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Indiana University, Purdue University, Indianapolis *

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Statistics

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Feb 20, 2024

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PBHL-B300 Homework #3 1. According to a 2017 Census Bureau survey, 12.6% of Americans are receiving Food Stamp assistance from the government, and 10.5% of Americans have no health insurance. Further, 7.7% of Americans fall into both categories. Using these rates, answer the following: (a) Are receiving Food Stamps and having no health insurance disjoint events? Explain. No, disjoint means they cannot coexist as a result (can’t pass and fail example). You can be both receiving food stamps and be uninsured- this is the 7.7% (b) What percent of Americans are receiving Food Stamps, but have health insurance? 12.6 - 7.7 = 4.9 Receiving food stamps - both categories = food stamps but insured (c) What percent of Americans are receiving Food Stamps or have no health insurance? (12.6 + 10.5)-7.7 (Food stamps+uninsured) - those who fall into both = 15.4% The question asks OR, so the both category cannot be included . (d) What percent of Americans are not receiving Food Stamps and have health insurance? 1-.126 = .874 1-.105 = .895 .895 x .875 = 78.2% aren’t receiving food stamps and are also insured (e) Is the event of some American receiving Food Stamps independent of the event of not having health insurance? 12.6% x 10.5% = 0.013% ≠ 7.7% They are dependent 2. A Gallup survey asks 1,500 Americans about their views regarding Legalization of Marijuana. The table below shows the proportions of responses by political ideology and whether they Favor or Oppose legalizing marijuana:

Favor Legalizatio n Oppose Legalization Tot al Conservative s 0.179 0.191 0.3 74 Political Moderates 0.278 0.095 0.3 76 Ideology Liberals 0.208 0.041 0.2 51 Total 0.664 0.327 1.0 00 (a) Are favoring legalization and being a liberal mutually exclusive? No, the 20.8% represent liberals who also favor legalization. They are non disjoint. (b) What is the probability that a randomly chosen respondent favors legalization or is a liberal? (66.4% + 25.1%) - 20.8% = 70.7% (Favoring legalization + liberals) - existing demographic of both (because question asks OR) (c) What is the probability that a randomly chosen respondent favors legalization given that he/she is a liberal? Favors legalization and is liberal / liberal 20.8/25.1 = 82.9% (d) What is the probability that a randomly chosen respondent favors legalization given that he/she is a conservative? (Favored legalization and is conservative ) / conservative 17.9/3.74 = 47.9% (e) Is whether or not a respondent favors legalization independent of their political ideology? Explain your answer. The opinions seem to vary from each political ideology (e.I. Liberals less than moderates), so it is probable that they’re dependent. (f) What is the probability that a randomly chosen respondent is a moderate given that he/she favors legalization? 3. In the 2016 presidential election Donald Trump received 46.4% of the popular vote. According to exit polls, of those who voted for Trump 19.9% had a high school education or less, while 17.2% of those who voted against Trump had a high school education or less. Suppose we randomly select a respondent from the exit poll and found that he/she has a high school education or less. What is the probability that this respondent voted for Trump? P(trump and high school ed) 0.199/0.464 = 0.429 4. Delta Airlines latest baggage fees for checked luggage are as follows: $30 for the first checked bag, $40 for the second checked bag, and $150 for the third checked bag. Suppose 53% of all Delta passengers have no checked luggage, 33% have one piece of checked luggage, 12% have

two pieces, and 2% have three pieces. We also suppose a negligible percent of passengers check more than three bags. (a) Build a probability model for 0, 1, 2, and 3 checked bags and use this to compute the average baggage revenue per passenger. bags x p(x) x p(x) no bags 0 0.53 0 1 30 0.33 9.9 2 40 0.12 4.8 3 150 0.02 3 (0 +9.9 + 4.8 +3) / 4 = 4.425 a passenger X P(X)/4 (b) About how much baggage revenue should Delta expect for a flight of 200 passengers? 4.425(200) —> $885 5. In all marathons run in the USA during 2010, the average (mean) time for men was 4:27 hours (or 267 minutes) with a standard deviation of 1:01 hours (61 minutes). For women running marathons, the average was 4:54 hours (294 minutes) with a standard deviation of 1:02 hours (62 minutes). Suppose the distributions for marathon times are approximately normal for both men and women. John ran the marathon in 5:06 hours (306 minutes), while his wife Mary finished the marathon in 5:09 hours (309 minutes). Obviously John finished the marathon faster than Mary, but they are curious about how each of them performed according to their genders. Remember: a better performance corresponds to a faster (shorter) finish time. (a) Write down the short-hand for the two normal distributions, for men and for women. Women marathon—> W(mn = 267; sd = 61) Men marathon—>M(mn = 294; sd = 62) (b) What are the Z-scores for John’s and Mary’s finishing times according to their genders? What do these Z-scores tell you when comparing them? (John score - mean)/SD 306-267/61 = 0.639 (Mary score - mean)/SD 309-294/62 = 0.242 Lower Z score means closer to mean, higher means farther from mean (c) Did John or Mary rank better within their own gender? Mary had a lower Z score, she ranked better within her gender (d) What percent of male marathon runners did John finish faster than? R results: [1] 0.2612995, so 26% faster

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What percent of female marathon runners did Mary finish faster than? R results: [1] 0.4044151, so 40.4% faster (e) If the distributions of finishing times are not nearly normal, would your answers to (b)-(e) be different? Yes, if the distribution was not normal I could not use Z scores to get the percentiles showing Mary and Jon’s ranks within their own gender (f) According to the above normal distribution for male marathon runners, what would be the cutoff time for the fastest 5% of the men? (that is, those who took the 5% shortest time) R results using qnorm: [1] -1.644854; -1.644854= (x-267)/61 —> 166.7 minutes (g) According to the above normal distribution for female marathon runners, what would be the cutoff time for the slowest 10% of the women? 100%-10%=90%, so qnorm(0.9) for top 90 percent [1] 1.281552 =(x-294)/62 —> 373.5 minutes 6. The average daily high temperature in July in Indianapolis is 85°F, with a standard deviation of 6°F. Suppose that July temperatures in Indianapolis closely follow a normal distribution. (a) What is the probability of observing a 90°F temperature or higher in Indianapolis during a randomly chosen day in July? I know I use pnorm but I think I’m overthinking how to enter this (b) How cool are the coolest 10% of the days (days with the lowest average high temperature) during July in Indianapolis? -1.28 = (observation - 85)/6 —> 77.32 degrees

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