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?
  1. getting a sum of 5?
  1. getting a sum of 12?

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.

poverty<- .146
lang_other<-.207
both<- .042

  1. Are living below the poverty line and speaking a foreign language at home disjoint? No, they it is possible to have both in the same family; 4.2% fall into both categories.

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

grid.newpage()
draw.pairwise.venn(14.6, 20.7, 4.2, category = c("Poverty", "Other Language"), lty = rep("blank",2), fill = c("blue", "red"), 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?
percent(poverty-both)
## [1] "10.4%"
  1. What percent of Americans live below the poverty line or speak a foreign language at home?
percent(poverty+lang_other-both)
## [1] "31.1%"
  1. What percent of Americans live above the poverty line and only speak English at home?
percent((1-poverty)*(1-lang_other))
## [1] "67.7%"
  1. Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home? *Yes, they are independent as one does not provide or force the outcome of the other.

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?
male_blue <- 114/204
female_blue <- 108/204
mf_blue <- 78/204

percent(male_blue+female_blue-(mf_blue))
## [1] "70.6%"
  1. What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?
percent(1*mf_blue)
## [1] "38.2%"

(c1) 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?

male_br_female_b<-19/54 

percent(1*male_br_female_b)
## [1] "35.2%"

(c2) What about the probability of a randomly chosen male respondent with green eyes having a partner with blue eyes?

male_g_female_b<-11/36 
percent(1*male_g_female_b)
## [1] "30.6%"
  1. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning. It does appear the eye colors are dependent. The value of partner with blue eyes varies based on values in rows (male eye color)

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.
Hard_a <- 28/95
Paper_a <-  67/94

percent (1*(Hard_a * Paper_a))
## [1] "21.0%"
  1. Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
Fiction_b <- 72/95
Hard_b <- 28/94

percent (1*(Fiction_b * Hard_b))
## [1] "22.6%"
  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.
Fiction_b <- 72/95
Hard_c <- 28/95
percent (1*(Fiction_b * Hard_c))
## [1] "22.3%"
  1. The final answers to parts (b) and (c) are very similar. Explain why this is the case. *with the same size is very small percentage of the total population, in this case a little over 1%, there will be little change from (b) to (c). The sample size will need to be bigger to see larger movement between the results.

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.

first <- 25
second <- 35
no_luggage <- .54
one_luggage <- .34
two_luggage <- .12

luggage <- c(0,1,2)
cost <- c(0,25, 25+35)
percent_pass <- c(.54,.34,.12)
  1. Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation.
fee <- data.frame(luggage, cost, percent_pass)
fee
##   luggage cost percent_pass
## 1       0    0         0.54
## 2       1   25         0.34
## 3       2   60         0.12
#calculate the average per passenger
avg_rev <- sum(cost*percent_pass)
avg_rev
## [1] 15.7
#calculate the variance
var_rev<- ((cost-avg_rev)^2)*percent_pass
total_var<- sum(var_rev)
total_var
## [1] 398.01
#calculate the SD
sd1<- round(sqrt(total_var),2)
  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.
expected_rev <- avg_rev * 120
expected_rev
## [1] 1884
var_120 <- 120*sd1^2
sd_120<- round(sqrt(var_120),2)
sd_120
## [1] 218.54

Income and gender. (3.38, p. 128) 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.

  1. Describe the distribution of total personal income.
income<- c("$1 to $9,999", "$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")
percent_income <- c(2.2,4.7,15.8,18.3,21.2,13.9,5.8,8.4,9.7)
males<- .59
females <- .41

Income_percent <- data.frame(income,percent_income)
Income_percent
##               income percent_income
## 1       $1 to $9,999            2.2
## 2 $10,000 to $14,999            4.7
## 3 $15,000 to $24,999           15.8
## 4 $25,000 to $34,999           18.3
## 5 $35,000 to $49,999           21.2
## 6 $50,000 to $64,999           13.9
## 7 $65,000 to $74,999            5.8
## 8 $75,000 to $99,999            8.4
## 9   $100,000 or more            9.7
barplot(Income_percent$percent_income, xlab = "Income", ylab = "percentage of income")

* Data is bimodal and Left Skewed

  1. What is the probability that a randomly chosen US resident makes less than $50,000 per year?
paste((2.2+4.7+15.8+18.3+21.2),"%",sep = "")
## [1] "62.2%"
  1. 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.
#Females account for 41% of the study.  If the probability of less than $50,000 per year is 62.2%, 
# we should be able to multiply that by the percentage of Females in the study to achieve the answer

paste(((2.2+4.7+15.8+18.3+21.2)*females),"%",sep = "")
## [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.
# adjusting the female's percentage with new data
females_under_50 <- .718
paste(((2.2+4.7+15.8+18.3+21.2)*females_under_50),"%",sep = "")
## [1] "44.6596%"