we get the following. P(Not A or Not B) = P(Not A) + P(Not B) P(Not A Not B) = 0.6 + 0.7 (0.6*0.7) = 0.88 b) If two events are mutually exclusive that does not necessarily imply that these would be independent also. This can be explained as shown below. The condition to be met for events to be categorised as mutually exclusive is that P(A and B) should be zero. However, in case of two independent events A and B, P(A and B) = P(A)*P(B) Thus, if A and B are both mutually exclusive and also independent, then the requisite condition is as follows (Gupta, 2008). P(A)*P(B) = 0 It is evident that the above condition would be satisfied only if atleast one of the terms amongst the above is zero. This is very rarely the case and therefore the independence of the events is not automatically ensured from the mutually exclusive nature of events and this would be true only in very specific case (Hillier, 2006). c) Mutually Exclusive Event (Example) Assume there is an company which has $ 100,000 surplus cash and is exploring two different investment choices which each would require $ 100,000 as the total investment. The company would consider the incremental costs and benefits of each of the projects and then narrow down on one of the projects. This is an example of mutually exclusive event since the company can pursue only one of the projects and not both since the money is limited only for one project. Hence, if one of the projects is chosen, the other is automatically rejected (Jackson, 2012). Independent Event (Example) Assume that a given company has $ 100 million budget for acquisitions. The company is currently carrying a due diligence on two potential targets A and B with requirement of $ 25 million and $ 75 million. Assume that the availability of due diligence staff is not a constraint the acquisition decision is independent since the company has enough resources to go ahead with acquisition of both the targets (Hastie, Tibshirani and Friedman, 2001). References Gupta, S. (2008), Research Methodology and Statistical Techniques, New Delhi: Deep Deep Publications Hastie, T., Tibshirani, R. and Friedman, J. (2001), The Elements of Statistical Learning, New York: Springer Publications Hillier, F. (2006), Introduction to Operations Research. New York: McGraw Hill Publications Jackson, S.L. (2012), Research Methods and Statistics: A Critical Thinking Approach, New York: Wadsworth Publishing