Assignment 2-Part 2 (Solution)

Assignment 2 - Part 2 ADM2303, Fall 2023 • The assignment is due on Friday, Oct 13 at 11:59 pm • The total point for this assignment is 34 points • For information regarding penalties, refer to the course outline "Part 2 Marking Scheme" • The solution file must be type-written and submitted on Brightspace in PDF format • You can use Microsoft Excel to carry out your calculations. However, you must show the detailed step by step calculations and process in your solution file. • Include a statement of academic integrity in your submission. • You are responsible for completing this assignment on your own. The integrity statement prohibits receiving assistance in answering questions through any form of service. Question 1. Oil Drilling (9 marks) An oil company is considering drilling for oil at a specific site. There are three possible outcomes if they decide to proceed: finding "No Oil," finding "Some Oil," or finding "Much Oil." Based on historical data and geological studies, the company estimates the following probabilities for these outcomes: • Probability of finding "No Oil": P(No Oil)=0.7 • Probability of finding "Some Oil": P(Some Oil)=0.2 • Probability of finding "Much Oil": P(Much Oil)=0.1 To gather more information before making a decision, the company has the option to conduct a seismic experiment. This experiment can produce one of three readings: "Low," "Medium," or "High." From past drilling sites, the company has gathered the following data related to the accuracy of a "Medium" reading: • Out of 100 past sites that yielded "No Oil," 5 gave a "Medium" seismic reading. • Out of 400 past sites that yielded "Some Oil," 376 gave a "Medium" seismic reading. • Out of 300 past sites that yielded "Much Oil," 9 gave a "Medium" seismic reading. a) Using the information provided, calculate the updated probabilities — also known as posterior probabilities — of finding "No Oil," "Some Oil," or "Much Oil," given that a "Medium" seismic reading is obtained. ( You can use tree diagram to organize the information. No need to depict that in your solution ) (6 marks) b) Suppose it costs $1 million to drill a site. The estimated revenue from finding "Some Oil" is $3 million, and from "Much Oil," it is $10 million. If the seismic experiment shows a "Medium" reading, should the company go ahead and drill? Justify your answer by calculating the expected monetary value of drilling based on the posterior probabilities you obtained in part b. (3 marks). Solution Part a)

P(No Oil)=0.7 P(Some Oil)=0.2 P(Medium | No Oil)=0.05 P(Medium | Some Oil)=0.94 P(Medium | Much Oil)=0.03 P(Low | No Oil) P(High | No Oil) P(Low | Some Oil) P(High | Some Oil) P(Low | Much Oil) P(High | Much Oil) P(Much Oil)=0.2 The tree diagram is for illustration only. Students are not required to depict that in their solution file. 𝑃(No Oil) = 0.7 𝑃(Some Oil) = 0.2 𝑃(Much Oil) = 0.1 𝑃(Medium|No Oil) = 5/100 = 0.05 𝑃(Medium|Some Oil) = 376/400 = 0.94 𝑃(Medium|Much Oil) = 9/300 = 0.03 𝑃(No Oil|Medium) = 𝑃(No Oil ∩ Medium) 𝑃(Medium) = (0.7)(0.05) (0.7)(0.05) + (0.2)(0.94) + (0.1)(0.03) = 0.1549 𝑃(Some Oil|Medium) = 𝑃(Some Oil ∩ Medium) 𝑃(Medium) = (0.2)(0.94) (0.7)(0.05) + (0.2)(0.94) + (0.1)(0.03) = 0.8318

d) Based on your calculations from parts (a) and (c), which estimate of the probability of failure ( p ) seems more likely to be accurate? Justify your answer. (1 mark) e) What should be the value of p such that the probability of experiencing no mission failures in 25 attempts is at least 0.999? (2 marks) f) Calculate the mean and standard deviation of mission failures for the next 25 missions under both the NASA estimate 1/60,000 and the Air Force estimate 1/35 (2 marks). Solution Part a) 𝑥 is the number of mission failures 𝑥 is binomial with 𝑛 = 25; ? = 1 60,000 ; ? = 59,999 60,000 𝑃(𝑥 = 0) = ( 25! 0!(25−0)! )( 1 60,000 ) 0 ( 59,999 60,000 ) 25 = 0.9996 1 mark: utilizing binomial formula 1 mark: correct final probability Part b) 𝑃(𝑥 ≥ 1) = 1 – 𝑃(𝑥 = 0) = 0.0004 Part c) 1 mark for correct calculation of each probability 𝑥 is binomial with 𝑛 = 25; ? = 1 35 ; ? = 34 35 𝑃(𝑥 = 0) = ( 25! 0!(25−0)! )( 1 35 ) 0 ( 34 35 ) 25 = 0.4845 𝑃(𝑥 ≥ 1) = 1 – 𝑃(𝑥 = 0) = 1 – 0.4845 = 0.5155 Part d) ? = 1 35 is more likely since the event did actually happen. Part e)

We perform inverse probability calculation here. set the probability to 0.999 and then solve for p and q : 𝑃(𝑥 = 0) = 25! 0!(25−0)! (?) 0 (?) 25 = 0.999 25! 25! (?) 0 (?) 25 = 0.999 (1) (1) ? 25 = 0.999 ? = (0.999) 1/25 = 0.999959981 ? = 1 – 0.999959981 = 0.000040019 At four decimal place the value of p should be zero. Part f) 0.5 mark for correct calculation of each part. Using NASA's Estimate μ = 25 × 1 60,000 = 25 60,000 = 1 2400 ≈ 0.0004167 𝜎 = √25 × 1 60,000 × (1 − 1 60,000 ) ≈ √ 25 × 59,999 3,600,000,000 ≈ 0.0204 Using Air Force's Estimate μ = 25 × 1 35 = 25 35 ≈ 0.7143 σ = √ 25 × 1 35 × (1 − 1 35 ) ≈ √ 25 35 × 34 35 ≈ 0.8330 Question 3. Investment in Mutual Funds (9 marks) You are an investor weighing your options for the coming year. You are considering investing in one of two different mutual funds: Fund 1 or Fund 2. The annual returns of these funds vary depending on the state of the economy. To guide your decision, you've estimated the probabilities for different economic scenarios for the next year. Your estimates for the economic conditions for the upcoming year are: 20% chance that the economy will be good, 50% chance that it will be fair, 30% chance that it will be poor.

?(?2) = 40 × 0.20 + 20 × 0.50 − 40 × 0.30 = 6% 𝜎2 = √(40 − 6) 2 × 0.20 + (20 − 6) 2 × 0.50 + (−40−6) 2 × 0.30 = √964 = 31.05 Part c) 1 mark for each CV ?𝑉1 = 11.14 6 ≅ 1.857 ?𝑉2 = 31.05 6 ≅ 5.175 The coefficient of variation measures the risk per unit of return. A higher coefficient of variation suggests higher risk for the expected return. Part d) If you are risk-averse, you would prefer to invest in Fund 1 because it has a lower standard deviation and coefficient of variation, indicating less risk for the expected return. Part e) 1 mark: mean 1 mark: standard deviation Fund 1: 12,000/30,000 = 0.4 Fund 2: 15,000/30,000 = 0.5 Cash: 3,000/30,000 = 0.1 E(Z) = 0.4 × E(R1) + 0.5 × E(R2) + 0.1 × 0 = 0.4 × 6% + 0.5 × 6% + 0.1 × 0% = 5.4% Expected annual return = 5.4% × 30,000 = $1,620 ??(?) = √(0.4 2 × 𝑉??(?1)) + (0.5 2 × 𝑉??(?2)) = √(0.4 2 × 11.14 2 ) + (0.5 2 × 31.05 2 ) = 16.14% Risk for the diversified portfolio = 16.14% × 30,000 = $4,842 Question 4. Poisson Distribution (6 marks) Let X be a random variable that follows a Poisson distribution with a mean (μ X ) of 2. a) Write down the formula for the Poisson probability distribution function (PDF) and describe the possible values that the random variable X can take. (1 mark)

b) Starting with the smallest possible value of X, calculate the probability p(X) for each value of X until the probability becomes smaller than 0.001. Summarize these results in a probability distribution table. (2 marks) c) Create a graph to represent the Poisson distribution using the probabilities calculated in Part B. Label your axes and probabilities clearly. (1 mark) d) Calculate the variance 𝜎 𝑥 2 of the distribution. Calculate the standard deviation 𝜎 𝑥 . (1 mark) e) Calculate the interval 𝜇 𝑥 ± 2𝜎 𝑥 . Using the probabilities from Part B, find the likelihood that X will fall within this interval. (1 mark) Solution Part a) 𝑥 is Poisson with 𝜇 = 2 ; ?(𝑥) = 𝑒 −2 (2) 𝑥 𝑥! for 𝑥 = 0, 1, 2, 3, … Part b) X=x P(X=x) 0 0.1353 1 0.2707 2 0.2707 3 0.1804 4 0.0902 5 0.0361 6 0.0120 7 0.0034 Part c) Part d) 0.0000 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0 1 2 3 4 5 6 7 P(X=x) X=x Poisson probability distribution with mean=2