2.6)

  1. P(sum of dice is 1) = 0
  2. P(sum of dice is 5) = 0.1111111
  1. P(sum of dice is 12) = 0.0277778

2.8)

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. Since you can live below the poverty line and speak a foreign language they are not disjoint.

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

library("VennDiagram", lib.loc="~/R/win-library/3.4")
## Loading required package: grid
## Loading required package: futile.logger
draw.pairwise.venn(20.7,14.6,4.2, c("         foreign language", "below poverty"))

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

10.4%

  1. What percent of Americans live below the poverty line or speak a foreign language at home?
14.6 + 20.7 - 4.2 
## [1] 31.1
  1. What percent of Americans live above the poverty line and only speak English at home?
100 - (14.6+20.7-4.2)
## [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?
indtest <- .146*.207
ifelse ((indtest == .042), "They are independent", "They are not independent")
## [1] "They are not independent"

2.20)

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

P(male = blue or female = blue) = P(male = blue) + P(female = blue) - P(male and female =blue)

P(male = blue or female = blue) = 114/204 + 108/204 - 78/204

P(male = blue or female = blue) = 0.7058824

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

P(blue eyed male has blue eyed partner) = 78/114

P(blue eyed male has blue eyed partner) = 0.6842105

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

P(brown eyed male has blue eyed partner) = 19/54

P(brown eyed male has blue eyed partner) = 0.3518519

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

P(green eyed male has blue eyed partner) = 11/36

P(green eyed male has blue eyed partner) = 0.3055556

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

P(blue eyed men with blue eyed women) =? P(blue eyed men) x P(blue eyed women)

78/204 =? 114/204 x 108/204

These values are not equal so the eye colors of male respondents and their partners are not independent.

2.30)

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

28/95 x 59/94 = 0.1849944

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

P(fiction chosen) x (P(hardcover|hardcover first) + P(hardcover|paperback first))

72/95 x ((28/94)x(59/72) + (27/94)x(13/72)) = 0.2243001

  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 x 28/95 = 0.2233795

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

The answers are similar because the sample size is a small sample of the population.

2.38)

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.
info <-  c("P(x)", "cost")
none <- c(.54, 0)
one <- c(.34, 25)
two <- c(.12, 60)

luggage <- data.frame (list(info, none, one, two), stringsAsFactors = FALSE)

colnames(luggage) <- c("info", "no pieces", "1 piece", "2 pieces")

luggage <- rbind(luggage, list("x*Prob", luggage[1,2]*luggage[2,2], luggage[1,3]*luggage[2,3], luggage[1,4]*luggage[2,4]))

avgoutcome <- sum(luggage[3,2:4], na.rm=TRUE)

dif1 <- luggage[2,2]-avgoutcome
dif2 <- luggage[2,3]-avgoutcome
dif3 <- luggage[2,4]-avgoutcome

sqr1 <- dif1^2
sqr2 <- dif2^2
sqr3 <- dif3^2

Psqr1 <- sqr1 * luggage[1,2]
Psqr2 <- sqr2 * luggage[1,3]
Psqr3 <- sqr3 * luggage[1,4]

variance <- sum(Psqr1,Psqr2,Psqr3, na.rm=TRUE)
stdev <- sqrt(variance)

luggage <- rbind(luggage, list("x - mean", dif1, dif2, dif3))
luggage <- rbind(luggage, list("(x - mean)^2", sqr1, sqr2, sqr3))
luggage <- rbind(luggage, list("(x - mean)^2 x P(x)", Psqr1, Psqr2, Psqr3))


luggage
##                  info no pieces 1 piece  2 pieces
## 1                P(x)    0.5400  0.3400    0.1200
## 2                cost    0.0000 25.0000   60.0000
## 3              x*Prob    0.0000  8.5000    7.2000
## 4            x - mean  -15.7000  9.3000   44.3000
## 5        (x - mean)^2  246.4900 86.4900 1962.4900
## 6 (x - mean)^2 x P(x)  133.1046 29.4066  235.4988
avgoutcome
## [1] 15.7
variance
## [1] 398.01
stdev
## [1] 19.95019

The mean is 15.7

The variance is 398.01

  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.

Revenue for 120 passengers = 1884

The standard deviation is 19.950188

I am assuming that the percentages of people that take 0 bags, 1 bag or 2 bags is independent of the number of people on the flight and so this would hold for a flight of 120 passengers. I would want to know the conditions that the original data was collected.

2.44) a) symmetric, unimodal

  1. 46.4%

  2. If you assume that gender and salary are independent P(making less than $50000 and being female) = .464 x .41 = 0.19024

  3. Gender and salary are not independent! You would expect 46.4% of people making under 50,000 dollars to be female. However the percentage is 71.8% of females make under $50000.