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 : The Probability is zero as the sum will never be 1

  1. getting a sum of 5?

Answer : The sum can be 5 in the outcomes ((4,1),(1,4),(2,3),(3,2)) among 36 combinations and hence the probablity P(4/36) = 0.11

  1. getting a sum of 12?

Answer : The sum can be 12 in the outcomes (6,6) among 36 combinations and hence the probablity P(1/36) = 0.027

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 : The two outcomes are not disjoint as they both can happen and 4.2% fall into this category

  1. Draw a Venn diagram summarizing the variables and their associated probabilities.
library(VennDiagram)
## Loading required package: grid
## Loading required package: futile.logger
venn.plot <- draw.pairwise.venn(area1      = 14.6,
                                area2      = 20.7,
                                cross.area = 4.2,
                                category   = c("Below Poverty Line", "Foreign Language"))

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

Answer : Roughly 14% of Americans live below Poverty line and speak only English.

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

Answer : 35.3% of Americans live below Poverty line or speak language other than English. 14.6%+20.7%

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

Answer : Roughly 60.5% of Americans live above the poverty line and speak only English at home.

  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 : These two events are dependent

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?

P(male respondent or partner) = P(Male Respondent with blue eyes) + P(partner with blue eyes) - P(Neither with blue eyes) = (114+108-18)/204

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

P(male respondent given partner with blue eyes) = (78/108)

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

P(male respondent with brown eyes given partner with blue eyes) = (19/108) P(male respondent with brown eyes given partner with blue eyes) = (11/108)

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

They are dependent as P(A | B) <> p(A)

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.

The probablility is 67/94

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

  2. 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.

The Probability is (28/94) - (13/71)

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

The Probability is (28/95) - (13/72). The difference in the events is very negligible i.e. 1 hardcover book which is fiction has created only little difference.

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. X P(X) X P(X) (X-E(X))^2 P(X)*(X-E(X))^2 First Bag 25 34% 8.50 (151.29) 51.43 Second Bag 35 12% 4.20 (497.29) 59.67 No Bag 0 54% 0 (161.29) 87.09 E(X)=12.70 V(X)=198.19 SD = sqrt(198.19) = 14.7 Average Revenue = 12.7$ Standard Deviation = 14.7$

  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.

        X      P(X)          X P(X)  (X-E(X))^2   P(X)*(X-E(X))^2

    First Bag 25 41/120 8.54 (146.16) 49.94 Second Bag 35 15/120 4.37 (487.96) 60.99 No Bag 0 64/120 0 (166.66) 88.88 E(X)=12.91 V(X)=199.81 SD = sqrt(199.81) = 14.13

Total Revenue = 12.91 * 120 = 1549.20 $ SD = 14.13 $

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.

Since Income is a continuous numeric variable the distribution is almost a smooth curve.

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

The Probablity is 0.62

  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.

P(US Resident income <50000 and Female) = 0.62 * 0.41 = 0.254

Assumption is 41% of females is distributed evenly for the income range.

  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.

P(US Resident income <50000 and Female) = 0.62 * 0.71 = 0.44 The assumption is invalid