This is chapter 2 homework for Data 606 for Fall 2018 semester. The chapter is related to Probability.

Following are the questions and their answers solved below:

2.6 Dice rolls. If you roll a pair of fair dice, what is the probability of (a) getting a sum of 1? Answer - The minimum sum can be 2. That means sum of 1 is not possible. Hence P(sum=1) = 0

  1. getting a sum of 5? Answer - P(sum=5) will be the probability the sum = 5. This can be achieved with the below combinations: 1,4 2,3 3,2 4,1

Hence 4 possible outcomes give the sum of 5. Total number of possible outcomes= 6 X 6 = 36 Hence P(sum=5) = 4/36

  1. getting a sum of 12? Answer - The only combination which gives a sum of 12 is: 6,6. That is only 1 possible outcome, hence P(sum=12) = 1/36

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. (a) Are living below the poverty line and speaking a foreign language at home disjoint? Answer - No, these are not disjoint, as there are more than 0 (4.2%) such people who are below the poverty line and speak a foreign language.

  1. Draw a Venn diagram summarizing the variables and their associated probabilities. Answer -
## install.packages("VennDiagram")
library(VennDiagram)
## Loading required package: grid
## Loading required package: futile.logger
grid.newpage()
draw.pairwise.venn(0.146, 0.207, 0.042, category = c("Below Poverty ;Line", "Foreign Language"), lty = rep("blank", 
                                                                                   2), fill = c("light blue", "pink"), alpha = rep(0.5, 2), cat.pos = c(0, 
                                                                                                                                                        0), cat.dist = rep(0.025, 2))

## (polygon[GRID.polygon.1], polygon[GRID.polygon.2], polygon[GRID.polygon.3], polygon[GRID.polygon.4], text[GRID.text.5], text[GRID.text.6], text[GRID.text.7], text[GRID.text.8], text[GRID.text.9])

  1. What percent of Americans live below the poverty line and only speak English at home? Answer - from the Venn diagram above, the plot in blue gives the people below poverty line and speaking only English. It is 0.104. In Percent = 10.4%

  2. What percent of Americans live below the poverty line or speak a foreign language at home? Answer - P(poverty=yes OR language=foreign) = P(poverty=yes) + P(language=foreign) - P(poverty=yes AND language=foreign)

= 0.146 + 0.207 - 0.042 = 0.311

In percent = 31.1%

  1. What percent of Americans live above the poverty line and only speak English at home? Answer - This population will be Poverty Line or speaking English only. This is total MINUS (povery=yes OR language=foreign) (poverty = yes OR language=foreign) is 31.1% from last question. So, Poverty Line or speaking English only = 100 - 31.1 = 68.9%

  2. Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home? Answer - P(poverty line=yes) = 0.146 P(language=foreign) = 0.207 P(poverty line=yes AND language=foreing) = 0.042

P(poverty line=yes) X P(language=foreign) = 0.146 X 0.207 = 0.0302 Hence P(poverty line=yes) X P(language=foreign) NOT equal to P(poverty line=yes AND language=foreign) That means the 2 events are not independent.

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.65 Partner (female) Blue Brown Green Total Blue 78 23 13 114 Self (male) Brown 19 23 12 54 Green 11 9 16 36 Total 108 55 41 204 (a) What is the probability that a randomly chosen male respondent or his partner has blue eyes? Answer - P(male=Blue OR female=Blue) = P(male=Blue) + P(female=Blue) - P(male=Blue AND female=Blue) = 114/204 + 108/204 - 78/204 = 144/204 = 0.71

  1. What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes? Answer -

P(female=Blue | male=Blue) = P(female=Blue AND male=Blue) / P(male=Blue) = (78/204) / (114/204) = 78/114 = 0.68

  1. 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? Answer - part-1: P(female=Blue | male=Brown) = P(female=Blue AND male=Brown) / P(male=Brown) = (19/204) / (54/204) = 19/54 = 0.35

patrt-2: P(female=Blue | male=Green) = P(female=Blue AND male=Green) / P(male=Green) = (11/204) / (36/204) = 11/36 = 0.31

  1. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning. Answer - From the contingency table, we can see the below relationship between the number of males and number of females with same eye color. From within the Blue eyed males, we see that the maximum probability is of the same colored females - 78/114 Similar trends are there for Brown - 19/54 and Green - 16/36 Also if we go vertically, Blue females having same colored eyes males has a probability of 78/108, for Green, it is 16/41. Both of these are the maximum. However for Brown females having Brown males as their partners is 23/55 which matches with Brown females having Blue eyed male partners, which is still the maximum. So, these are certainly not independent, while there is no causal relationship between the color of eyes combination for a couple.

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. Format Hardcover Paperback Total Type Fiction 13 59 72 Nonfiction 15 8 23 Total 28 67 95 (a) Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement. Answer - (28/95) X (59/94) = 0.185

  1. Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement. Answer - There are 2 scenarios here:
  1. If the first book drawn is a hardcover

  2. If the first book drawn is a paperback

  3. Probability of drawing a fiction hardcover book first and then a hardcover book second = (13/95) X (27/94) = 0.0393

  4. Probability of drawing a fiction paperback book first and then a hardcover book second = (59/95) X (28/94) = 0.185

As either of the 2 events can happen, these 2 above values need to be added together: 0.0393 + 0.185 = 0.2243

  1. 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.

Answer -

(13/95) X (28/95) This comes from first scenario (59/95) X (28/95) This comes from second scenario

Adding the 2 (13 X 28 + 59 X 28) / (95 X 95) = 0.2234

  1. The final answers to parts (b) and (c) are very similar. Explain why this is the case. Answer - This is beacuse as the sample drawn us a very small fraction of the total population. Hence the first scenario of drawing 2 books without the replacement of the first after the first book drawn is not going to make a big difference even if the 2 events would have been independent, that is scenario 2.

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. (a) Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation. Answer - i 0 1 2 Total xi $0 $25 $60 P(X=xi) 0.54 0.34 0.12 xi * P(X=xi) 0 8.5 7.2 15.7 xi - u -15.7 9.3 44.3
(xi - u)^2 246.49 86.49 1962.49
(xi - u)^2 * P(X=xi) 133.105 29.407 235.499 398.011

E(X) = x0 X P(X=x0) + x1 X P(X=x1) + x2 X P(X=x2)

Average revenue per passenger = (0 X 0.54) + (25 X 0.34) + (60 X 0.12) = $15.7

Variance = ???(xi - u)^2 * P(X=xi) = 398.011 SD = ???Variance = ???398.011 = 19.95

  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. Answer - For 120 passengers, the revenue would be 120 X (average revenue per passenger) = 120 X 15.7 = $1884

If a data set is multiplied by a constant, the standard deviation also gets multiplied by the same constant value. Hence, SD for 120 passengers flight = 120 X SD for 1 passenger = 120 X 19.95 = 2394

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.69 (a) Describe the distribution of total personal income. Answer -

library(ggplot2)

annual.personal.income.dist <- data.frame(income = c("1-9,999", "10,000-14,999", "15,000-24,999", "25,000-34,999", "35,000-49,999", "50,000-64,999", "65,000-74,999", "75,000-99,999", "100,000 0r more"), percent = c(2.2, 4.7, 15.8, 18.3, 21.2, 13.9, 5.8, 8.4, 9.7))

ggplot(annual.personal.income.dist, aes(x=annual.personal.income.dist$income, y=annual.personal.income.dist$percent)) + geom_bar(stat = "identity") + labs(x="income ranges", y="per centage") + theme(axis.text.x = element_text(angle = -90))

  1. What is the probability that a randomly chosen US resident makes less than $50,000 per year? Answer - To get this, we need to add up the probability of all the disjoint / mutually exclusive income ranges up to $50,000, which will be: 0.022 + 0.047 + 0.158 + 0.183 + 0.212 = 0.622

  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. Answer - Assuming that the ratio of the males and females are same throughout the various income ranges, probability of a resident - having income less than 50,000 and is female is: 0.622 X 0.41 = 0.255

  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. Answer - This value defies the assumption used in the above question. If we take the assumption used in c above, then the females form the below percentage values in each of the salary ranges as shown below:

$1 to $9,999 or loss 2.2 X .41 = 0.902 $10,000 to $14,999 4.7 X .41 = 1.927 $15,000 to $24,999 15.8 X .41 = 6.478 $25,000 to $34,999 18.3 X .41 = 7.503 $35,000 to $49,999 21.2 X .41 = 8.692 $50,000 to $64,999 13.9 X .41 = 5.699 $65,000 to $74,999 5.8 X .41 = 2.378 $75,000 to $99,999 8.4 X .41 = 3.444 $100,000 or more 9.7 X .41 = 3.977

So, the total is 41%, while the female form 25.5% of total population which is under 50,000 range. Hence 25.5 / 41 = 62.19% wich is not the same as given in this question d. Hence the assumption used in c will be false if this is true.