Question 2.8 - Poverty and Language
belowPL <- 0.146
nonEnglishSpeaking <- 0.207
bothCategories <- 0.042
grid.newpage()
draw.pairwise.venn(
area1 = belowPL, area2 = nonEnglishSpeaking,
cross.area = bothCategories, category = c("Below Poverty Line", "Foreign Language Speakers")
, lty = rep("blank", 2), fill = c("light blue", "yellow"), 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])
Question 2.20 - Assortative Mating
assortiveMatingStats <- read.csv("https://raw.githubusercontent.com/jbryer/DATA606Fall2016/master/Data/Data%20from%20openintro.org/Ch%202%20Exercise%20Data/assortive_mating.csv")
table(assortiveMatingStats)
## partner_female
## self_male blue brown green
## blue 78 23 13
## brown 19 23 12
## green 11 9 16
# (a) What is the probability that a randomly chosen male respondent or his partner has blue eyes?
# P(Male Blue or Female Blue) = P(Male Blue) + P(Female Blue) - ( P(Male Blue total) + P(Female Blue Total) )
((sum(assortiveMatingStats$self_male =="blue")/nrow(assortiveMatingStats)) + (sum(assortiveMatingStats$partner_female =="blue")/nrow(assortiveMatingStats)))
## [1] 1.088235
- (sum(assortiveMatingStats$self_male =="blue" & assortiveMatingStats$partner_female=="blue")/nrow(assortiveMatingStats))
## [1] -0.3823529
# (b) What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?
# P(Male Blue and Female Blue)
(sum(assortiveMatingStats$self_male =="blue" & assortiveMatingStats$partner_female=="blue")/nrow(assortiveMatingStats))
## [1] 0.3823529
# (c) What is the probability that a randomly chosen male respondent with brown eyes has a partner with blue eyes?
# P(male Brown and Female Blue) = P(Male Brown) + P(Female Blue)
( sum(assortiveMatingStats$self_male =="brown" & assortiveMatingStats$partner_female == "blue") / nrow(assortiveMatingStats) )/( sum(assortiveMatingStats$self_male =="brown") / nrow(assortiveMatingStats) )
## [1] 0.3518519
# What about the probability of a randomly chosen male respondent with green eyes having a partner with blue eyes?
( sum(assortiveMatingStats$self_male =="green" & assortiveMatingStats$partner_female == "blue") / nrow(assortiveMatingStats) )/( sum(assortiveMatingStats$self_male =="green") / nrow(assortiveMatingStats) )
## [1] 0.3055556
# (d) Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.
(sum(assortiveMatingStats$self_male =="blue" & assortiveMatingStats$partner_female=="blue")/nrow(assortiveMatingStats)) # Blue Male with Blue Female
## [1] 0.3823529
(sum(assortiveMatingStats$self_male =="blue")/nrow(assortiveMatingStats)) # Blue Male ratio
## [1] 0.5588235
# No, they are not independent. Probability of men with blue eyes are not equal to the female partners.
Question 2.30 - Books on a bookshelf
books <- read.csv("https://raw.githubusercontent.com/jbryer/DATA606Fall2016/master/Data/Data%20from%20openintro.org/Ch%202%20Exercise%20Data/books.csv")
table(books)
## format
## type hardcover paperback
## fiction 13 59
## nonfiction 15 8
# (a) Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement.
(sum(books$format == "hardcover")/nrow(books)) * (sum(books$format == "paperback" & books$type == "fiction")/(nrow(books)-1) )
## [1] 0.1849944
# (b) Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
(sum(books$type == "fiction")/nrow(books)) * (sum(books$format == "hardcover")/(nrow(books)-1) )
## [1] 0.2257559
# (c) 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.
(sum(books$type == "fiction")/nrow(books)) * (sum(books$format == "hardcover")/nrow(books) )
## [1] 0.2233795
# (d) The final answers to parts (b) and (c) are very similar. Explain why this is the case.
# Because the difference in the number of books is only one which is very small number compared to the total books size 95
Question 2.38 - Baggage fees
baggageCharge <- c(0, 25, 70)
probBags <- c(0.54, 0.34, 0.12)
baggageDF <- data.frame(baggageCharge, probBags)
names(baggageDF) <- c("Charge", "Probability")
baggageDF
## Charge Probability
## 1 0 0.54
## 2 25 0.34
## 3 70 0.12
# (a) Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation.
# Average Revenue per Passenger
averageRevenue <- (sum((baggageDF$Charge * baggageDF$Probability)))
averageRevenue
## [1] 16.9
Question 2.44 - Income and gender
# 2.44 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.
annualIncome <- 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"
)
populationTotal <- c(
2.2
,4.7
,15.8
,18.3
,21.2
,13.9
,5.8
,8.4
,9.7
)
incomePopulationDF <- data.frame(annualIncome, populationTotal)
incomePopulationDF
## annualIncome populationTotal
## 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 $100,000 or more 9.7
barplot(incomePopulationDF$populationTotal, main="Income Distribution", xlab="% of Population")
# Probability that a randomly chosen US resident makes less than $50,000 per year
(2.2 + 4.7 + 15.8 + 18.3 + 21.2) / 100
## [1] 0.622