2023.11.11. Assignment No. 4 - Answer Key

ECON 245: Descriptive Statistics and Probability Dr. Paola Beneras P. Due: Nov. 21, 2023 Assignment No. 4 - Answer Key Instructions: Ensure your assignment is legible and that you staple your assignment before handing it in. Show your work. Question 1. Education and Household Income. The following table shows the number of households (in 1000s) and the household income by the highest level of education for the head of household. Only households in which the head has a high school diploma or more are included. Household Income Highest Level of Education Under $ 25,000 $ 25,000-49,999 $ 50,000-99,999 $ 100,000+ High School Degree 9880 9970 9441 3482 Bachelor’s Degree 2484 4164 7666 7817 Master’s Degree 685 1205 3019 4094 Doctoral Degree 79 160 422 1076 a. Develop a joint probability table. Household Income Highest Level of Education Under $ 25,000 $ 25,000-49,999 $ 50,000-99,999 $ 100,000+ Total High School Degree 0.1505 1519 0.1438 0.0530 0.4993 Bachelor’s Degree 0.0378 0.0634 0.1168 0.1191 0.3371 Master’s Degree 0.0104 0.0184 0.0460 0.0624 0.1371 Doctoral Degree 0.0012 0.0024 0.0064 0.0164 0.0265 Total 0.2000 0.2361 0.3130 0.2509 1.0000 b. What is the probability that the head of one of those households has a master’s degree or more education? Identify what kind of probability you are calculating. - This is the sum of two marginal probabilities: P (master’s degree or more education) = 0 . 1371 + 0 . 0265 = 0 . 1636 c. What is the probability that a household headed by someone with a high school diploma earns $ 100,000 or more? Identify what kind of probability you are calculating. - This is a conditional probability: P (100 k or more | HS) = P (100 k or more ∩ HS) P (HS) = 0 . 0530 0 . 4993 = 0 . 1016 d. What is the probability had one of these households has an income below $ 25,000? Identify what kind of probability you are calculating. - This is a marginal probability: P (under 25) = 0 . 2000 1

e. What is the probability that a household headed with someone with a bachelor’s degree earns less than $ 25,000? Identify what kind of probability you are calculating. - This is a conditional probability: P (under 25 | BD) = P (under 25 ∩ BD) P (BD) = 0 . 0378 0 . 3371 = 0 . 1121 f. Is household income independent of education level? - No. Household income is not independent of education level. People who have a HS degree have a lower probability of having a higher household income: P (100 k or more | HS) = 0 . 1016 ̸ = P (100 k or more) = 0 . 2509 Question 2. Finding Gold in the Artic. A mining company is in negotiations for a contract in Northern Canada. Preliminary geologic studies have computed the following probabilities for the site: P (high-grade gold) = 0 . 50 P (low-grade gold) = 0 . 20 P (no gold) = 0 . 30 Using these probabilities, calculate the following: a. What is the probability of finding gold? - P (AU) = 0 . 50 + 0 . 20 = 0 . 70 After exploring the first site, soil samples are taken noting the concentration of gold in soil. The probabilities of finding a particular type of soil identified by the sample are: P (soil | high-grade gold) = 0 . 20 P (soil | low-grade gold) = 0 . 80 P (soil | no gold) = 0 . 20 b. How should the mining company interpret these samples? How would you revise probabilities (explain in plain English, step by step)? What is the new probability of finding gold? Would you make a bid to mine this particular site? - Let, S = soil test results; AU i = quality of gold, where: i = 1: represents high-grade and i = 2: represents low-grade, and i = 3: represents no gold: Events P ( AU i ) P ( S | AU i ) P ( AU i ∩ S ) P ( AU i | S ) High-grade ( AU 1 ) 0.50 0.20 0.10 0.3125 Low-grade ( AU 2 ) 0.20 0.80 0.16 0.5000 No gold ( AU 3 ) 0.30 0.20 0.06 0.1875 1.00 P(S) = 0.32 1 - The prior probabilities are P ( AU i ). - The quality of the soil depends and varies with the grade of gold found, which gives the conditional proba- bilities P ( S | AU i ). - The probability of finding a particular type of soil is given by P ( S ). – This probability is found by looking at the probability of each experimental outcome, i.e., at the sum of the three joint probabilities. - Trying to identify the type of gold given a particular soil sample becomes looking for the posterior probabilities P ( AU i | S ) 2

b. What is the probability that the player will make at least one shot? - There are three ways in which the player can make at least one shot: – make the first shot and miss the second shot; – miss the first shot and make the second shot; – make both shots. - Event “Miss the Shot” is the complement of the event “Make the Shot,” so: P (Miss the Shot) = 1 − P (Make the Shot) = 1 − 0 . 93 = 0 . 07 • So the probability of making at least one shot is 99.51%: P (Miss the Shot) P (Miss the Shot) = (0 . 93)(0 . 07) = 0 . 0651 P (Miss the Shot) P (Miss the Shot) = (0 . 07)(0 . 93) = 0 . 0651 P (Miss the Shot)(Miss the Shot) = (0 . 93)(0 . 93) = 0 . 864 0 . 0651 + 0 . 0651 + 0 . 864 = 0 . 9951 c. What is the probability that the player will miss both shots? - This can be found in two ways. Directly: P (Miss the Shot) P (Miss the Shot) = (0 . 07)(0 . 07) = 0 . 0049 - Or can recognize that the event “Miss both Shots” is the complement of the event “Make at Least One of the Two Shots” so: P (Miss the Shot) P (Miss the Shot) = 1 − 0 . 9951 = 0 . 0049 d. Late in a basketball game, a team often intentionally fouls an opposing player in order to stop the clock. The strategy tends to be intentionally fouling the other team’s worst free-throw shooter. Assume that the team’s worst free-throw shooter makes 58% of his free-throw shots. Calculate the probabilities for this player for parts (a), (b), and (c) again, assuming two shots. Is intentionally fouling this player (who makes 58%) of the free throws a better strategy than intentionally fouling the player who makes 93% of the free throws? - For the player who makes 58% of free throws: P (Make the Shot) = 0 . 58 for each foul shot. – So probability that this player will make two consecutive foul shots is: P (Make the Shot) P (Make the Shot) = (0 . 58)(0 . 58) = 0 . 3364 . – Again, there are ways to make at least one shot: the player can make the first shot and miss the second shot, miss the first shot and make the second shot, or make both shots. – Because the event “Miss the Shot” is the complement of the event “Make the Shot”: P (Miss the Shot) = 1 − P (Make the Shot) = 1 − 0 . 58 = 0 . 42 . – Therefore: the probability of making at least one shot is 82.36%: P (Make the Shot) P (Miss the Shot) = (0 . 58)(0 . 42) = 0 . 2436 P (Miss the Shot) P (Make the Shot) = (0 . 42)(0 . 58) = 0 . 2436 P (Make the Shot) P (Make the Shot) = (0 . 58)(0 . 58) = 0 . 3364 - Can again find the probability the 58% free-throw shooter will miss both shots in two ways, directly: P (Miss the Shot) P (Miss the Shot) = (0 . 42)(0 . 42) = 0 . 1764 – Or recognize that the event “Miss Both Shots” is the complement of the event “Make at Least One of the Two Shots,” so P (Miss the Shot) P (Miss the Shot) = 1 − 0 . 9951 = 0 . 1764 – Therefore, intentionally fouling the 58% free-throw shooter is a better strategy than intentionally fouling the 93% shooter. 4

Question 6. Mortgage Defaults. A local bank reviewed their mortgage portfolio given the current interest rates. In the past, approximately 5% of households with mortgages defaulted. Management, thus, established a prior probability of 0.05 that any particular household will default on their mortgage. The bank also discovered that for customers who do not default, the probability of missing a monthly payment is 0.20. The probability of missing a monthly payment for households who default on their mortgage is 1. a. Assume a household missed one or more monthly payments, what is the probability they will default on their mortgage? Let, M = missed payment D 1 = customer defaults D 2 = customer does not default So, P ( D 1 ) = 0 . 05 P ( D 2 ) = 0 . 95 P ( M | D 2 ) = 0 . 20 P ( M | D 1 ) = 1 . 00 Therefore, P ( D 1 | M ) = P ( D 1 ) P ( M | D 1 ) P ( D 1 ) P ( M | D 1 ) + P ( D 2 ) P ( M | D 2 ) = (0 . 05)(1 . 00) (0 . 05)(1 . 00) + (0 . 95)(0 . 20) = 0 . 21 b. The bank is considering changing the policies calling the loans and demanding full payment if the probability that a household defaults is greater than 0.20. If the policy changes, would the bank recall mortgages of households that miss a monthly payment? Why or why not? - Yes, the probability of default is greater than 0.20. Question 7. Random Variables. List 5 experiments (that you experience on your daily life) and associate them to a random variable. In each example, identify the values that the random variable can assume. Identify if the random variable is discrete or continuous and ensure you include at least two continuous variables. Example: Experiment: completing this 10-question assignment. Random Variable (x): number of questions an- swered correctly. Values: 0, 1, 2, ..., 10 and it is a discrete random variable. Question 8. Operating Room Use. The following data were collected by the number of operating rooms in the Royal Jubilee Hospital over a 20-day period. On three of the days, only one operating room was used, on five of the days two rooms were used, on eight of the days three rooms were used, and on four days all four of the hospital’s operating rooms were used. a. Use the relative frequency approach to construct an empirical discrete probability distribution for the number of operating rooms in use on any given day. x f(x) 1 3/20 = 0.15 2 5/20 = 0.25 3 8/20 = 0.40 4 4/20 = 0.20 Total 1.00 b. Draw a graph of the probability distribution. 5

- E ( y ) = 2 . 0447. The expected value or mean number of cups per day for adults that drink at least one cup of coffee on an average day is 2.05, or approximately a mean of two cups per day. As expected, the mean is somewhat higher when we only take into account adults who drink at least one cup of coffee per day. Question 10. Random Variable. The following table provides a probability distribution of random variable, x . x f(x) 3 0.25 6 0.50 9 0.25 For the following, show all steps and explain, with words, what is being calculated in each step. a. Compute the expected value of x . - The expected value is the sum of the values of the random variable using the probabilities of each (given by f ( x )) as weights. E ( x ) = 6 b. Compute the variance of x . - The variance is calculated by computing the squared differences between each value of the random variable and its expected value (mean) and averaging these squared differences, again, using the probabilities of each value as weights. V ar ( x ) = 4 . 5 c. Compute the standard deviation of x . - The standard deviation is simply the square root of the variance. Since we use squared deviations when computing the variance, the units for it are squared, the standard deviation allows us to have the same units as the variable. σ ( x ) = 2 . 12 - Calculations: x f ( x ) x · f ( x ) x − µ ( x − µ ) 2 ( x − µ ) 2 · f ( x ) 3 0.25 0.75 -3 9 2.25 6 0.50 3 0 0 0 9 0.25 2.25 3 9 2.25 1 .00 µ = 6 σ 2 ( x ) = 4.5 σ ( x ) = 2 . 12 7