Data 606 Homework 2

2.6, 2.8, 2.20, 2.30, 2.38, 2.44

2.6 Dice rolls. If you roll a pair of fair dice, what is the probability of (a) getting a sum of 1? Ans: None because minimun of a pair of fair dice is 2.

  1. getting a sum of 5? Ans: There are 4 Combination of dice: (1,4), (2,3), (3,2), (4,1), the probability of getting a sum of 5 is 4/12 = 1/3

  2. getting a sum of 12? Ans: One.

2.8 Poverty and language. The American Community Survey is an ongoing survey that provides data every year to give communities the current information they need to plan investments and services. The 2010 American Community Survey estimates that 14.6% of Americans live below the poverty line, 20.7% speak a language other than English (foreign language) at home, and 4.2% fall into both categories.

  1. Are living below the poverty line and speaking a foreign language at home disjoint? Ans: No because there are 4.2% fall into both living below the proverty line and speaking a foreign language.

  2. Draw a Venn diagram summarizing the variables and their associated probabilities.

Ans:

library(VennDiagram)
## Loading required package: grid
## Loading required package: futile.logger
library(ggplot2)

venn.plot <- draw.pairwise.venn(area1=0.146,area2= 0.27,cross.area =0.042, c("Below proverty line", " Speaker Foreign Lang"))
grid.draw(venn.plot)

grid.newpage()

  1. What percent of Americans live below the poverty line and only speak English at home? Ans: Only speak english: 1 - 20.7% = 79.3%. P(below poverty line and speak english) = 14.6% x 79.3% = 11.5%

  2. What percent of Americans live below the poverty line or speak a foreign language at home? Ans: P(below proverty line or speak foreign language) - P(both below proverty line and speak foreign language) = 14.6% + 20.7% - 4.2% = 31.1%

  3. What percent of Americans live above the poverty line and only speak English at home? Ans: P(above proverty line and only speak english) = (1-14.6%) * 79.3% = 67.7%

  4. Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home? Ans: No.

2.20 Assortative mating. Assortative mating is a nonrandom mating pattern where individuals with similar genotypes and/or phenotypes mate with one another more frequently than what would be expected under a random mating pattern. Researchers studying this topic collected data on eye colors of 204 Scandinavian men and their female partners. The table below summarizes the results. For simplicity, we only include heterosexual relationships in this exercise

  1. What is the probability that a randomly chosen male respondent or his partner has blue eyes? Ans: P(self male with blue eyes) + P(partner female with blue eyes) - P(both male female with blue eyes) = 114/204 + 108/204 - 78/204 = 70.59%

  2. What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes? Ans: P(female with blue eyes| male with blue eyes) = 78/114 = 68.4%

  3. What is the probability that a randomly chosen male respondent with brown eyes has a partner with blue eyes? What about the probability of a randomly chosen male respondent with green eyes having a partner with blue eyes? Ans: P(female with blue eyes| male with brown eyes ) = 19/54 = 35.2% P(female with blue eyes| male with green eyes ) = 11/36 = 30.6%

  4. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning. Ans: No, from the observation above, males with blue eyes favorite females with blue eyes.

2.30 Books on a bookshelf. The table below shows the distribution of books on a bookcase based on whether they are nonfiction or fiction and hardcover or paperback.

  1. Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement. Ans: P(Hardcover first) * P(paperback fiction) = ( 28/95 ) * (59 / 94) = 18.50%

  2. Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement. Ans: P(first:hard cover fiction and second: hard cover) + P(first: papaerback fiction and second: hard cover) = (13/95)(27/94) + (59/95)(28/94) = 22.43%

  3. Calculate the probability of the scenario in part (b), except this time complete the calculations under the scenario where the first book is placed back on the bookcase before randomly drawing the second book. Ans: P(first: fiction and second: hardcover) = (72/95) * (28/95) = 22.338%

  4. The final answers to parts (b) and (c) are very similar. Explain why this is the case. Ans: Because the difference of (b) and (c) is replacement of the book which changes the denominator from 1/(9594) to (1/9595). The impact of the overall percentage number is small.

2.38 Baggage fees. An airline charges the following baggage fees: $25 for the first bag and $35 for the second. Suppose 54% of passengers have no checked luggage, 34% have one piece of checked luggage and 12% have two pieces. We suppose a negligible portion of people check more than two bags.

  1. Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation.

Ans: i: a = passengers with no checked bag, b= with one check bags, c=two checked bag x(i) as Revenue: a = $0, b=$25, c=$35 P(i): a= 0.54, b=0.34, c=0.12 expected value( (i) * P(i) ): a=0.540, b= 0.3425= 8.5, c= 0.12(25+35)=7.2 => total: 15.7 x(i) - expected value: a=-15.7, b=-7.2, c=-8.5 (x(i) - expected value)^2: a=246.49, b= 51.84, c=72.25 (x(i) - expected value)^2 P(i): a=133.10, b=17.63, c=8.67 => total: 159.4 standard deviation: (159.4)^(1/2) = 12.63

Average revenue per passenger is 15.7 Standard deviation: 12.63

  1. About how much revenue should the airline expect for a flight of 120 passengers? With what standard deviation? Note any assumptions you make and if you think they are justified. Ans: 120*(15.7) = $1884. From the probability given above, the passengers with bags are not normal distribution and it is left skewed. It may not be a good idea to use standard deviation to estimate the revenue.

2.44 Income and gender. The relative frequency table below displays the distribution of annual total personal income (in 2009 inflation-adjusted dollars) for a representative sample of 96,420,486 Americans. These data come from the American Community Survey for 2005-2009.

This sample is comprised of 59% males and 41% females (a) Describe the distribution of total personal income. Ans: Right Skewed.

  1. What is the probability that a randomly chosen US resident makes less than $50,000 per year? Ans: P(make lesser than $50k) = (2.2 + 4.7 + 15.8 + 18.3 + 21.2) = 62.2%

  2. What is the probability that a randomly chosen US resident makes less than $50,000 per year and is female? Note any assumptions you make. Ans: Assume the samples of males and females are similar across all salary categories. P(make lesser than $50k and female) = 62.2% * 41% = 25.50%

  3. The same data source indicates that 71.8% of females make less than $50,000 per year. Use this value to determine whether or not the assumption you made in part (c) is valid. Ans: the assumption in (c) is not correct. If 71.8% of females make lesser than $50k: Samples of making lesser than 50K is 62.2% * 96,420,486 = 59,973,542 And, overall number of females in the samples is 41% * 96,420,486 = 39,532,399
    And, number of female who make lesser than $50k is 71.8% * 39,532,399 = 28,384,262

    Therefore, the probability that a randomly chosen US resident makes less than $50,000 per year and is female: 28,384,262 / 96,420,486 = 29.44%.