Dice rolls. (3.6, p. 92) If you roll a pair of fair dice, what is the probability of

  1. getting a sum of 1? There is zero chance of getting a sum of 1 with 2 die because the lowest you can get is 1, 1, which totals 2. 0/36
  2. getting a sum of 5?
    Die1:1,2,3,4 4/6 DIe2:4,3,2,1 4/6 4/6 * 4/6 =16/36
  3. getting a sum of 12? Die1:6 1/6 Die2:6 1/6
    1/6*1/6=1/36

Poverty and language. (3.8, p. 93) 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? 4.2%
  2. Draw a Venn diagram summarizing the variables and their associated probabilities.
library(VennDiagram)
## Loading required package: grid
## Loading required package: futile.logger
pov <- 14.6

forLang <- 20.7

both <- 4.2

povOnly <- pov - both

forLangOnly <- forLang - both

venn.plot <- draw.pairwise.venn(pov, forLang, cross.area=both,  
c("Poverty", "Foreign Language"), fill=c("lightblue", "lightgreen"),
cat.dist=-0.08, ind=FALSE)

grid.draw(venn.plot)

  1. What percent of Americans live below the poverty line and only speak English at home? 0.146 * 0.793= 0.116 11.6 % of Americans live below the poverty line and only speak English at home
  2. What percent of Americans live below the poverty line or speak a foreign language at home? 0.854 * .207 * 100= 17.7 17.7 percent of Americans live below the poverty line or speak a foreign language at home 14.6+ 20.7-4.2= 31.1%

  3. What percent of Americans live above the poverty line and only speak English at home? 100-14.6= 85.4 100-20.7=79.3 0.8540.793 100= 67%

  4. Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?

Yes, the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home.

Assortative mating. (3.18, p. 111) 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?
(108+114-78)/204
## [1] 0.7058824
  1. What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?
(78/114)
## [1] 0.6842105
  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?
#a randomly chosen male respondent with brown eyes has a partner with blue eyes
(19/54)
## [1] 0.3518519
#a randomly chosen male respondent with green eyes having a partner with blue eyes
(11/36)
## [1] 0.3055556
  1. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.

Yes the eye colors of male respondents and their partners are independent because the eye color of the males dont affect the outcome of their partners eye colors.

Books on a bookshelf. (3.26, p. 114) 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.
(28/95) * (59/94)
## [1] 0.1849944
  1. Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
(72/95) *(28/94)
## [1] 0.2257559
  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.
(72/95)*(28/95)
## [1] 0.2233795
  1. The final answers to parts (b) and (c) are very similar. Explain why this is the case.

Because as the sample size increases, taking out one book would not have as big an impact on the probabilities on the second draw.

Baggage fees. (3.34, p. 124) 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. The average revenue is $15.7 per passenger. Standard deviation is 12.6.
fees<-c(0, 25, 25+35)
prob<-c(0.54, 0.34, 0.12)
#average revenue
avg_rev<-sum(fees*prob)
avg_rev
## [1] 15.7
#standard deviation
sd<-round(sqrt(sum((fees-avg_rev)^2 * prob)), 3)
sd
## [1] 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. For 120 passenger, the revenue is $1884.
passgr<-120
avg_rev120<-avg_rev*passgr
avg_rev120
## [1] 1884
sd_120<-round(sqrt(120*sd^2), 3)
sd_120
## [1] 218.541
income<-c("$1-$9,999 or less", "$10,000 to $14,999", "15,000 to $24,999", "25,000 to $34,999", "$35,000 to $49,999", "$50,000 to $64,999", "$65,000 to $74,999", "$75,000 to $99,999", "$100,000 or more")
total<-c(0.022, 0.047, 0.158, 0.183, 0.212, 0.139, 0.058, 0.084, 0.097)
income_df<-data.frame(income, total)
income_range<-c(9999,14999,24999,34999,49999,64999,74999, 99999, 100000)
df_incomes<-data.frame(income,total)
barplot(df_incomes$total, xlab="Income Ranges", ylab="Probability")

sum(df_incomes$total[1:5])
## [1] 0.622

  1. Describe the distribution of total personal income. Looks like a rightskewed bimodial distribution.

  2. What is the probability that a randomly chosen US resident makes less than $50,000 per year? 62.2.%

  3. 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.

25.5%

62.2 * 0.41
## [1] 25.502
  1. 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.

This value will deem my assumption invalid.