HW2
Qixing Li
February 17, 2019
2.6 Dice rolls
If you roll a pair of fair dice, what is the probability of
(a) getting a sum of 1?
The probability is 0, because the smallest sum is 2.(b) getting a sum of 5?
The probability of getting P(1, 4) = P(1) x P(2) = 1/6 x 1/6 = 1/36
The probability of getting P(2, 3) = P(2) x P(3) = 1/6 x 1/6 = 1/36
The probability of getting P(3, 2) = P(3) x P(2) = 1/6 x 1/6 = 1/36
The probability of getting P(4, 1) = P(4) x P(1) = 1/6 x 1/6 = 1/36
Therefore the probability of getting a sum of 5 is 1/36 + 1/36 + 1/36 + 1/36 = 4/36 = 1/9(c) getting a sum of 12?
The probability of getting a sum of 12 is P(6) x P(6) = 1/6 x 1/6 = 1/362.8 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. Some of the families who both speak a foreign language at home and also fall below the poverty line.(b) Draw a Venn diagram summarizing the variables and their associated probabilities.
library(VennDiagram)
draw.pairwise.venn(area1 = 14.6, area2 = 20.7, cross.area = 4.2,category = c('Poverty', 'Foreign Language'), cat.dist=-0.1, fill = c("red", "blue"))
## (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])(c) What percent of Americans live below the poverty line and only speak English at home?
14.6 - 4.2 = 10.4%(d) What percent of Americans live below the poverty line or speak a foreign language at home?
20.7 + 14.6 - 4.2 = 31.1%(e) What percent of Americans live above the poverty line and only speak English at home?
1-16.5%-4.2%-10.4% = 68.8%(f) Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?
No, they are not indepentdent. Because P(Below Poverty Line and speak Foreign Language) != P(Below Poverty Line) * P(speak Foreign Language)
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?
P(male or parner blue) = (114+108-78)/ 204 = 70.59%(b) What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?
P(male and parner blue) = 78/204 = 38.25%(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 and partner blue) = 19/204 = 9.31%
P(male green and partner blue)= 11/204 = (d) Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.
The eye colos of male respondents and their partners are not independent.
The probability of a couple have the same color of eyes is: (78+23+16)/204=57.35%
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.
P(hardcover & paperback fiction)= (28/95) * (59/94)= 18.50%(b) Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
P(Hardcover Fiction & Hardcover) = (13/95) * (27/94) = 3.93%
P(Paperback Fiction & Hardcover) = (59/95) * (28/94) = 18.50%
P(Fiction first & Hardcover second) = 3.93% + 18.50% = 22.43%(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) = 72/95 = 75.79%
P(Hardcover) = 28/95 = 29.47%
P = 75.79% * 29.47% = 22.34%(d) The final answers to parts (b) and (c) are very similar. Explain why this is the case.
Because we only take 2 times of the book it may not make so much difference.
2.38 Baggage fees
An airline charges the following baggage fees: $25 for the first bag and $35 for the second. Suppose 54% of passengers have no checked luggage, 34% have one piece of checked luggage and 12% have two pieces. We suppose a negligible portion of people check more than two bags.
(a) Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation.
library(tidyverse)
baggage <- tribble(~bag, ~fees, ~prob,
#--|--|----
0, 0, 0.54,
1,25,0.34,
2,60, 0.12)
baggage$weightRev <- baggage$prob * baggage$fees
# Compute the average revenue
avgRev <- sum(baggage$weightRev)
avgRev## [1] 15.7# Compute Variance
baggage$DiffMean <- baggage$fees - avgRev
baggage$DiffMeanSqrd <- baggage$DiffMean ^ 2
baggage$DiffMeanSqrdTimesProb <- baggage$DiffMeanSqrd * baggage$prob
as.data.frame(baggage)## bag fees prob weightRev DiffMean DiffMeanSqrd DiffMeanSqrdTimesProb
## 1 0 0 0.54 0.0 -15.7 246.49 133.1046
## 2 1 25 0.34 8.5 9.3 86.49 29.4066
## 3 2 60 0.12 7.2 44.3 1962.49 235.4988# Compute standard deviation
varRevPerPax <- sum(baggage$DiffMeanSqrdTimesProb)
sdRevPerPax <- sqrt(varRevPerPax)
sdRevPerPax## [1] 19.95019The average revenue per passenger is 15.7 and the standard deviation is 19.95.(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 (15.7)= 1884
standard deviaiton=120(19.95)=23942.44 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.
The distribution is multimodal, skewed to the right.(b) What is the probability that a randomly chosen US resident makes less than $50,000 per year?
p(x<50,000)= 0.022+0.047+0.158+0.183+0.212= 62.2%(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.
Assuming that the relationship is independent between the income and the gender,
then P(less than 50,000 and female)= P(less than $ 50,000) * P(female)
= (0.622) * (0.41)= 25.50%(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.
The same data source indicates that 71.8% of females make less than $50,000, which is very different from the result above. Therefore, the assumption is not valid.