Graded: 2.6, 2.8, 2.20, 2.30, 2.38, 2.44

2.6 Dice rolls

If you roll a pair of fair dice, what is the probability of:

a) getting a sum of 1?

There is no probability of getting a sum of 1 due to the lowest number on a pair of fair dice is 2.

b) getting a sum of 5?

# (2,3),(1,4), (3,2),(4,1)
(4/36)*100
## [1] 11.11111

c) getting a sum of 12?

#(6,6)
(1/36)*100
## [1] 2.777778

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 that English at home, and 4.2% fall into both categories.

a. Are living below the poverty line and speaking a language other than English at home disjoint?

  • Two outcomes are called disjoint if they cannot both happen. Living below the poverty and speaking a language other than English are not disjoint outcomes, since they both can happen.

b. 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 Poverty Line", "Speak Foreign Language"), 
                   lty = rep("blank", 2), fill = c("yellow", "lightpink"), 
                   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])

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

  • 10.4% live below the poverty line and only speak English at home

d. What percent of Americans live below the poverty line or speak a language other than English at home?

(14.6+20.7)- 4.2
## [1] 31.1

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

100-31.1
## [1] 68.9

f. Is the event that someone lives below the poverty line independent of the event that the person speaks a language other than English at home?

  • The event that someone lives below the poverty line is not independent of the event that the person speaks a foreign language at home.

Poverty probability is 14.6%. Foreign speakers probability is 20.3%.

20.3%-14.6% = 5.7% which is relatevely large difference.

2.20 Assortative mating

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.

library(knitr)
Male <- c("Blue", "Brown", "Green")
Blue <- c(78, 19, 11)
Brown <- c(23,23,9)
Green <- c(13,12,16)
df = data.frame(Blue, Brown, Green, row.names = Male) # Do not incldue Male yet:
df$Totals <- apply(df, 1, sum)
df["Total" ,] <- colSums(df)
kable(df)
Blue Brown Green Totals
Blue 78 23 13 114
Brown 19 23 12 54
Green 11 9 16 36
Total 108 55 41 204

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

(114+19+11)/204
## [1] 0.7058824

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

P(female=blue|male=blue) = 78/114 = 0.68

#P(female=blue|male=blue) 

78/114 
## [1] 0.6842105

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

#P(female=blue|male=brown) 

19/54 
## [1] 0.3518519

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

#P(female=blue|male=green)
11/36
## [1] 0.3055556

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

P(female=Blue|male=Brown) =19/54 = 0.35.

P(female=Blue) = 108/204 = 0.53.

P(female=Blue|male=Brown) not equal P(female=Blue) - the eye colors of male respondents and their partners are dependent.

2.30 Books on a bookshelf

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

(28/95) * (59/94)
## [1] 0.1849944

b) Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.

(72/95) * (28/94)
## [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.

(72/95) * (28/95)
## [1] 0.2233795

d)The final answers to parts (b) and (c) are very similar. Explain why this is the case.

If we will take one book out of 95 books it will not make a huge difference, therefor the final answers are very similar.

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.

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

BAGS <- c(0,1,2)
COST <- c(0,25,60)
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 the average revenue per passenger

Avg<-sum(COST*PERCENT)
Avg
## [1] 15.7

Average revenue per passenger is $15.7.

#compute the corresponding standard deviation

var <- ((COST - Avg)^2) * PERCENT
totalvar <- sum(var)
totalvar
## [1] 398.01
std <- round(sqrt(totalvar),2)
std
## [1] 19.95

Standard Deviation is 19.95.

b) 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 revenue for a flight
revenue <-(Avg*120)
revenue
## [1] 1884

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

(a) Describe the distribution of total personal income.

  • Distribution is Right Skewed. Median Income is between 35K and 50K.

b) What is the probability that a randomly chosen US resident makes less than $50,000 per year?

  • P(make lesser than $50k) = (2.2 + 4.7 + 15.8 + 18.3 + 21.2) = 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.

Assume the distribution of male and female remained similar across the income levels.

  • P(make lesser than $50k and female) = 62.2% * 41% = 25.50%

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.

The assumption made in part (c) is not valid.

If 71.8% of females make lesser than $50k the distribution of male and female does not remained similar across the income levels.