Solution2_Winter 2024 V5 PDF

STAT 151 1 STAT 151 Assignment 2 Solutions Due date : refer to the Course Outline Purposes This assignment has two parts. The following questions assess your ability to identify the sample space of a chance experiment, calculate probabilities using the equally likely outcome model, and the addition, complementation, conditional probability, and multiplication rules, determine if two events are independent by calculation, and apply counting rules. This assignment also assesses your understanding of discrete probability distributions, your ability to find the mean (expected value) and the standard deviation of a discrete random variable and your ability to identify and work with binomial random variables. The second part assesses your ability to use R commander to compute the probabilities listed above. Instructions For every assignment in this course, you are required to complete the questions or tasks in Part A by hand. This means that to do any calculation or drawing, you will NOT use R commander or any computer application. That is, you are meant to do the calculations manually with a non- programmable scientific calculator and use a pen or pencil to draw figures or build a distribution table on paper (or on an iPad/tablet). Then you will submit a photo of your written solution using the appropriate submission box on the corresponding Crowdmark submission page. Before you complete Part B using R commander, you should read and practice the R commander steps by following the related examples in the Lab Manual and the Demos, which you can download via a link in the Course Content folder on mêskanâs. Where appropriate, units should be included in your answer. Show your calculations fully and provide a concluding sentence to your problems. Part A 1. The handedness of a group of 3 people are Ambidextrous (A), Lefthanded(L), or Righthanded(R). Suppose two people are randomly selected with replacement (that is, a person is selected and their handedness is observed. This person then returns to the group). Let A be the event of selecting an ambidextrous person, L be the event of selecting a lefthanded person, and R be the event of selecting a righthanded person. (a) List all possible outcomes. (2 marks) There are nine possible outcomes S = {AA, AL, AR, LL, LA, LR, RR, RA, RL}. (b) List all possible outcomes for each of the following events and find the corresponding probabilities. (8 marks)

STAT 151 2 i. E1 = {Exactly 1 person of type L is drawn} ii. E2 = {The second person is right handed} iii. E3 = {At MOST one person is ambidextrous} iv. E4 = {Both people have the same handedness} i. E1= {AL, LA, LR, RL} . Based on the 𝑓 𝑁 rule, ?(?1) = 4 9 ≃ 0.444 ii. E2 = {AR, LR, RR} . Based on the 𝑓 𝑁 rule, ?(?2) = 3 9 ≃ 0.333 iii. E3 = {AL, AR, LL, LA, LR, RR, RA, RL} . Based on the 𝑓 𝑁 rule, ?(?3) = 8 9 ≃ 0.889 iv. E4 = {AA, LL, RR} . Based on the 𝑓 𝑁 rule, ?(?4) = 3 9 ≃ 0.333 (c) List all possible outcomes and find the probabilities of the following events. (12 marks: 2 for each) i. Not E3 ii. E1 & E3 iii. E1 & E4 iv. E2 & E3 v. E2 & E4 vi. E3 or E4 i. Not E3 = {outcomes in S but not in E3} = {AA}. Based on the 𝑓 𝑁 rule, ?(??? ?3) = 1 9 ≃ 0.111 . Or by the complement rule, ?(??? ?3) = 1 − ?(?3) = 1 − 8 9 = 1 9 ≃ 0.111 . ii. E1 & E3 ={ elements common to both E1 and E3 }= {AL, LA, LR, RL} . Based on the 𝑓 𝑛 rule, ?(?1&?3) = 4 9 ≃ 0.444 . iii. E1 & E4 = {elements common to both E1 and E4} = ∅ . Therefore, ?(?1&?4) = 0 . iv. E2 & E3 = {elements common to both E2 and E3} = { AR, LR, RR }. Therefore, ?(?2&?3) = 3 9 ≃ 0.333 v. E2 & E4 = {elements common to both E2 and E4} = {RR} . Therefore, ?(?2&?4) = 1 9 ≃ 0.111 . vi. E3 or E4 = {elements in either E3 or E4, or both E3 and E4} = { AA, AL, AR, LL, LA, LR, RR, RA, RL } . Based on the 𝑓 𝑛 rule, ?(?3 ?? ?4) = 9 9 = 1.000 . Or based on the general addition rule ?(?3 ?? ?4) = ?(?3) + ?(?4) − ?(?3&?4) = 8 9 + 3 9 − 2 9 = 9 9 = 1.000 . Note that the previous calculation required P(E3&E4)=P({ LL, RR }) = 2 9 . (d) Identify all possible pairs of events defined in part (b) that are mutually exclusive. (3 marks) The intersection of E1 and E2 is E1 & E2={LR}. Hence, E1 and E2 are not mutually exclusive. The intersection of E1 and E3 is E1 & E3 ={ AL, LA, LR, RL }. Hence, E1 and E3 are not mutually exclusive. The intersection between E1 and E4 is empty. Hence, E1 and E4 are mutually exclusive.

STAT 151 4 Later in the course we will see how these figures compare to the figures found in Table 3 in the 2016 research paper “Footedness Is Associated with Self -reported Sporting Performance and Motor Abilities in the General Population”, by Ulrich S. Tran and Mar tin Voracek https://www.frontiersin.org/articles/10.3389/fpsyg.2016.01199/full , but for now we will limit ourselves to questions regarding sampling choices. (a) How many different samples are possible? (2 marks) Since we are choosing 9 students from a group of size 1000, the number of possible samples of size 9 is 1000? 9 = 2 ,658 ,017 ,764 ,500 ,203 ,964 ,000 . (b) How many different samples of size 9 are possible subject to the constraint that no 2 students may have the same handedfooted result? (2 marks) 600 are RHRF, 265 are RHMF, 31 are RHLF, 3 are MHRF, 17 are MHMF, 4 of them are MHLF, 14 of them are LHRF, 19 of them are LHMF, and 47 of them are LHLF. Since each student must have a different handedfooted result we must choose exactly 1 student from each of the 9 handedfooted results. The number of possible ways for this to occur is: 600? 1 × 265? 1 × 31? 1 × 3? 1 × 17? 1 × 4? 1 × 14? 1 × 19? 1 × 47? 1 = 12,570,961,032,000 (c) What is the probability that a random sample of 9 students from this group has no two students with the same handedfooted result? (2 marks) There are 2,658,017,764,500,203,964,000 possible samples of size 9 and 12,570,961,032,000 samples with no two students having the same handedfooted result. Therefore, the probability of obtaining a sample of 9 students with no two students having the same handedfooted result is: ?(?? ??? ???????? ?ℎ𝑎?? ?𝑎?? ℎ𝑎?????????? ??????) = ? ? = 12570961032000 2658017764500203964000 ≃ 0.0000000047294495920585994121723656222123 4. A certain lottery sells 10 million tickets for $2 each. Let X denote your winnings upon purchasing 1 ticket, and suppose X has the following probability distribution ? ?(? = ?) $2,000,000 1 10,000,000

STAT 151 5 $100,000 10 10,000,000 $1,000 20 10,000,000 $0 9,999,969 10,000,000 (a) What is the expected value of X. (2 marks) The expected value (mean) of ? is given by 𝜇 = ∑??(? = ?) = 2000000 ⋅ 1 10000000 + 100000 ⋅ 10 10000000 + 1000 ⋅ 20 10000000 + 0 ⋅ 9,999,969 10000000 = 2000000 + 1000000 + 20000 + 0 10000000 ≃ 0.302 Interpretation: if you purchase a single $2 ticket, you are expected to receive a return of about 30.2¢. (b) Since each ticket costs $2, define Y=X-2 to be your profit from purchasing 1 ticket. Compute and interpret the expected value of Y. (2 marks) Since the expected value of ? is 0.302, it follows that the expected value of ? is 0.302 – 2 =-1.698 Interpretation: You are expected to lose about $1.698 for each ticket you purchase. c) What is the probability that you have no winnings if you purchase a ticket? (So you win $0 on your ticket) (1 mark) probability of winning $0 on 1 ticket is ?(? = 0) = 9,999,969 10,000,000 = 0.9999969 d) Suppose you purchase 30 tickets. Let T be the total winnings among all 30 tickets. What is the probability that you have 30 losses? (So no ticket wins any money). For simplicity, assume independence. (2 marks) It follows from the special multiplication rule (for independent events) that the probability of 30 losses is ?(? = 0) = ?(? = 0) 30 = ( 9,999,969 10,000,000 ) 30 = (0.9999969) 30 ≃ 0.999907004

STAT 151 7 = 20(0.4) + 15(0.3) + 10(0.3) = 8 + 4.5 + 3 = 15.5 pairs ?(?????𝑖?) = 16 ⋅ ?(????) + 12 ⋅ ?(???𝑖??)+ 8 ⋅ ?(?????) = 16(0.4) + 12(0.3) + 8(0.3) = 6.4 + 3.6 + 2.4 = 12.4 pairs b) Will the Queen have enough gloves for 30 days, if she needs a new pair every day? (2 marks) Yes, even in the worst case scenario, she will have exactly 30 pairs of gloves (12 + 10 + 8). c) Compute the standard deviation for Eugenie’s gloves. (3 marks) Students can compute the standard deviation using either the computing formula or the defining formula. Defining formula: ??(?????𝑖?) = √ 𝛴(? − 𝜇) 2 ?(? = ?) = √(16 − 12.4) 2 ?(????) + (12 − 12.4) 2 ?(???𝑖??) + (8 − 12.4) 2 ?(?????) = √(3.6) 2 ⋅ 0.4 + (−0.4) 2 ⋅ 0.3 + (−4.4) 2 ⋅ 0.3 = √5.184 + 0.048 + 5.808 = √11.04 = 3.323 ?𝑎𝑖?? Computing formula: ??(?) = √𝛴? 2 ?(? = ?) − 𝜇 2 = √16 2 ?(????) + 12 2 ?(???𝑖??) + 8 2 ?(?????) − 12.4 2 = √256 ∙ 0.4 + 144 ∙ 0.3 + 64 ⋅ 0.3 − 153.76 = √102.4 + 43.2 + 19.2 − 153.76 = √11.04 = 3.323 pairs 6.Linus Ullmark of the Boston Bruins had the best save percentage of any goalie in the 2022- 2023 season. We take this to mean, loosely, that Ullmark has a 93.8% chance of making a save when a puck is shot at his goal. In a practice in the 2023-2024 season, a fellow Bruin makes 20 shots on Ullmark.

STAT 151 8 a) What is the probability that Ullmark will make saves on more than 18 shots on goal? (4 marks) Let ? be the number of goals that Ullmark will save. Then ? follows a binomial distribution with ? = 20 and ? = 0.938 with ?(? = ?) = ?? 𝑥 ? 𝑥 (1 − ?) 𝑛−𝑥 = 20? 𝑥 0.938 𝑥 (1 − 0.938) 20−𝑥 ,? = 0,1, ⋯,20 . Thus, ?(? > 18) = ?(? ≥ 19) = ?(? = 19) + ?(? = 20) = 20? 19 0. 938 19 (1 − 0.938) 20−19 + 20? 20 0.938 20 (1 − 0.938) 20−20 = 20(0.938) 19 (0.062) + (1)(0.938) 20 (1) = 0.367515461 + 0.278007663 = 0.645523124 Thus the probability that Ullmark makes saves on more that 18 shots on goal is 0.645523124 b) What is the probability that he will miss stopping (allow) at least 2 of the shots on goal? (4 marks) Let ? be the number of goals that Ullmark will miss stopping (allow). Then ? follows a binomial distribution with ? = 20 and ? = 0.062 with ?(? = ?) = ?? 𝑥 ? 𝑥 (1 − ?) 𝑛−𝑥 = 20? 𝑥 0.062 𝑥 (1 − 0.062) 20−𝑥 ,? = 0,1, ⋯,20 . ?(? ≥ 2) = 1 − ?(? ≤ 1) = 1 − ?(? = 1) − ?(? = 0) = 1 − 20? 1 0.062 1 (1 − 0.062) 20−1 − 20? 0 0.062 0 (1 − 0.062) 20−0 = 1 − 20(0.062)(0.938) 19 − 0. 938 20 = 1 − 0.367515461 − 0.278007663 = 0.354476876 Thus, the probability that he will miss stopping (allow) at least 2 of the shots on goal is 0.354476876. c) How many shots on goal do you expect Ullmark to stop? (2 marks) Since X follows a binomial distribution with n=20 and p=0.938 it follows that the mean is: 𝜇 = ?? = 20 × 0.938 = 18.76 goals. Hence, we expect Ullmark to stop 18.76 shots on goal out of 20 shots on goal. d) Obtain the standard deviation of the number of shots on goal that Ullmark will stop. (2 marks)

STAT 151 10 ?(?𝐻) = 1026 12720 = 0.081 (c) Find, by hand, the probability that a randomly selected person is left handed and right footed. (2 marks) ?(?𝐻&??) = 177 12720 = 0.014 (d) Find, by hand, the probability that a randomly selected person is left handed or right footed. (3 marks) ?(?𝐻 ?? ??) = ?(?𝐻) + ?(??) − ?(?𝐻&??) = 1026 12720 + 7838 12720 − 177 12720 = 8687 12720 = 0.683 (e) Find, by hand, the probability that a randomly selected left-handed person is mixed footed. That is, find the probability that a randomly selected person is mixed-footed given they are left-handed. (3 marks) ?(??|?𝐻) = ?(??&?𝐻) ?(?𝐻) = 247/12720 1026/12720 = 247 1026 = 0.241. (f) Find, by hand, the probability that a randomly selected mixed-footed person is left-handed. (3 marks) ?(?𝐻|??) = ?(??&?𝐻) ?(??) = 247/12720 3839/12720 = 247 3839 = 0.064 (g) Are events MF and LH independent? Explain your answer by hand. (2 marks) ?(?𝐻) = 0.081 , and ?(?𝐻|??) = 0.064 → ?(?𝐻|??) ≠ ?(?𝐻) MF and ?𝐻 are not independent. Alternatively, ?(?𝐻&??) = 247 12720 = 0.019, and ?(?𝐻)?(??) = 1026 12720 ⋅ 3839 12720 = 0.024 → ?(?𝐻&??) ≠ ?(?𝐻)?(??). Once again, MF and ?𝐻 are not independent. (h) Are events ?𝐻 and ?? mutually exclusive? Explain your answer by hand. (2 marks) No, since ?(?𝐻&??) = 247 12720 = 0.019 ≠ 0 , they are not mutually exclusive. 2.The data set HEARTFAILUREPREDICTION looks at several variables that play a role in heart failure prediction for 918 people in the United States. Please download this dataset from the

STAT 151 11 data files for Assignment 2 that can be found online. ( For the curious, this set of open data can be found at https://www.kaggle.com/datasets/fedesoriano/heart-failure-prediction .) Columns found in the dataset are as follows. 1. Age: age of the patient [years] 2. Sex: sex of the patient [M: Male, F: Female] 3. ChestPainType: chest pain type [TA: Typical Angina, ATA: Atypical Angina, NAP: Non- Anginal Pain, ASY: Asymptomatic] 4. RestingBP: resting blood pressure [mm Hg] 5. Cholesterol: serum cholesterol [mm/dl] 6. FastingBS: fasting blood sugar [1: if FastingBS > 120 mg/dl, 0: otherwise] 7. RestingECG: resting electrocardiogram results [Normal: Normal, ST: having ST-T wave abnormality (T wave inversions and/or ST elevation or depression of > 0.05 mV), LVH: showing probable or definite left ventricular hypertrophy by Estes' criteria] 8. MaxHR: maximum heart rate achieved [Numeric value between 60 and 202] 9. ExerciseAngina: exercise-induced angina [Y: Yes, N: No] 10. Oldpeak: oldpeak = ST [Numeric value measured in depression] 11. ST_Slope: the slope of the peak exercise ST segment [Up: upsloping, Flat: flat, Down: downsloping] 12. HeartDisease: output class [1: heart disease, 0: Normal] Use R commander to obtain the most appropriate table to answer the following questions. For each question, copy or take a screenshot of the output in R commander and submit it with your solutions to each question. To save space, you only need to copy and paste what is asked for in the questions and should adjust the size of the image when appropriate. Hint: construct a frequency distribution in part (a), and contingency tables in parts b through f. (a) If we randomly select a person, what is the probability that they are a female? (2 marks) COMMANDS: Statistics > Summaries > Frequency Distributions Variable: Sex OK

STAT 151 13 (e) If we randomly select a person, what is the probability they are asymptomatic in chest pain or have no exercise-induced angina, given they are male? (4 marks) There are a number of ways to solve this question. The following solution is obtained by constructing a multi-way table with ChestPainType as the row variable, ExerciseAngina as the column variable, and Sex as the control variable. ?(????????𝑎?𝑖? ?? ??|?𝑎??) = 162 + 264 + 101 + 104 + 30 162 + 264 + 101 + 12 + 104 + 46 + 30 + 6 = 661 725 = 0.912 (f) If we randomly select a male with no exercise-induced angina, what is the probability that they have asymptomatic chest pain(4 marks) There are a number of ways to solve this question. The following solution is obtained by constructing a multi-way table with ChestPainType as the row variable, ExerciseAngina as the column variable, and Sex as the control variable. P(ASY|No & Male) = 162 162+101+104+30 = 162 397 = 0.408

STAT 151 14 3.Linus Ullmark of the Boston Bruins had the best save percentage of a goalie in the 2022-2023 season. We take this to mean, loosely, that Ullmark has a 93.8% chance of making a save when a puck is shot at his goal. In a practice for the 2023-2024 season, a fellow Bruin will make 20 shots on Ullmark. a) What is the probability that Ullmark will miss saving (and hence allow) at most 4 of the shots on goal? (2 marks) Let ? be the number of shots on goal that Ullmark will miss saving and hence allow a goal. Then ? follows a binomial distribution with ? = 20 and ? = 0.062 with ?(? = ?) = ?? 𝑥 ? 𝑥 (1 − ?) 𝑛−𝑥 = 20? 𝑥 0.062 𝑥 (1 − 0.062) 20−𝑥 ,? = 0,1, ⋯,20 . Find P(X < = 4) using the commands below. COMMANDS: Distributions>Discrete Distributions>Binomial Distribution>Binomial Tail Probabilities Variable Value: 4 Binomial Trials 20 Probability of Success 0.062 Lower Tail, OK Therefore, the probability that Ullmark will allow at most 4 goals is 0.9935306 . b) What is the probability that he will save (stop) no more than 17 of the shots on goal? (3 marks) Let ? be the number of shots on goal that Ullmark will save (stop). Then ? follows a binomial distribution with ? = 20 and ? = 0.938 with ?(? = ?) = ?? 𝑥 ? 𝑥 (1 − ?) 𝑛−𝑥 = 20? 𝑥 0.938 𝑥 (1 − 0.938) 20−𝑥 ,? = 0,1, ⋯,20 . Find P(X <=17) using the commands below.