HW3_SOHAIB_KHURAM_09202023

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Mechanical Engineering

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Dec 6, 2023

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BIOS 500 Homework 3 Name: Sohaib Khuram Although you are encouraged to work together in groups, collaboration does not mean copying a classmate’s response to a question. What you turn in must represent your understanding of the concepts learned. 1. Interpreting box plots . Do Exercise 3.133 on p. 126 in your textbook. In general, HB SS patients have lower hemoglobin levels while HB SC and HB ST tend to have higher levels. However, the variation in the values are much more spread out in the HB ST group compared to the SC group as shown by the larger whiskers. 2. Mutually exclusive events . Do Exercise 4.52 on p. 160. (a) (Not A) contains all people older than 20 years old (inclusive) (b) (B or D) contains all people older than 30 years old (inclusive) (c) (A & C) contains all people younger than 20 years old and older than 64 years, which is impossible so this is an empty set (d) A, B, and D are mutually exclusive while the rest are not 3. Conditional probability . Do Exercise 4.106 on p. 178. (a) P (B) = 2/4 = ½; there is a 1 in 2 possibility that the 2 nd toss is going to be heads (b) P (B | A) = P (B & A) / P (A) = ¼ divided by 2/4 = ½ = 50% probability that the 2 nd toss is heads given that the first toss is heads (c) P (B | C) = P (B & C) / P(C) = 2/4 divided by ¾ = 2/3 = 66% probability that the second toss is heads given that at least one toss is heads (d) P (C) = ¾ = 75% probability that at least one toss is heads (e) P (C | A) = P (C & A)/ P (A) = 2/4 divided by 2/4 = 1 = 100% probability that at least one toss is heads given that the 2 nd toss is heads (f) P (C | (not B)) = P (C & (not B)) / P (not B) = ¼ divided by 2/4 = ½ = 50% probability that at least one toss is heads given that the second toss is not heads 4. Decide whether each of the statements below is true or false. 1

(a) The mean of the first test scores is 81. This means that 50% of the class got a score less than 81. False (b) A distribution is skewed to the left. The mean is less than the median. True (c) The mean does not have to be one of the observations in a data set, while the median has to be one of the observations in the data set. TRUE (d) The mean is resistant to outliers; the median is not. FALSE (e) The sum of the deviations from the mean reflects the amount of variability in a data set. FALSE (f) The standard deviation of a variable (i.e., not a constant) is always positive, and it cannot be zero or negative. true (g) If all the observations have the same value, the range, the variance, the standard deviation and the IQR are all equal to zero. TRUE (h) The range, the standard deviation, the IQR (but not the variance) are all expressed in the same units as the measurements used in the data set. TRUE (i) Adding a constant number to each observation in the data set does not change the range, variance, standard deviation and IQR. TRUE (j) If the weights of females measured in pounds are converted to kilograms, the numeric value of the sample standard deviation does not change. FALSE 5. (a) The mean homework score of 20 students is 82. Is it possible to find the sum of the 20 homework scores? If yes, what is it? Yes, because we calculate the mean by adding up all the events and dividing by the number of entries, in this case students. Multiply the mean by the number of students so 82*20 = 1640 (b) The median homework score of 20 students is 82. Is it possible to find the sum of the 20 homework scores? If yes, what is it? No we can’t since median is not related to the sum of events, it’s just the most common one 6. Study the table below which I took from the CDC on incidence of prostate cancer in different age groups among male seniors (70 years old or older) in 2014. The first two columns provide an age distribution for male seniors who did not have prostate cancer before 2014; the third column gives the percentage of new cases of prostate cancer in 2014 in each age group. 2

Age Group (year) % Seniors % Prostate Cancer 70-74 39.0% 0.52% 75-79 30.4% 0.44% 80-84 18.3% 0.34% 85+ 12.3% 0.31% 100.0% The incidence of prostate cancer decreases as age increases. What must we bear in mind when interpreting this statement? The table provides population proportions and does not translate into some connection of causality. What the table is really saying is that as in the older age groups, the proportion of people with prostate cancer is lower but that does not mean that the risk of prostate cancer is lower in older age groups. There’s also survivorship bias to account for as the older age groups might not be diagnosed as they might not be going to the hospital or that the disease could be linked to people not getting to the older age brackets. 7. In an open access data set on 828 COVID-19 patients worldwide, 59% were male and 8% had diabetes. Suppose 3% of the COVID-19 patients were diabetic males. A person is selected at random in this sample. P (Male) or P(M) = probability of being male = .59 P(Diabetes) or P(D) = .08 P(M ∩ D) = Probability of being diabetic and male = .03 (a) What is the probability that the COVID-19 patient is diabetic or male? • P(D) +P(M) – P(D ∩ M) = .08 + .59 - .03 = .64 (b) What is the probability that the COVID-19 patient is a diabetic female? • P(D ∩ F) = P(D) – P(D ∩ M) = .08 - .03 = .05 (c) What is the probability that the COVID-19 patient is a nondiabetic male? • P(M) – P(D ∩ M) = .59 - .03 = .56 (d) What is the probability that the COVID-19 patient in this sample is a non- diabetic female? • P(F) – P(D ∩ F) = 1- P(M) – P(D ∩ F) = 1 - .59 - .05 = .36

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(e) If the COVID-19 patient is diabetic, what is the probability that this person is male? • P(M | D) = P(M & D) \ P(D) = .03/.08 = .375 (f) What percentage of diabetic COVID-19 patients in this sample is male? • (828 * .03)/(828*.08) = .375 (g) What percentage of nondiabetic COVID-19 patients in this sample is male? • 828*.56 / 828 * .92 = .6087 (h) Of the nondiabetic COVID-19 patients, what percentage is female? • 828* .36 / 828 * .92 = .3913 Show your work to receive full credit. (You may use a rule of probability or draw a Venn diagram.) * Source: https://www.diabetesresearchclinicalpractice.com/action/showPdf?pii=S0168-8227%2820%2930545-3 8. Download the HEALTH data set. Report the following statistics on we I ght. (You don’t have to interpret them, but make sure you understand what each statistic is saying.) (a) N: 80 (b) Mean: 159.39 (c) Median: 161 (d) Mode: 135 (e) Variance: 1216.40 (f) standard deviation: 34.88 (g) range (Here we typically write min-max): 94.3(min) – 255.9(max) or 161.6 (h) IQR (Here we typically write Q 1 -Q 3 ) 135 – 179.6 or 44.6

(i) lower limit (Q 1 - 1.5 IQR): 135 – 66.9 = 68.1 (j) upper limit (Q 3 + 1.5 IQR): 179.6 + 66.9 = 246.5 (k) coefficient of variation: 21.88 (l) 10th percentile: 112.3 (m) 90th percentile: 207 (n) Skewness: .39 (o) Kurtosis: -0.115 9. we I ght was measured in pounds (lbs). If you calculate the mean, you should find that it is 159 . 4 lbs. If you convert all the values of we I ght from pounds to kilograms, you should find that the mean weight is going to be 72 . 3 kg. Notice that 159 . 4 = 72 . 3 , but you know that 159 . 4 lbs = 72 . 3 kg. That is, in order to interpret a number that represents the mean weight you necessarily have to attach the unit of measurement. You cannot simply say that the mean is 159.4 or 72.3, for example, and expect people to understand the statistic you are reporting. In other words, when reporting the mean weight, bear in mind that you are not reporting a pure number. You are reporting a statistic that requires a unit of measurement. Which of the following statistics does not require a unit of measurement? In other words, when calculating this statistic bear in mind you are reporting a pure number. (Note that there is more than one answer.) (a) standard deviation (b) range (c) first quartile (d) variance (e) maximum (f) median (g) mode (h) coefficient of variation (i) frequency (j) relative frequency The highlighted values are either proportions or percentages so they do not require units of measurement. 10. Body temperature is not one of the variables in the data set you just analyzed, but suppose this variable is present in the data set and it is measured in ◦ F. Suppose further that the mean temperature is 98 . 17 ◦ F. If you convert the values of temperature from Fahrenheit to Centigrade, you should find that the new mean temperature is 36 . 76 ◦ C. Again, notice that 98 . 17 = 36 . 76 , but you know that 98 . 17 ◦ F = 36 . 76 ◦ C. In other words, just like mean weight, mean temperature cannot be interpreted without attaching a unit of measurement. In the previous question, you should have said that when reporting the coefficient of variation you do not need to ̸̸ ̸̸

attach a unit of measurement – because the coefficient of variation is unitless. (Yes, I am telling you one of the answers to question 9.) However, when you change the temperature values from Fahrenheit to Centi- grade, you will find that the coefficient of variation, albeit a pure number, will change its value. Why is this happening? Because the coefficient of variation is based around the mean and standard deviation that changes during the unit conversion, which usually isn’t a problem if you change a value by a scalar (cm to mm) but since the values are also shifted after scaling, the center and spread change as well, therefor the CV is no longer the same 11. Did you use an artificial intelligence (AI) software to complete this assignment? If you did, describe how you used the tool and write the prompts you used to generate results. (a) I used AI to get the SAS code to show which options I need to put to get the requested statistics (IQR, std dev, frequency)

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