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

  1. getting a sum of 1?

Ans: Zero probability of gettig a sum 1 with 2 fair dice.

  1. getting a sum of 5??

Ans: Possible combinations for sume of 5

1 + 4
2 + 3
3 + 2
4 + 1

paste0('The probability of getting a sum of 5 is ', round((4/36),3))
## [1] "The probability of getting a sum of 5 is 0.111"
  1. getting a sum of 12?

Ans: There is only 1 combination that will lead to a sum of 12, (6,6).

paste0('The probability of getting a sum of 12 is ', round((1/36),3))
## [1] "The probability of getting a sum of 12 is 0.028"

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.

  1. Are living below the poverty line and speaking a foreign language at home disjoint?

Ans: No.

  1. Draw a Venn diagram summarizing the variables and their associated probabilities.
library(VennDiagram)
## Loading required package: grid
## Loading required package: futile.logger
#below data in percentile
below_poverty <- 14.6
foreign_language <- 20.7
both_cat <- 4.2

vennDgm <- draw.pairwise.venn(below_poverty, foreign_language, both_cat, category = c("BelowPoverty People", "ForeignLanguage"))
grid.draw(vennDgm)

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

Ans: It is evident from the diagram that only .104% people livebelow the poverty line and only speak english at home.

poverty_line <- 0.146
foreignlanguage <- 0.207
both_class <- 0.042
poverty_line - both_class
## [1] 0.104
  1. What percent of Americans live below the poverty line or speak a foreign language at home?

Ans: 

poverty_line + foreignlanguage - both_class
## [1] 0.311
  1. What percent of Americans live above the poverty line and only speak English at home?

Ans: 

1- (poverty_line + foreignlanguage - both_class)
## [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?

Ans: 

(poverty_line * foreignlanguage) 
## [1] 0.030222
both_class
## [1] 0.042

Poverty having English with other language (.0302), does not equal to poverty and English with other lanaguage (0.042), so the events are dependent.

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.

  1. What is the probability that a randomly chosen male respondent or his partner has blue eyes?
108/204 + 114/204 - 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?
78/114
## [1] 0.6842105
  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?
19/54
## [1] 0.3518519
11/36
## [1] 0.3055556
  1. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.
#blue
78/204
## [1] 0.3823529
114/204 * 108/204
## [1] 0.2958478
#brown
23/204
## [1] 0.1127451
54/204 * 55/204
## [1] 0.07136678
#green
11/204
## [1] 0.05392157
36/304 * 41/204
## [1] 0.02380031

Ans: The events are not independent according to the multiplication rule

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.

  1. Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement.
round(28/95 * 59/94, 3)
## [1] 0.185
  1. Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
round(72/95 * 28/94, 3)
## [1] 0.226
  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.
round(72/95 * 28/95, 3)
## [1] 0.223
  1. The final answers to parts (b) and (c) are very similar. Explain why this is the case.

Ans: One book difference will not impact the probability as much as if there were a lot less books since there are a lot of books (94 vs. 95)

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.
luggage_0 <- 0
luggage_1 <- 25
luggage_2 <- 25+35

baggage_fee <- c(luggage_0, luggage_1, luggage_2)
luggage_percent <- c(0.54, 0.34, 0.12)

luggage_rev <- baggage_fee * luggage_percent

expected_val <- sum(luggage_rev)

# sqrt((x1 - u)^2 * P(X = x1) + .... + (xk - u)^2 * P(X = xk))
baggage_sd <- sqrt((luggage_0 - expected_val)^2 * (luggage_percent[1]) + (luggage_1 - expected_val)^2 * (luggage_percent[2]) + (luggage_2 - expected_val)^2 * (luggage_percent[3]))

paste0("The Expected value is:", expected_val)
## [1] "The Expected value is:15.7"
paste0('The baggage sd value is:', round(baggage_sd,4))
## [1] "The baggage sd value is:19.9502"
  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.
expected_120 <- 120 * expected_val
variance_120 <- 120 * baggage_sd^2
variance_120
## [1] 47761.2
sd_120 <- sqrt(variance_120)
sd_120
## [1] 218.5434
round(expected_120,2)
## [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.

  1. Describe the distribution of total personal income.

Ans: The distribution is skewed to the right. The distribution is skewed probably because of the fewer people with very high salaries ($100k or more)

  1. What is the probability that a randomly chosen US resident makes less than $50,000 per year? P(less than $50K)
less_50k <- 2.2 + 4.7 + 15.8 + 18.3 + 21.2
less_50k
## [1] 62.2
  1. What is the probability that a randomly chosen US makes less than $50,000 per year and is female? Note any assumptions you make.
females <- 0.41
femaleless_50k <- less_50k * females
femaleless_50k
## [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.
less_50k
## [1] 62.2
femaleless_50k
## [1] 25.502

Income and gender are dependent. The percentage of females who made less than 50k a year (71.8%) does not equal the % of all people who made less than 50K a year (62.2%)