Stat_300_Fa23_Midterm_Exam (1)

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Stat 300 — Introduction to Probability and Statistics Midterm Exam ARC Fall 2023 Chima Sanchez Instructions This exam is designed to assess your learning in the first half of Stat 300 for the Fall 2023 semester. Please complete all questions, showing your work as needed. If you are asked to perform a calculation, show all work and/or explicitly state what functions you used on your TI-84 calculator. For all probability calculations, please express your answer as an exact fraction or decimal number rounded to four places after the decimal, as appropriate, for full credit. You are allowed to use your personal notes and a graphing calculator of your choice, but you may not use the textbook, consult the Internet, use a math helper app such as PhotoMath, use AI such as ChatGPT, or collaborate with others. This exam has 12 questions, totalling 100 points toward your midterm exam grade category. Honesty Statement Before you begin, please read the following academic honesty statement and sign as indi- cated on the line below. Failure to sign will result in a zero grade . I certify that I am submitting my own original work and did not receive assistance from others nor used any unauthorized aids. I understand that if I am found to have violated these rules, my exam grade is automatically assigned a zero and may be reported to the Office of Student Conduct. Signature Date 1

1. Describe the type of data and level of measurement in the variable given: the heights of 21–65 year-old women. 2. Please refer to the following scenario and Table 1: a study was performed by the American Medical Association to measure the effectiveness of a new software program designed to help stroke patients regain their problem-solving skills. Patients were asked to use the software program twice a day, once in the morning and once in the evening. The study observed 200 stroke patients recovering over a period of several weeks. Group Showed Improvement No Improvement Deterioration Used program 105 74 19 Did not use program 89 99 12 Table 1: Summarized data collected from stroke patients at end of study The study sought informed consent from the participants, in which they were warned of the risks of no improvement or deterioration in their condition, and subjects were selected for the treatment group by randomization. What kind of observational study or experimental design was used here? Be as descriptive as possible. 2 The variable "heights of 21-65-year-old women" is quantitative data as it involves numerical measurements and interval level of measurement as it allows for differences to be taken A prospective study involves collecting data moving forward from groups that share common factors (cohorts). This scenario, stroke patients were tracked over time as they used the software program, and their progress was observed to measure the impact of the intervention on their problem- solving skills. The study focused on observing the outcomes as the patients continued using the software over the study period, making it a prospective study or a cohort study.

3. Table 2 contains daily high temperatures for a Northern California town in the month of April: 61 61 62 64 66 67 67 67 68 69 70 70 70 71 71 72 74 74 74 75 75 75 76 76 77 78 78 79 79 95 Table 2: Daily high temperatures for a Northern California town in the month of April. Construct a relative frequency distribution for the data in Table 2 using a class width of 5 and first lower class limit of 60. 4. Table 3 contains the distances between 20 retail stores and a large distribution center. The distances are in miles. 29 37 38 40 58 67 68 69 76 86 87 95 96 96 99 106 112 127 145 150 Table 3: Distances (in miles) between 20 retail stores and a large distribution center. Calculate the mean ¯ x of these distances. Round to the nearest tenth of a mile. 3 class tally of temperatures frequency relative frequency 60-64 65-69 70-74 80-84 85-89 90-94 61, 61, 62, 64 66,67,67,67,68,69 70,70,70,71,71,72,74,74,74,75,75,75 none none 95 4 6 12 9 75-79 75,75,76,76,77,78,78,79,79 0 0 1 4/30 ≈0.133 6/30 ≈ 0.2 12.30 ≈ 0.4 9/30 ≈ .3 0 0 1/30 ≈ 0.0 add up all the distances = 1488 x =1488/20 =74.4 Rounded to the nearest tenth of a mile, the mean distance is 74.4 miles.

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5. For the data in Table 3, calculate the standard deviation s of the distances. Round to the nearest hundredth of a mile. 6. For the data in Table 3, use the Strong Range Rule of Thumb to identify any signif- icant values in the data set. 7. Two balanced dice are thrown and the sum of the rolls is observed. Find the probability that the sum was not 7. 4 squared differences from the mean for each distance total: 12433.6 s ≈. 12433.6/20-1 s ≈. 654.4 s ≈ 25.57 Rounded to the nearest hundredth of a mile, the standard deviation of the distances is approximately 25.57 miles Significantly low values for a population data set: μ −2 σ= 74.4 − 2 × 25.57 =74.4 − 51.14 ≈23.26 Significantly high values for a population data set: μ+2σ= 74.4 + 2 × 25.57 = 74.4 + 51.14 ≈ 125.54 All values in the dataset ranging from 29 miles to 150 miles fall within the range of 23.26 to 125.54 miles. Therefore, based on this definition of the Strong Range Rule of Thumb, there are no significantly low or significantly high values in the dataset. Outcomes that result in a sum of 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) Therefore, the number of combinations that do not sum up to 7 is 36 −6=30. The probability that the sum of the rolls is not 7 is the number of outcomes that do not sum up to 7 divided by the total number of outcomes: Probability = Total number of outcomes/Number of outcomes not summing to 7 = 30/36 = 5/6 Therefore, the probability that the sum of the rolls is not 7 is 5/6

8. Please refer to the table below. Registered to vote Not registered Total Aged 18-40 23 16 39 Aged 41+ 44 12 56 Total 67 28 95 Table 4: Contingency table detailing the voter registration status and age of residents in a community census. Let A represent the event in which a person selected at random is aged between 18 and 40 and let B represent the event in which a person selected at random is registered to vote. a) Find P ( A ∪ ¯ B ) . b) Find P ( A | B ) . c) Are A and B independent as probabilistic events? Justify your answer. 5 P(A ∪ B)=P(A)+P( B )−P(A∩B) P(A) is the probability of being aged between 18-40. P( B) is the probability of not being registered, which is 1 − P(B). P(A ∩B) is the probability of both being aged 18-40 and registered to vote. P(A) ≈ 0.4105 P(B) ≈ 0.7053 P(B)=1 − P(B)=1 −0.7053= 0.2947 P(A ∩B)≈ 0.2421 P(A ∪B)=0.4105+0.2947−0.2421≈ 0.4631 P(A ∪B)≈ 0.4631, which is the probability of being either aged 18-40 or not being registered to vote. P(A ∩B)≈0.2421 (probability of being both aged 18-40 and registered to vote) P(B) ≈0.7053 (probability of being registered to vote) P(A ∣B)= P(B)/P(A∩B) = 0.7053/0.2421 ≈0.3431 P(A ∣B)≈0.3431, which is the probability of a person being aged between 18-40 given that the person is registered to vote. In the case of independent events, if P(A ∩B)=P(A)×P(B), the events are independent. P(A ∩B)≈0.2421 is not equal to P(A)×P(B) ≈0.2893, which indicates that A and B are not independent events.

9. It’s estimated that 12% of the U.S. population has myopia, or nearsightedness. Suppose a random sample of 15 people is drawn from the U.S. population using telephone polling and X represents the number of people in the sample who have myopia. a) What is the name and parameter(s) of the distribution of X ? Specify the numerical values of the parameters. b) Find the probability that between 2 and 4 people in the sample have myopia. 10. For each part, find the probability and sketch the normal distribution density curve along with the shaded area corresponding to the probability. a) Find P ( Z ≤ 2 . 72) b) Find P ( X ≥ 230) when X is normally distributed with a mean of 200 and standard deviation of 15. 6 n = the number of trials or sample size p = the probability of success in a single trial In this case: n = 15 (the sample size) p = 0.12 (the probability of a person having myopia) The distribution of X, the number of people in the sample who have myopia, follows a binomial distribution with parameters n = 15 and p = 0.12. n is the number of trials (sample size) = 15 p is the probability of success (probability of having myopia) = 0.12 P(X<2)=P(X=0)+P(X=1) P(X<2) =(15/0)× 0.12 × (1 −0.12) ^15+( 15/1)×0.12^1×(1−0.12)^14 P(X<2)=0.0423+0.1897 ≈0.232 P(X ≤4)= (15/0)×0.12^0×(1−0.12)^15+(15/1)×0.12^1×(1−0.12)^14+(15/2)×0.12^2×(1−0.12)^13 + (15/3)×0.12^3×(1 −0.12)^12+(15/4)×0.12^4 ×(1−0.12)^11 P(X ≤4)=0.0423+0.1897+0.3115+0.2824+0.1468≈0.9727 P(2 ≤X≤4)=P(X≤4)−P(X<2) P(2 ≤X≤4)≈0.9727−0.232≈0.7407 The probability that between 2 and 4 people in the sample have myopia is approximately 0.7407. P(Z ≤2.72)≈0.9968 P(Z ≤2.72)≈0.9968 z= 230 −200/15 = 30/15 =2 P(X ≥230) by looking up the area to the right of z=2 in the standard normal distribution table or by using a calculator that provides cumulative probabilities for the standard normal distribution. standard normal distribution table or a calculator, the probability P(X ≥230) is approximately 0.0228 or 2.28%. 230 mean 200

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11. The age of Twitch stream viewers is normally distributed with a mean of 19 years old and a standard deviation of 3 years. Find the 90th percentile of viewer age. 12. In an observational study, the number of daily spam calls received to the cell phones of 50 randomly selected participants was recorded and found a mean of 6 and standard deviation of 2.4. What is the probability that in a similar sample of 50 people, the mean number of daily spam calls is at least 7? 7 X is the value you want to find (the 90th percentile in this case). μ is the mean (19 years). σ is the standard deviation (3 years). 1.28= X-19/3 X −19=1.28×3 X −19=3.84 X ≈22.84 the 90th percentile of the viewer age on Twitch is approximately 22.84 years old. X = value you're interested in (in this case, 7) μ = population mean σ = population standard deviation n = sample size Given: X=7 μ=6 (sample mean) σ=2.4 (sample standard deviation) n=50 Z= 7 −6/2.6/ 50 Z=1/2.4/ 50 Z= 2.4/7.071 Z ≈ 1/1.76875 Z ≈0.5658 the probability that in a similar sample of 50 people, the mean number of daily spam calls is at least 7 is approximately 28.61%. Z=0.5658 is approximately 0.2861 or 28.61%. Z=X- μ n σ n

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