Prelim_Exam1_Solutions

Math 147 B - Prelim Exam 1, October 6th, 2023 Honor Code: I have neither given nor received aid on this exam. I understand that violation of this statement or the Binghamton code of academic integrity is punished at minimum with failure of the course. Printed Name: Signature: Student ID: Instructions: You may bring in a single sheet of paper with formulas etc., and a basic calculator. No cell phones, computers or other devices with a communication capability. The exam consists of 10 sides (5 sheets), which includes this side of instructions, 7 sides of questions, 2 sides with the standard Normal tables. For numeric answers give either decimals with three significant figures or simplified fractions unless otherwise specified in the question. Hand in all pages of the exam, but not your formula sheet or scrap paper. Question Points Out of 1 9 2 11 3 5 4 6 5 14 6 3 7 6 8 6 9 8 1

1. (9 points) A company interested in the health of its employees started a health pro- gram including monitoring blood pressure. Based on age, employees were categorized according to ranges of blood pressure by age intervals. Data are shown in the table below. i) What percent of the employees who have high blood pressure are over 50 years old? 72 / 154 = 0 . 4675324675 = 46 . 75% ii) What percent of all employees have normal or high blood pressure? 80% (384 / 480). iii) What is the conditional distribution of blood pressure type within the Under 30 age interval? Do you think that an employee’s blood pressure type is independent of their age interval? Use statistics to justify your reasoning. Under 30 Low 27/98 = 27.55% Normal 48/98 = 48.98 % High 23/98 = 23.47 % Among employees who are under 30 years old, the percentage who have high blood pressure is 23.47%. It is less frequent than high blood pressure among all employees: the percentage of all employees who have high blood pressure is 154 / 480 = 32% significantly larger. Thus Blood Pressure type does not seem independent of age interval. 2. (11 points) Students taking an intro stats class reported the number of credit hours that they were taking that quarter. Summary statistics are shown in the table. ¯ x 16.65 Median 16 Min 5 Q 1 15 Q 3 19 Max 28 Sd 2.96 i) Suppose that the college charges 73 per credit hour plus a flat student fee of 35 per quarter. For example, a student taking 12 credit hours would pay 35 + 73(12) = 911 for that quarter. 2

3. (5 points) Minus-36 degrees Fahrenheit : That’s how cold it got in Old Forge, New York, on February 4, 2023 according to the NWS, during an arctic cold front. 116 degrees Fahrenheit : That’s how hot it was on June 22, 2023 in Tahoka, Texas — breaking a 110-year-old record, according to the NCEI — during a three-week heat wave. The low temperature in Old Forge, in February, has mean 4 ◦ F and a standard deviation of 10 ◦ F. The high temperature in Tahoka, in June, has mean 91 ◦ F and a standard deviation of 7 ◦ F. Which weather event was the most unusual? z OF = − 36 − 4 10 = − 4 and z Taho = 116 − 91 7 = 3 . 5714. Old Forge! 4. (5 points) The graphs below shows the return (top) and the log return (bottom) com- pared to January 1st 2018 for the closing price of Google (GOOG) in 2018. Financial analysts often use log returns rather than simple returns for various mathematical benefits. Describe how the re-expression of returns using the logarithm changed the histograms. The shape of the original percent returns data was slightly skewed right and somewhat unimodal. Re-expressing the percent returns using the logarithm made the histogram much more symmetric and unimodal, and it reduced the spread substantially. 5. (14 points) You recently acquired refined glass beads from the Corning Museum of Glass. For jewelry purpose, it is recommended to drill holes of diameter between 0.99 4

and 1.01 millimeters (mm), into these 8mm diameter beads. Your power drill pro- duces holes with diameter following a Normal Model with mean 1 . 00mm and standard deviation 0 . 07mm. i) Using your power drill, what percent of the holes drilled would have a diameter greater than the recommended sizes? The z -score is z = (1 . 01 − 1) / 0 . 07 = 0 . 14. We are looking for the area under the standard Normal curve to the right of 0.14, that is P ( Z > 0 . 14) = 1 − P ( Z ≤ 0 . 14). From the standard Normal table, the area under the curve to the left of 0.14 is P ( Z ≤ 0 . 14) = 0 . 5557 and hence P ( Z > 0 . 14) = 0 . 4443. This percentage is 44 . 43%. ii) Using your power drill, what percent of the holes drilled would have a diameter of the recommended size, that is, between 0.99 and 1.01mm? We are looking for the area under the standard normal curve between z 1 = (0 . 99 − 1) / 0 . 07 = − 0 . 14 and z 2 = (1 . 01 − 1) / 0 . 07 = 0 . 14. From the standard Normal table, the area under the curve to the right of − 0 . 14 is P ( Z ≤ − 0 . 14) = 0 . 4443. The area between is P ( Z ≤ 0 . 14) − P ( Z ≤ − 0 . 14) = 0 . 5557 − 0 . 4443 = 0 . 1114. The percentage is only 11 . 14%. iii) If the recommended range of diameter size was (1 − c, 1 + c ), for what value of c would 99.7% of the holes drilled have a diameter of recommended size? We want c satisfying ”Area between 1 − c and 1 + c = 0 . 95”. According to the 68 − 95 − 99 . 7 rule we must have (1 − c − 1) / 0 . 07 = − 3 and (1 + c − 1) / 0 . 07 = 3, which is possible if c = 0 . 21. iv) Assume that you can borrow a power drill which produces holes with diameter following a Normal model with same mean µ = 1 . 00mm but with a smaller stan- dard deviation σ . For what values of σ would you get at least 95% of all drilled beads having a diameter of size between 0.99 and 1.01mm? We would like at least 95% of the z -scores lie within (0 . 99 − 1) /σ and (1 . 01 − 1) /σ . According to the 68 − 95 − 99 . 7 rule we must have (0 . 99 − 1) /σ = − 2 and (1 . 01 − 1) /σ = 2, which is possible if σ ≤ 0 . 005. 6. (3 points) For the following scatterplot, 5

¯ x = 6, ¯ y = 3, s x = 1, s y = 2, r = − 0 . 5 9. (8 points) Doctors are interested in the relationship between the dosage of a medicine and the time required for a patient’s recovery. The following table shows summary statistics of the dosage levels (in grams) and recovery times (in hours) for a sample of 10 patients. Dosage Level Recovery Time Mean 1.34 24.1 Standard Deviation 0.212 10.723 Correlation − 0 . 7306 Suppose we want to predict Recovery Time from Dosage: i) Find the slope estimate, b 1 . What does it mean, in this context? The slope is b 1 = r s y s x = − 0 . 7306 10 . 723 0 . 212 = − 36 . 954 It means that each additional grams of dosage level reduces the recovery time, on average, by 36.954 hours. ii) Find the intercept estimate, b 0 . What does it mean, in this context? The intercept is b 0 = ¯ y − b 1 ¯ x = 24 . 1 − ( − 36 . 954)1 . 34 = 73 . 61836 ≈ 73 . 62 It means that, on average, the recover time for a patient who is not administered that medecine is 73.62 hours. iii) What percent of the variability in the recovery time can be explained by the medecine dosage? R 2 = ( − 0 . 7306) 2 = 0 . 53377636 = ≈ 53 . 4% 7

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