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?
  2. getting a sum of 5?
  3. getting a sum of 12?

ANSWER: There are 36 possible outcomes from rolling a pair of dice.

  1. There are no outcomes where a pair of dice sume to 1. There for the probability is 0/36=0

b)There are 4 possible outcomes of getting a sum of 5: (1,4), (4,1), (2,3), (3,2). So the probability is 4/36.

4/36
## [1] 0.1111111
  1. There is one possible outcome (6,6) getting a sum of 12 so the probability is 1/36.
1/36
## [1] 0.02777778

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?

ANSWER: The sets are not disjoint and in fact overlap with 4.2% falling in both categories.

  1. Draw a Venn diagram summarizing the variables and their associated probabilities.
library(VennDiagram)
## Loading required package: grid
## Loading required package: futile.logger
grid.newpage()                                        # Move to new plotting page
draw.pairwise.venn(area1 = 20.7,                        # Create pairwise venn diagram
                   area2 = 14.6,
                   cross.area = 4.2)

## (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])
#x<-list(A=1:146, B=105:311)
#ggVennDiagram(x, label=count)
  1. What percent of Americans live below the poverty line and only speak English at home?

ANSWER: 10.4%

  1. What percent of Americans live below the poverty line or speak a foreign language at home?

ANSWER: (10.4+4.2+16.5)/100

(10.4+4.2+16.5)/100
## [1] 0.311
  1. What percent of Americans live above the poverty line and only speak English at home?

ANSWER: (100-10.4-4.2-16.5)/100

(100-10.4-4.2-16.5)/100
## [1] 0.689
  1. Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?

P(<pov)andP(foreign)=4.2%

vs Indep - P(<pov)P(foreign)=.145.207

.145*.207
## [1] 0.030015
#not equal to
.042
## [1] 0.042
#So not independent

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?
(114+108-78)/204
## [1] 0.7058824
  1. What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?
78/204
## [1] 0.3823529
  1. What is the probability that a randomly chosen male respondent with brown eyes has a partner with blue eyes?
19/204
## [1] 0.09313725

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

11/204
## [1] 0.05392157
  1. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.
#If independent the following three probabilities would be equal:

#P(Fblue|Mblue)=78/114
78/114
## [1] 0.6842105
#P(Fblue|Mgreen)=11/36
11/36
## [1] 0.3055556
#P(Fblue|Mbrown)=19/54
19/54
## [1] 0.3518519
#not independent

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.

ANSWER: (28/95)*(59/94)

(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.

ANSWER: (72/95)*(28/94)

(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.

ANSWER: (72/95)*(28/95)

(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.

ANSWER: Sampling from a larger population has less of an impact whether with or without replacement. Notice the denominator in the second fraction changes by 1.

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.
luggage<-data.frame(Event=c('No luggage','One piece','Two pieces'),
                    X=c(0,25,60), pX=c(.54,.34,.12))
luggage$XpX<-luggage$X*luggage$pX
names(luggage)
## [1] "Event" "X"     "pX"    "XpX"
lug2<-luggage$pX
names(lug2)<-c("0","1","2")
barplot(lug2, main='Probability of carrying # of luggage')

ANSWER: Average revenue per passenger is Expected Value, and Standard deviaton computed

# Expected Value
m1<-sum(luggage$XpX)
m1
## [1] 15.7
#Variance Calculation
v1<-(m1-luggage$XpX)^2
v2<-sum(v1)
sd<-sqrt(v2)
sd
## [1] 19.25045
  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.
#Revenue for 120 passgeners
R<-m1*120
R
## [1] 1884
#Var for 120 passengers
#Var(X1+X2+X3+    X120)
#Var(X1)+Var(X2)+ Var(X3)+....=120(Var(X))
V<-120*v2
V
## [1] 44469.6
#Standard deviation for 120 passengers
sd2<-sqrt(V)
sd2
## [1] 210.8782

ANSWER: In calculating the variance, its was assumed that each passenger is independent of each other. This may not be true, for example a family of 4 travelling by share bags and are not independent of each other.

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<-data.frame(Event=c('$1 to $9,999 or loss','$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','$100,000 or more'),p=c(.022,.047,.158,.183,.212,.139,.058,.084,.097))
income
incomep<-income$p
names(incomep)<-income$Event

barplot(incomep, main='Probability distribution of personal income')

ANSWER: SEE PLOT - not symmetric, possible bimodal

  1. What is the probability that a randomly chosen US resident makes less than $50,000 per year?
.212+.183+.158+.047+.022
## [1] 0.622
  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.
#If income and gender are independent then....

.622*.41
## [1] 0.25502
#However, it is important to note that gender and income are typically not independent.
  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.
#Given P(<50|F)=.718

#In part C, it was assumed that gender and income were independent, subject to the calculation:

.622*.41
## [1] 0.25502
#.78 does not equal .255 
#SO NOT INDEPENDENT