Question 2.6

Dice rolls. If you roll a pair of fair dice, what is the probability of

  1. getting a sum of 1? 0 Probability of getting a 1

  2. getting a sum of 5? Outcome 1 = P(Y=1) * P(Y=4) = 1/6 * 1/6 = 1/36

Outcome 2 = P(Y=2) * P(Y=3) = 1/6 * 1/6 = 1/36

Outcome 3 = P(Y=3) * P(Y=2) = 1/6 * 1/6 = 1/36

Outcome 4 = P(Y=4) * P(Y=1) = 1/6 * 1/6 = 1/36

Probability of getting a sum of 5 = 4/36

  1. getting a sum of 12? Outcome 1 = P(Y=6) * P(Y=6) = 1/6 * 1/6 = 1/36

Probability of gettng a sum of 12 = 1/36

Question 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.

Define Variables:

P(PL) = 14.6% P(FL) = 20.7% P(PL and FL) = 4.2%

PL <- 14.6/100
FL<- 20.7/100
PLandFL <- 4.2/100

  1. Are living below the poverty line and speaking a foreign language at home disjoint? Living below the povery line and speaking a foreign language are not disjoint as both can occur together (P(A and L)).

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

library(VennDiagram)
## Loading required package: grid
## Loading required package: futile.logger
grid.newpage()
draw.pairwise.venn(area1 = 14.6, area2 = 20.7, cross.area = 4.2, category = c("Below PL", 
    "ForeignLang"))

## (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?
PLEng <- paste(round((PL - PLandFL)*100, digits = 1), "%", sep="")
PLEng
## [1] "10.4%"

10.4 % Americans live below the poverty line and speak English at home.

  1. What percent of Americans live below the poverty line or speak a foreign language at home? Use General Addition rule P(PL) or P(FL) = P(PL) + P(FL) - P(PL AND FL)
#General Addition Rule
PLorFL <- paste(round((PL + FL - PLandFL)*100, digits = 1), "%", sep="")
PLorFL
## [1] "31.1%"

31.1% of Americans live below the poverty line and speak english at home

  1. What percent of Americans live above the poverty line and only speak English at home?

Find complement of P(PL) and P(FL) and use multiplication rule for independent processes

P(PL or PLc) = P(PL) + P(PLc) = 1 1 - P(PL) = P(PLc)

P(FL or FLc) = P(FL) + P(FLc) = 1 1 - P(FL) = P(FLc)

#compliment of PL and FL
PLc <- 1 - PL
PLc
## [1] 0.854
FLc <- 1 - FL
FLc
## [1] 0.793
#multiplication rule for independent processes
PLcFLc <- paste(round((PLc * FLc)*100, digits=1), "%", sep="")
PLcFLc
## [1] "67.7%"

67.7% of Americans live above the poverty line and only speak english at home

  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 event that someone lives below the poverty line is indepent of the event that someone speaks a foreign language at home because knowing the outcome of one provides not information about outcome of the other.

Question 2.20

  1. What is the probability that a randomly chosen male respondent or his partner has blue eyes?

General Addition Rule:

P(MBlue) + P(FBLue) - P(MBlue AND FBlue)

MBlue <- 114/204
FBlue <- 108/204
FMBlue <- 78/204
ForMblue <- paste(round((MBlue + FBlue - FMBlue)*100, digits=1), "%", sep = "")
ForMblue
## [1] "70.6%"

70.6% probability that a male or female has blue eyes

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

Probability that a male respondent has a partner with blue eyes is 78/204

FandMblue <- paste(round((FMBlue)*100, digits = 1), "%", sep = "")
FandMblue
## [1] "38.2%"
  1. What is the probability that a randomly chosen male respondent with brown eyes has a partner with blue eyes?

19/204

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

11/204

  1. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.

Question 2.30 (a)Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement.

A <- paste(round(((28/95) * (67/94))*100, digits = 2), "%", sep = "")
A
## [1] "21.01%"
  1. Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
B <- paste(round(((72/95) * (28/94))*100, digits = 2), "%", sep = "")
B
## [1] "22.58%"
  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.
C <- paste(round(((72/95) * (28/95))*100, digits = 2), "%", sep = "")
C
## [1] "22.34%"
  1. The final answers to parts (b) and (c) are very similar. Explain why this is the case.

The results are similar because when the sample size is only a small fraction of the population (under 10%), observations are nearly independent even when sampling without replacement.

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.

  1. Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation.
bags <- c(0,1,2)
cost <- c(0,25,35+25)
percent <- c(.54, .34, .12)
baggagefees <- data.frame(bags, cost, percent)
baggagefees
##   bags cost percent
## 1    0    0    0.54
## 2    1   25    0.34
## 3    2   60    0.12
#compute expected value
Avgrev <- sum(cost * percent)
Avgrev
## [1] 15.7

The average revenue per customer is $15.70.

#compute variance
variability <- ((cost - Avgrev)^2) * percent
totalvar <- sum(variability)
totalvar
## [1] 398.01
#compute standard deviation (square root of variance)
sd <- round(sqrt(totalvar),2)
sd
## [1] 19.95

The standard deviation is 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.
expectedrev <- Avgrev * 120
expectedrev
## [1] 1884

The expected revenue for 120 passengers is $1884.

var120 <- (120 * sd^2)
var120
## [1] 47760.3
sd120 <- round(sqrt(var120),2)
sd120
## [1] 218.54

The standard deviation for the expected revenue of 120 passengers is 218.54

Question 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

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

d <- data.frame(income, total)
d
##               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   $100,000 or more   9.7
barplot(d$total, names.arg = income, xlab = "Personal Income", ylab = "Total")

The distribution is unimodal

  1. What is the probability that a randomly chosen US resident makes less than $50,000 per year?
Pless50 <- paste(round(((2.2 + 4.7 + 15.8 + 18.3 + 21.2)/ sum(total))*100, digits = 2), "%", sep ="")
Pless50
## [1] "62.2%"

**The probability of less that 50,000 is 62.2%.

(c)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.

femless50 <- ((2.2 + 4.7 + 15.8 + 18.3 + 21.2)/ sum(total)) * 41/100
femless50
## [1] 0.25502

The probability that a US resident makes less than $50,000 per year and is female is 26%.I do not want to have to assume this, but men tend to habe higher salaries than women, so the actually percentage of women below $50,000 maye actually be much higher

(d)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. If 71.8% of females actually make less than $50,000 per year than that would mean men tend to make more money than money do and women’s salararies are towards the lower end of the personal income distribution.