CUNY 606 Homework Assignment 2

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

  1. getting a sum of 1?

Zero percent chance. The lowest sum that can be achieved is 2.

  1. getting a sum of 5?

There are four possible combinations that can lead you to a sum of 5. (2,3), (3,2), (1,4), (4,1). Hence, it is 4/36 = 1/9.

round(1/9,3)
## [1] 0.111
  1. getting a sum of 12?

There is only 1 combination that will lead to a sum of 12, (6,6). Therefore the answer is 1/36.

round(1/36,3)
## [1] 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?

No. There are 4.2% of people who are in poverty and speak a language other than English.

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

2.8b ** Above represents the Venn Diagram.**

  1. What percent of Americans live below the poverty line and only speak English at home?
.042
## [1] 0.042
  1. What percent of Americans live below the poverty line or speak a foreign language at home?
.104 + .042 + .165
## [1] 0.311
  1. What percent of Americans live above the poverty line and only speak English at home?
1 - (.104 + .042 + .165)
## [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(Poverty) x P(English with other language) = .0302, which does not equal to P(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?
19/54
## [1] 0.3518519

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

11/36
## [1] 0.3055556
  1. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.

P(Blue Eye Males) x P(Blue Eye Females) = (114/204) x (108/204) = .296 (assuming independence). However, we know that P(Blue Eye males and Blue Eye Females) in this data set = (78/204) = .382, which is not equal. Thus this is dependent.

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.
(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.

This depends on the circumstance. If the first book is a hardcover fiction book first and then the second book is a hardcover, the probability is:

(72/95) * (27/94)
## [1] 0.2176932

If was a paperback fiction book first and then the second book was a hardcover, the probability is:

(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.
(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.

The sample sizes are large enough where if you take one book out, it would be negligible in the overall calculations.

2.38 Baggage fees. An airline charges the following baggage fees: 25 dollars for the first bag and 35 dollars 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.
BagFee <- c(0, 25, 60)
Prob <- c(.54, .34, .12)

Expected.Value <- (BagFee[1] * Prob[1]) + (BagFee[2] * Prob[2]) + (BagFee[3] * Prob[3])

Variance <- 0
i <- 1
while (i <= length(BagFee)){
  temp <- ((BagFee[i] - Expected.Value)^2 * Prob[i])
  Variance <- Variance + temp
  i <- i + 1
}

round(Expected.Value,2)
## [1] 15.7
round(Variance, 2)
## [1] 398.01
SD <- sqrt(Variance)
round(SD,2)
## [1] 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.
EX120 <- 120 * Expected.Value
Var120 <- 120 * SD^2
SD120 <- sqrt(Var120)
round(EX120,2)
## [1] 1884
round(Var120,2)
## [1] 47761.2
round(SD120,2)
## [1] 218.54

There are 120 different discrete passengers on this airline. The assumption is that each passenger paid a 15.70 dollars on average for bag fees.

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.

It appears that this is unimodal with slight right skew.

  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.

The female population comprises of 41% of the population. We assume that being female and making money are independent factors. P(Less than 50k & female) =

.622 * .41
## [1] 0.25502
  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.

Part C states that .255 of the population are female and makes less than 50k. Here, in part D, this states that 71.8% of females made less than 50k. There is a huge discrepancy in the numbers, and thus it is unlikely to be independent.