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?

Answer

With a pair of dice the minimum sum is 2 so events(E) of getting sum=1 is 0, so Probability is

P(E) = n(E)/n(S) = 0/36= 0

  1. getting a sum of 5?

Answer

E= {(1,4),(2,3),(3,2),(4,1)}= 4

P(E)=4/36=1/9

  1. getting a sum of 12?

Answer

E={(1,1)}

P(E)=1/36

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?

Answer

Since there is population which is below pverty line and also speaks foreign language, they are not disjoint.

  1. Draw a Venn diagram summarizing the variables and their associated probabilities.
library(VennDiagram)
library(grid)
library(futile.logger)


vennDiag <- draw.pairwise.venn(area1 = 14.6, area2 = 20.7, cross.area = 4.2, category = c("BPL", 
    "Foreign language"),fill = c("light blue", "pink"), cat.pos = c(0, 
    0), cat.dist = rep(0.025, 2))
grid.draw(vennDiag);

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

Answer

10.4 %

  1. What percent of Americans live below the poverty line or speak a foreign language at home?

Answer

Total in below poverty line + total speaking foreign language - the total that meet both criteria.

(14.6 + 20.7) - 4.2 = 35.3-4.2=31.1%

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

100 - 31.1 = 68.9%

  1. Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?

Answer

If the events are independent then P(A ∩ B) = P(A)P(B).

P(A ∩ B) = 0.042 P(A)P(B) = 0.146 x 0.207 =0.030222

Not equal, so they are not independent.

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?

Answer

P(A) = Probability of blue eyed males P(B) = Probability of blue eyed partners

P(A or B) = P(A)+P(B)-P(A and B) =(114/204) + (108/204)-(78/204) =0.7059 70.59%

  1. What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?

Answer

P(A|B) = P(A and B)/P(B)

P(A and B) = Blue eyed male with blue eyed partner = 78 P(B) = Blue eyed males = 114 so, 78/114=0.6842 or 68.42%

  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?

P(A|B) = P(A and B)/P(B)

P(A and B) = brown eyed male with blue eyed partner = 19 P(B) = brown eyed males = 54 so, 19/36=0.3518 or 35.18%

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

11/36 = 30.56%

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

Answer

From the above calculated probabilities we see - Eye colors do not seem independent, probabilities calculated above are different and show that the values are not independent.

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.

Answer

(28/95)x(59/94)= 0.1849=18.49%

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

Answer

(72/95)x(28/94)= 0.2257=22.57%

  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.

Answer

(72/95)x(28/95)= 0.2233=22.33%

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

Answer

Larger the samples, smaller the difference. When the possible events are considerable large, the outcome will not be affected by much when there’s no replacement in random drawings.

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.

Answer

bags <- c('0-Bag','1-Bag','2-Bags')
fees <-c(0,25,60)
pass<-c(0.54,0.34,0.12)
sample <- data.frame(bags, fees, pass)
avg_rev_pp <- sum(sample$pass*sample$fees)/sum(sample$pass)
paste("Average revenue per passenger is ",avg_rev_pp)
## [1] "Average revenue per passenger is  15.7"
sd<- sqrt(0.54*(0-avg_rev_pp)^2 + 0.34*(25-avg_rev_pp)^2 + 0.12*(60-avg_rev_pp)^2)
paste("Standard deviation is ",round(sd,digits = 2))
## [1] "Standard deviation is  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.

Answer

For 120 passengers

for120 <- 120*15.7
paste("For 120 passengers the revenue is ",for120)
## [1] "For 120 passengers the revenue is  1884"
paste("Standard deviation is ",round(sd,digits = 2))
## [1] "Standard deviation is  19.95"

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.

Answer

As per the below plot, it is a continuous, multi-modal distribution

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

Answer

Less than 50,000 are 2.2+4.7+15.8+18.3+21.2 = 62.2% probability

  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.

Answer

Since we don’t have the information of females having less than 50k income we assume the events are independent :,

P(less than 50k income and female)=P(less than 50k income)xP(female) = 0.6220.41 =0.2550=25.50%*

  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.

Answer

P(less than 50k income and female)=P(less than 50k income)xP(female) 0.718 = 0.622*0.41 0.718 ≠ 0.255, This shows that the events are not independent and the assumption in case (c) doesn’t hold good.