Exercise 2.6 Dice Rolls If you roll a pair of fair dice, what is the probability of…

#Create a vector entitled fair dice
#There are 36 different outcomes
fair_dice <- c(1,2,3,4,5,6,5,4,3,2,1)/36
#Display findings
fair_dice
##  [1] 0.02777778 0.05555556 0.08333333 0.11111111 0.13888889 0.16666667
##  [7] 0.13888889 0.11111111 0.08333333 0.05555556 0.02777778

(A)Getting a sum of 1? 0 (B)Getting a sum of 5? Occurs with 1:4, 2:3, 3:2, and 4:1 the outcome is 0.11111111 (C)Getting a sum of 12? Occurs once with two sixes the outcome is 0.02777778

Exercise 2.8 Poverty and Language (A)Are living below the poverty line and speaking a foreign language at home disjoint? Disjoint outcomes cannot happen at the same time, therefore the answer is no since both outcomes are independent of each other.

(B)Draw a Venn diagram summarizing the variables and their associated problems?

#Incomplete (Having difficulties loading package)

(C)What percent of Americans live below the poverty line and only speak English at home?

#Create variables
Below_PovLine <- c(0.146)
Other_English <- c(0.207)
Both <- c(0.042)
Below_PovLine_Only <- Below_PovLine - Both
#Display results
Below_PovLine_Only
## [1] 0.104
  1. What percent of Americans live below the poverty line or speak a foreign language at home?
#Use the General Addition Rule
(Below_PovLine + Other_English) - Both
## [1] 0.311
  1. What percent of Americans live above the poverty line and only speak English at home?
(100 - (0.104+0.042+0.165))/100
## [1] 0.99689
  1. Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home? The results are independent

Exercise 2.20 Assortative Mating Import data into R

am<-read.csv(url("https://raw.githubusercontent.com/jbryer/DATA606Fall2017/master/Data/Data%20from%20openintro.org/Ch%202%20Exercise%20Data/assortive_mating.csv"))
#Use Table function to display total of am
table(am)
##          partner_female
## self_male blue brown green
##     blue    78    23    13
##     brown   19    23    12
##     green   11     9    16
#Obtain total of partner_female
(sum(am$partner_female=="blue"))
## [1] 108
(sum(am$partner_female=="brown"))
## [1] 55
(sum(am$partner_female=="green"))
## [1] 41
#Obtain total of self_male
(sum(am$self_male=="blue"))
## [1] 114
(sum(am$self_male=="brown"))
## [1] 54
(sum(am$self_male=="green"))
## [1] 36
  1. What is the probability that a randomly chosen male respondent or his partner has blue eyes?
((sum(am$self_male =="blue")/nrow(am)) +
 (sum(am$partner_female =="blue")/nrow(am)))- (sum(am$self_male =="blue" & am$partner_female=="blue")/nrow(am))
## [1] 0.7058824
  1. What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?
(sum(am$self_male =="blue" & am$partner_female=="blue")/nrow(am))
## [1] 0.3823529
  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?
#Brown Eye Male Blue Eye Partner
(sum(am$self_male =="blue" & am$partner_female=="blue")/nrow(am))/(sum(am$self_male=="blue")/nrow(am))
## [1] 0.6842105
(sum(am$self_male =="green" & am$partner_female=="blue")/nrow(am))/(sum(am$self_male=="green")/nrow(am))
## [1] 0.3055556
  1. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.

They are not independent because their outcomes are not equivalent

Exercise 2.30 Books on a Bookshelf

books<-read.csv(url("https://raw.githubusercontent.com/jbryer/DATA606Fall2017/master/Data/Data%20from%20openintro.org/Ch%202%20Exercise%20Data/books.csv"))
#Use Table function to display total of books
table(books)
##             format
## type         hardcover paperback
##   fiction           13        59
##   nonfiction        15         8
#Find the sum of hardcover
sum(books$format=="hardcover")
## [1] 28
#Find the sum of paperback
sum(books$format=="paperback")
## [1] 67
#Find the sum of fiction
sum(books$type=="fiction")
## [1] 72
#Find the sum of nonfiction
sum(books$type=="nonfiction")
## [1] 23
  1. Find the probability of drawing a hardcover book ???rst then a paperback ???ction book second when drawing without replacement.
#Obtain the total of hardcover books and divide by the total of books
#Multiply the total of paperback boks which are divided by the total of books minus 1 since drawings are conducted without replacement
(sum(books$format=="hardcover"))/sum(table(books))*
(sum(books$format=="paperback"))/(sum(table(books))-1)
## [1] 0.2100784
  1. Determine the probability of drawing a ???ction book ???rst and then a hardcover book second, when drawing without replacement.
#As in the first example drawings are conducted without replacement
(sum(books$type=="fiction"))/sum(table(books))*
(sum(books$format=="hardcover"))/(sum(table(books))-1)
## [1] 0.2257559
  1. Calculate the probability of the scenario in part (b), except this time complete the calculations under the scenario where the ???rst book is placed back on the bookcase before randomly drawing the second book.
#In this example drawings are being replaced
(sum(books$type=="fiction"))/sum(table(books))*
(sum(books$format=="hardcover"))/sum(table(books))
## [1] 0.2233795
  1. The ???nal answers to parts (b) and (c) are very similar. Explain why this is the case.

There is not much difference because the denominator changed by 1 only. If there was an exponential change then the disparity would be greater.

Exercise 2.38 Baggage Fees

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

  2. About how much revenue should the airline expect for a ???ight of 120 passengers? With what standard deviation? Note any assumptions you make and if you think they are justi???ed.

Exercise 2.44 Income and Gender

#Create vectors to mirror exercise 2.44

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")

total <-c(2.2,4.7,15.8,18.3,21.2,13.9,5.8,8.4,9.7)

layout <- data.frame(income,total)
layout
##               income total
## 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   $100,000 or more   9.7
  1. Describe the distribution of total personal income.
barplot(layout$total, horiz=T)

  1. What is the probability that a randomly chosen US resident makes less than $50,000 per year?
prob_50less <-sum(layout[1:5,]$total)
prob_50less
## [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.
#Sample is comprised of 59% males and 41% females
prob_fem <-0.41 * prob_50less
prob_fem
## [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.
fem50 <- 0.718 * 0.41
fem50
## [1] 0.29438