Title: CUNY SPS MDS Data606_HW3"

Author: Charles Ugiagbe

Date: “9/17/2021”

Exercise 1

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?

Solutions 1:

Total number of outcomes in rolling 2 dice = 6 X 6 = 36 Probability = no of required outcome/no of Total outcome

  1. Prob(getting a sum of 1) = 0 (the least possible outcome it rolling 2 dice is 2)

  2. Prob(getting a sum of 5) possible outcomes = (1,4); (2,3); (4,1); (3,5)

getting_a_sum_of_5 <- 4/36
round(getting_a_sum_of_5,2)
## [1] 0.11

(c) Prob(getting a sum of 12)

possible outcome = (6,6)

getting_a_sum_of_12 <- 1/36
round(getting_a_sum_of_12,2)
## [1] 0.03

Exercise 2:

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?
  2. Draw a Venn diagram summarizing the variables and their associated probabilities.
  3. What percent of Americans live below the poverty line and only speak English at home?
  4. What percent of Americans live below the poverty line or speak a foreign language at home?
  5. What percent of Americans live above the poverty line and only speak English at home?
  6. Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?

Solutions 2:

(a) No, they are not Disjoint because some American lives below poverty line and also speaks foreign language.

  1. Let A = American live below the poverty line, P(A) = 0.146

    Let B = Speak language other than English at home, P(B) = 0.207

    P(A⋂B) = 0.042

    P(A⋂B′) = 0.146-0.042 = 0.104

    P(B⋂A′) = 0.207-0.042 = 0.165

library(VennDiagram)
## Warning: package 'VennDiagram' was built under R version 4.1.1
## Loading required package: grid
## Loading required package: futile.logger
## Warning: package 'futile.logger' was built under R version 4.1.1
grid.newpage()
draw.pairwise.venn(area1 = 14.6, area2 = 20.7, cross.area = 4.2,
                   category = c("Living below Poverty Line", "Speak a Forign Language"), 
                   lty = rep("blank", 2), fill = c("#6495ED", "#ffb84d"), 
                   alpha = rep(0.4, 2), cat.pos = c(0,  0), cat.dist = rep(0.02, 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])

  1. percentage that live below poverty and only speaks English

    P(A⋂B′) = 0.146-0.042 = 0.104

    percentage = 10.4%

  2. percentage that live a below a poverty line or speak a foreign languge.

    P(A⋃B) = P(A) + P(B) - P(A⋂B)

    P(A⋃B) = 0.146 + 0.207 - 0.042 = 0.311

    percentage = 31.1%

  3. percentage that lives above the poverty line and only speak English at home

    100% - 31.1% = 68.9%

  4. The events A and B are not independent from each other because P(A) ≠ P(A|B).

Exercise 3:

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?
  2. What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?
  3. Whatistheprobabilitythatarandomlychosenmalerespondentwithbrowneyeshasapartner with blue eyes? What about the probability of a randomly chosen male respondent with green eyes having a partner with blue eyes?
  4. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.

Solution 3:

(a) (108 + 114 - 78)/204 = 0.7059

(b) 78/114 = 0.6842

(c) Probability that male respondent with brown eyes has partner with blue eyes:

19/54 = 0.3519.

Probability that male respondent with green eyes has partner with blue eyes:

11/36 = 0.3056

(d) Eye colour of male respondent and their partner are not independent

Reason:

P(male=Blue|female=Brown)≠ P(male=Blue)

P(male=Blue|female=Brown) = P(male=Blue & female=Brown)/P(female=Brown)
23/204 / 55/204 = 23/55 = 0.4182

P(male = Blue) = 114/204 = 0.5588

since P(male=Blue|female=Brown)≠P(male=Blue), we say they are not independent

Exercise 4:

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.
  2. Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
  3. 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.
  4. The final answers to parts (b) and (c) are very similar. Explain why this is the case.

Solutions 4:

(a) P(X) = P(Hardcover Book) x P(Paperback Fiction Book) = 28/95 * 59/94 = 0.1850

(b) P(X) = P(Fiction Book) x P(Hardcover Book) = 72/95 * 28/94 = 0.2258

(c) P(X) = P(Fiction Book) x P(Hardcover Book) = 72/95 * 28/95 = 0.2234

(d) The final answer to parts (b) and (c) are very similar because in part (b), book is drawn without replacement and the total outcome reduces to 94 after the first pick while in part (c) books is drawn with replacement and the total outcome remains the same after the first pick

Exercise 5:

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.
  2. 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.

Solution 5:

library(tinytex)

(a)

The exptected value E(X) = ∑x∗P(X=x) E(X) = (0 * 0.54) + (25 * 0.34) + (60 * 0.12) = 15.7

The average revenue per passenger is $15.7.

The variance V(X) = E(X^2) - (E(X))^2

V(X) = [(0^2)0.54 + (25^2)0.34 + (60^2)*0.12] - (15.7)^2

V(X) = 644.5 - 246.49 = 398.01.

variance V(X) = 398.01

The standard deviation SD(X)= √Var(X)

SD(X) = √398.01 = √19.95.

Standard Deviation (SD) = √19.95.

(b)

The expected revenue E(x) for a flight of 120 passengers

= 120 * E(X)

= 120 * $15.7 = $1,884.

Exercise 6:

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.
  2. What is the probability that a randomly chosen US resident makes less than $50,000 per year?
  3. 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.
  4. 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.

Solution 6:

(a)

library(ggplot2)
Income <- c("$1 to $9,999 or less", "$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")
Income_Percent <- c(.022, .047, .158, .183, .212, .139, .058, .084, 0.097)

IncomeTable <- cbind(Income,Income_Percent)

IncomeTable <- data.frame(IncomeTable)

IncomeTable$Percentage <-  as.double(levels(IncomeTable$Income_Percent))[IncomeTable$Income_Percent]

ggplot(IncomeTable, aes(x=factor(Income),y=Income_Percent,fill=factor(Income))) + geom_bar(stat="identity") + labs(title="Distribution of Total Personal Income") + ylab("Percentage") + theme(legend.position = "none", axis.title.x = element_blank(), axis.text.x=element_text(angle=45)) + theme(plot.title = element_text(hjust=0.5)) + theme(axis.text.x = element_text(margin = margin(t = 25, r = 20, b = 0, l = 0)))

(a) From the plot, Income looks normally distributed but it increases towards the end.

(b) P(X < $50,000) = 0.022+0.047+0.158+0.183+0.212 = 0.622.

(c) Assuming that income and female are independent.

Then P(X<$50,000 and X=females) = P(X<$50,000) * P(X=females)

= 0.622 * 0.41 = 0.26.

(d) In (c), we made the assumption that the probability of US resident income and female are independent. That is 0.26. However, the observed data source shows that the 71.8% of female makes less than $50,000. Therefore, the assumption made in part (c) is not valid. In other words, income and female are not independent.