1 Pre-Requistes : Setup R environment for DATA 606.

See https://data606.net/assignments/homework/ for more information. Chapter 2 - Probability Practice: 2.5, 2.7, 2.19, 2.29, 2.43 Graded: 2.6, 2.8, 2.20, 2.30, 2.38, 2.44

2 Exercises

2.1 Question#2.6 Dice rolls.

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

2.1.1 (a) getting a sum of 1?

P(1) = P(1)*P(0) + P(0)*P(1) = (1/6)*(0) + (0)*(1/6) = 0

2.1.2 (b) getting a sum of 5?

P(5) = P(1)*P(4) + P(2)*P(3) + P(3)*P(2) + P(4)*P(1) = (1/6)*(1/6) + (1/6)*(1/6) + (1/6)*(1/6) + (1/6)*(1/6) = 4(1/36) = 1/9

2.1.3 (c) getting a sum of 12?

If numerical, identify as continuous or discrete. If categorical, indicate if the variable is ordinal.

P(12) = P(6)*P(6) = (1/6)*(1/6) = 1/36

2.2 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.

2.2.1 (a) Are living below the poverty line and speaking a foreign language at home disjoint?

P (below poverty line) = P(<pl) = 14.6/100 = 0.146
P (speak foreign language) = P(sfl) = 20.7/100 = 0.207
P(below poverty line + speak foreign language) = P(<pl + spl) = 4.2/100 = 0.042
P(!below poverty line + !speak foreign language) = 0, since 4.2% do overlap

2.2.2 (b) Draw a Venn diagram summarizing the variables and their associated probabilities.

library(VennDiagram)
## Loading required package: grid
## Loading required package: futile.logger
Venn <- draw.pairwise.venn(14.6, 
                           20.7,
                           cross.area=4.2, 
                           c("Below Poverty Line", "Speak Foreign Language"), 
                           fill=c("red", "blue"),
                           cat.dist=-0.10,
                           ind=TRUE)
grid.draw(Venn)

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

P(below poverty line + speak english language) = P(<pl + !spl) =  P(<pl) - P(<pl + spl) = 0.146-0.042 = 0.104
Percent(below poverty line + speak english language) = 0.104*100 = 10.4%

2.2.4 (d) What percent of Americans live below the poverty line or speak a foreign language at home?

P(below poverty line | speak foreign language) = P(<pl | spl) = P(<pl) + P(spl) - P(<pl + spl) = 0.146+0.207-0.042 = 0.311
Percent(below poverty line | speak foreign language) = 0.311*100 = 31.1%

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

P(above poverty line + speak english language) = P(>pl + spl) = 1-(0.146+0.207-0.042) = 0.689
Percent(below poverty line | speak english language) = 0.689*100 = 68.9%

2.2.6 (f) Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?

P(!below poverty line + !speak foreign language) = 0 - It is not disjoint or independent instead it is joint or dependent

2.3 Question#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.

|Self (male) / Partner (female)| Blue |Brown |Green |Total| |Blue |78 |23 |13 |114| |Brown |19 |23 |12 |54| |Green |11 |9 |16 |36| |Total |108 |55 |41 |204|

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

P(Male Blue_eyes) = 114/204
P(Female Blue_eyes) = 108/204
P(Male Blue_eyes + Female Blue_eyes) = 78/204
P(Male | Blue_eyes) = P(Male Blue_eyes) + P(Female Blue_eyes) + P(Male Blue_eyes + Female Blue_eyes) = 114/204 + 108/204 - 78/204 = 144/204 = 0.706

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

P(Male Blue_eyes / Female Blue_eyes) = P(Male Blue_eyes + Female Blue_eyes)/P(Male Blue_eyes) = 78/114 = 0.684

2.3.3 (c) 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?

P(Male Brown_eyes / Female Blue_eyes) = P(Male Brown_eyes + Female Blue_eyes)/P(Male Brown_eyes) = 19/54 = 0.352
P(Male Green_eyes / Female Blue_eyes) = P(Male Green_eyes + Female Blue_eyes)/P(Male Green_eyes) = 11/36 = 0.306

2.3.4 (d) Does it appear that the eye colors of male respondents and their partners are independent?

Explain your reasoning.

Using the last example, the eye colors of male respondents and their partners would be independent if
P(Male Green_eyes / Female Blue_eyes) = P(Female Blue_eyes)

However, that formula is not true
P(Male Green_eyes / Female Blue_eyes) != P(Female Blue_eyes) => 11/36 != 108/204

Hence, they are dependent.

2.4 Question#2.30 Books on a bookshelf

The table below shows the distribution of books on a bookcase based on whether they are nonfiction or fiction and hardcover or paperback. Format |Type |Hardcover |Paperback |Total| |Fiction |13 |59 |72| |Nonfiction |15 |8 |23| |Total |28 |67 |95|

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

P(Hardcover) = 28/95
P(Fiction/Paperback-1) = 59/94
P(Hardcover + Fiction/Paperback) = P(Hardcover)*P(Fiction/Paperback-1) = (28/95)???(59/94)=0.185

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

P(Hardcover) = (13+15)/95 = 28/95
P(Fiction/Hardcover) = 13/95
P(Fiction + Hardcover-1) = Scenario1 + Scenario2 = P(Fiction/Paperback + Hardcover-1) + P(Fiction/Hardcover + Hardcover-1) = P(Fiction/Paperback)*P(Hardcover-1) + P(Fiction/Hardcover)*P(Hardcover-1) = ((59/95)*(28/94)) + ((13/95)*(27/94)) = 0.2243

2.4.3 (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.

P(Fiction + Hardcover) = P(Fiction)*P(Hardcover) = (72/95)*(28/95) = 0.2234

2.4.4 (d) The final answers to parts (b) and (c) are very similar.

Explain why this is the case.

Given the fact that dividing by 95 will result in a smaller number than dividing by 94, scenario (c) will have a smaller probability than scenario (b). However in scenario (b) we have to factor in the slight probablity event that the first book was both hardcover and fiction. When taking this into account it reduces the overall probabilty of (b) and by coincidence makes it similar to (c).

2.5 Question#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.

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

baggage_fees = c(0, 25, 60)
prob_of_ppl_checked_in_bag = c(0.54, 0.34, 0.12)
no_of_checked_in_bag = c(0, 1, 2)
airline_fees <- data.frame(baggage_fees, prob_of_ppl_checked_in_bag, no_of_checked_in_bag)
airline_fees$weighted <- airline_fees$baggage_fees * airline_fees$prob_of_ppl_checked_in_bag
revenue_per_passenger <- sum(airline_fees$weighted)
revenue_per_passenger
## [1] 15.7
airline_fees$variance <- (airline_fees$baggage_fees - revenue_per_passenger)^2 * airline_fees$prob_of_ppl_checked_in_bag
std_dev_per_passenger <- sqrt(sum(airline_fees$variance))
std_dev_per_passenger
## [1] 19.95019

2.5.2 (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.

Revenue = 120*revenue_per_passenger
Revenue
## [1] 1884
Standard_Deviation = sqrt(120*std_dev_per_passenger^2)
Standard_Deviation
## [1] 218.5434

2.6 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. Income Total $1 to $9,999 or loss 2.2% $10,000 to $14,999 4.7% $15,000 to $24,999 15.8% $25,000 to $34,999 18.3% $35,000 to $49,999 21.2% $50,000 to $64,999 13.9% $65,000 to $74,999 5.8% $75,000 to $99,999 8.4% $100,000 or more 9.7%

2.6.1 (a) Describe the distribution of total personal income.

library(ggplot2)
personal_income <- data.frame("Income" = c("1-9999", "10000-14999", "15000-24999", "25000-34999", "35000-49999", "50000-64999", "65000-74999", "75000-99999", "1000000+"), "Total" = c(0.022, 0.047, 0.158, 0.183, 0.212, 0.139, 0.058, 0.084, 0.097))
#names(personal_income) <- c("Income", "Total")
personal_income$Income <- factor(personal_income$Income, levels = c("1-9999", "10000-14999", "15000-24999", "25000-34999", "35000-49999", "50000-64999", "65000-74999", "75000-99999", "1000000+"))
personal_income
##        Income Total
## 1      1-9999 0.022
## 2 10000-14999 0.047
## 3 15000-24999 0.158
## 4 25000-34999 0.183
## 5 35000-49999 0.212
## 6 50000-64999 0.139
## 7 65000-74999 0.058
## 8 75000-99999 0.084
## 9    1000000+ 0.097
ggplot(personal_income, aes(y = Total, x = Income)) +
geom_bar(stat = "identity") +
theme(axis.text.x = element_text(angle = 90))

The data has the highest frequency for the income range of $35,000 to $49,999 with a left skew. However, the distribution is bimodal with a second smaller peak at the $100,000+ range.

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

sum(personal_income$Total[1:5])
## [1] 0.622

2.6.3 (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.

P(<$50,000 | Female) = P(<$50,000 & Female) / P(Female)
P(<$50,000 & Female) = P(<$50,000 | Female) * P(Female)

Assuming that females and males have same proportion of earning less than $50,000 as the overall sample, we can assume:
P(<$50,000 | Female) = P(<$50,000 | Male) = 0.622
P(<$50,000 & Female) = 0.622???0.41 = 0.255

2.6.4 (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.

P(<$50,000 | Female) = P(<$50,000 & Female) / P(Female)
P(<$50,000 & Female) = P(<$50,000 | Female) * P(Female)
P(<$50,000 & Female) = 0.718???0.41 = 0.294