Dice rolls. (3.6, p. 92) If you roll a pair of fair dice, what is the probability of
## 1.a The probability of rolling a sum of 1 is 0%. The minimum observable sum is 2 {}
## 1.b The probability of rolling a sum of 5 is 1/9 (approimately 11.11%).
## {(1, 4) (2,3), (3, 2), (4, 1)}
## 1.c The probability of rolling a sum of 12 is 1/36 (approximately 2.78%) {(6, 6)}
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.
## 2.a Living below the poverty line and speaking a foreign language at home are
## NOT disjoint. It is fully possible to live below poverty and also speak a
## foreign language.
## 2.b
## 2.c The percent of Americans living below the poverity line and only speak English
## at home is P(Below_Poverty) - P(Both) = 14.6% - 4.2% = 10.4%
## 2.d The percent of Americans living below the poverty line or speak a foreign
## language at home is P(Below_Poverty) + P(Foreign_Language) - P(Both)
## = 14.6% + 20.7% - 4.2% = 31.1%
## 2.e The percent of Americans living above the poverty line and only speak English
## at home is 100 - (P(Below_Poverty) + P(Foreign_Language) - P(Both))
## = 100% - 31.1% = 68.9%
## 2.f Using the Mulitiplication Rule for Independent Processees, we can show that
## unfortunately, no, the event that someone lives below the poverty line is
## independent of the event that the person speaks a foreign language at home
## are NOT independent.
##
## If the events were independent then
## P(Poverty AND Language) = P(Poverty) x P(Language)
## It was given that
## P(Poverty AND Language) = 4.2%
## P(Poverty) = 14.6%
## P(Language) = 20.7%
## P(Poverty) x P(Language) = 14.6% x 20.7% = 3.02% which is NOT equal to 4.2%.
## Thus the two events are not fully 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.
## 3.a The probability a randomly chosen male respondent or his partner has blue eyes
## can be found by P(male_blue) + P(partner_blue) - P(both)
## = 114/204 + 108/204 - 78/204 = 144/204 = 70.59%
## 3.b The probability a randomly chosen male respondent with blue eyes has a partner
## with blue eyes is 78/114 = 68.42%.
## 3.c The probability a randomly chosen male respondent with brown eyes has a partner
## with blue eyes is 19/54 = 35.19%
## The probability a randomly chosen male respondent with green eyes has a partner
## with blue eyes is 11/36 = 30.56%
## 3.d No, it does not appear that eye colors of male respondents and their partners
## are independent. The distribution of observed pairs does not follow the
## Mulitiplication Rule for Independent Processees.
## The probablity of male having blue is 114/204 = 56%
## The probablity of partner haing blue is 108/204 = 53%
## The observed probablity of both male and partner blue is 78/204 = 38%.
## P(male_blue) x P(partner_blue) = 56% x 53% = 30%
## Since P(male_blue) x P(partner_blue) is NOT equal to P(male_blue AND partner_blue)
## then these events 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.
## 4.a P(hardcover) = 28/95 = 29.5%
## P(paperback fiction without replacement) = 59/94 = 62.8%
## P(hardcover then paperback fiction without replacement)
## = 28/95 x 59/94 = 1652/8930 = 18.5%
## 4.c P(fiction) = 72/95 =83.2%
## P(hardcover) = 28/95 = 29.5%
## P(fiction then hardcover) = 72/95 x 28/95 = 2016/9025 = 22.3%
## 4.d The results are very similar because the scenarios differ by only one book.
## The results would differ more if the number of selections was greater.
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.
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.
## 6.a The distribution is right skewed
## 6.b The probability that a randomly chosen US resident makes less than $50,000
## per year is 62.2%
## 6.c The probability that a randomly chosen US resident makes less than $50,000
## per year and is female is 62.2% x 41% = 25.5%
## Assuming income is independent of gender
## 6.d Given that 71.8% of females make less than $50,000, it is clear the assumption
## that income is independent of gener is invalid.