DATA606 - HW 2

“OpenIntro Statistics”

library(VennDiagram)

## Loading required package: grid

## Loading required package: futile.logger

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

(a) getting a sum of 1?

No possible outcomes have a sum of 1 from a pair of fair dice, thus

Prob(sum of 1) = 0

(b) getting a sum of 5?

Outcomes that have a sum of 5: (1, 4) (2, 3) (3, 2) (4, 1)

Total number of outcomes: 36

Prob(sum of 5) = 4/36 = 1/9 = .11

(c) getting a sum of 12?

Outcomes that have a sum of 12: (6, 6)

Total number of outcomes: 36

Prob(sum of 12) = 1/36 = .03

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.

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

No, they are not disjoint because 4.2% fall into both categories.

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

grid.newpage()
draw.pairwise.venn(area1 = .146, area2 = .207, cross.area = .042, category = c("Poverty", "Foreign Language"), fill = c("light blue", "pink"), cat.pos = c(0, 0))

## (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])

# cat.pos - position of category titles, distanced from the center of circle

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

10.4% of Americans live below the poverty line and only speak English at home.

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

.165 + .042 + .104

## [1] 0.311

31.1% of Americans live below the poverty line or speak a foreign language at home.

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

\[ P(\textrm{above poverty and only English speaking}) = 1- P(\textrm{poverty or speak foreign language}) \]

1 - .311

## [1] 0.689

68.9% of Americans live below the poverty line or speak a foreign language at home.

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

If they are independent, then \[ P(\textrm{below poverty}) * P(\textrm{speaks foreign language}) = P(\textrm{below poverty & speak foreign language}) \]

.146 * .207 #prob. of below poverty multiplied by prob. of speaks foreign language

## [1] 0.030222

.042 #prob. of those fall into both category

## [1] 0.042

.146 * .207 == .042 #Are they equal?

## [1] FALSE

Since they are not equal, they are not independent of each other.

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.

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

(114 + 108 - 78) / 204

## [1] 0.7058824

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

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

19 / 54 #chosen male with brown eyes has a partner with blue eyes

## [1] 0.3518519

11 / 36 #chosen male with green eyes has a partner with blue eyes

## [1] 0.3055556

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

Eye colors of male respondents and their partners are independent if \[ P(\textrm{eye color of male}) * P(\textrm{eye color of female}) = P(\textrm{same color for both})\]

114/204 * 108/204

## [1] 0.2958478

78/204

## [1] 0.3823529

114/204 * 108/204 == 78/204

## [1] FALSE

Prob. of male with blue eye multiplied by prob. of their partners with blues does NOT equal to the prob. of both partners having blue eyes. Thus, they are not independent.

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.

(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.

2 cases:

draw a hardcover fiction book first (prob. = 13/95), then a hardcover book second (prob. = 27/94), without replacement

draw a paperback fiction book first (prob. = 59/95), then a hardcover book second (prob. = 28/94), without replacement

Sum the probabilities of these 2 cases

(13/95) * (27/94) + (59/95) * (28/94)

## [1] 0.2243001

(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.

The answers are similar because the sample space is large; drawing 1 book out of 95 books doesn’t effect the probability of drawing next book much.

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.

mean <- 0*.54 + 25*.34 + (25+35)*.12 #average revenue
sd <- sqrt(0^2*.54 + 25^2*.34 + 60^2*.12) #standard deviation
paste("Average revenue per passenger is $", mean)

## [1] "Average revenue per passenger is $ 15.7"

paste("Standard deviation is $", round(sd, 2))

## [1] "Standard deviation is $ 25.39"

(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.

Assume that 120 passengers are independent,

then the mean of this flight = 120 * individual mean

the variance of this flight = 120 * individual variance

paste("For a flight of 120 passengers we expect $", 120 * mean)

## [1] "For a flight of 120 passengers we expect $ 1884"

paste("with standard deviation $", round(sqrt(120 * sd^2), 2))

## [1] "with standard deviation $ 278.1"

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.

income_gender <- data.frame(Income = c("1-9,999 or loss", "10,000-14,999", "15,000-24,999", "25,000-34,999", "35,000-49,999", "50,000-64,999", "65,000-74,999", "75,000-99,999", "100,000 or more"), Total = c(.022, .047, .158, .183, .212, .139, .058, .084, .097))

bp <- barplot(income_gender$Total)
axis(1, at = bp, labels = income_gender$Income)

Personal income distribution is unimodal and symmetric centered at 35,000-49,999.

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

.02 + .047 + .158 + .183 + .212

## [1] 0.62

(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.

.62 * .41

## [1] 0.2542

Assuming 59% males and 41% females distributed evenly across all income levels, the percentage of female making less than $50,000 per year is 25.42%.

(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.

71.8% is a big difference from 25.42% I obtained from part (c), therefore the assumption that male and female proportions are distributed evenly across all income levels is not valid.

DATA606 - HW 2

Mia Chen

February 17, 2019

“OpenIntro Statistics”

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

(a) getting a sum of 1?

(b) getting a sum of 5?

(c) getting a sum of 12?

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

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

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

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

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

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

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

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

(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?

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

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.

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

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

(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.

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

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.

(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.

(a) Describe the distribution of total personal income.

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

(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.

(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.