Problems

2.6

Dice rolls. If you roll a pair of fair dice, what is the probability of

(a) getting a sum of 1?

The probablility of rolling a sum of 1 with a pair of dice is 0. The lowest sum you could roll with a pair of dice is 2.

(b) getting a sum of 5?

There are 4 ways of getting a sum of 5: \((1,4),(2,3),(3,2),(4,1)\) out of \(6^2\) outcomes. Therefore the probability is \(\frac{1}{36}\) = 0.111

(c) getting a sum of 12?

*There is only one way to get a sum of 12, and that is with both dice rolling a 6.

P(sum of 12 ) = \(\frac{1}{36}\) or 0.03*

2.8

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. There are 4.2% of people who are in poverty and speak a language other than English.

(b) Draw a Venn diagram summarizing the variables and their associated probabilities.

library (VennDiagram)
poverty <-14.6
foreign.language<-20.7
both<-4.2
poverty.only<-foreign.language-both

venn.diag <- draw.pairwise.venn(poverty, foreign.language, cross.area = both, c("Poverty", "Foreign Language"), fill=c("red","blue"), cat.dist=-0.10, ind=FALSE)
grid.draw(venn.diag)

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

We subtract the portion of Americans who speak a foreign language from those who are below the poverty line. So, It is 0.104

(d) What percent of Americans live below the poverty line or speak a foreign language at home?

Only speak english: 1 - 20.7% = 79.3%. P(below poverty line and speak english) = 14.6% x 79.3% = 11.5%

(e) What percent of Americans live above the poverty line and only speak English at home?

Answer: 4.2%

(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. These events are not independent. An independent event is defined as events where this statement is true: \(P(A)\cap P(B)=P(A)P(B)\). The Proff : \(0.146 * 0.207 \neq 0.042\).

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 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?

Because a male and female eye colors are independent from one another, P(self male with blue eyes) + P(partner female with blue eyes) - P(both male female with blue eyes)= \(\frac{114}{204} + \frac{108}{204} - \frac{78}{204}\) or approximately 0.71.

(b) What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?

The probability is \(\frac{78}{114}\) or approximately 0.68.

(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(female with blue eyes| male with brown eyes ) = \(\frac{19}{54}\) or about 0.35.

P(female with blue eyes| male with green eyes ) = \(\frac{11}{36}\) or approximately 0.31.

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

The eye colors are not independent

2.30

Books on a bookshelf. The table 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 first) x P(paperback fiction) = \(\frac{28}{95} * \frac{59}{94}\) or approximately 0.18.

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

Since we are drawing without replacement, it is possible that we draw a hardcover fiction book first, thus altering the probability of getting a hardcover book on the second draw. In other words, our second draw is not independent so we cannot simply multiply their individual probabilities.

P(Hardcover fiction) = \(\frac{13}{95}\), P(Softcover fiction) = \(\frac{59}{95}\), P(Hardcover|Hardcover fiction) = \(\frac{27}{94}\), P(Hardcover|Softcover fiction) = \(\frac{28}{94}\).

P(fiction and hardcover) = \((\frac{13}{95} \times \frac{27}{94}) + (\frac{59}{95} \times \frac{28}{94}) =\) 0.224.

(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 and hardcover) = \(\frac{72}{95} \times \frac{28}{95} =\) 0.223.

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

It is only because non-replacement only impacted 1 of 95 books, The impact of the overall percentage number is small.

2.38

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.

num_bags = c(0,1,2)
cost_baggage = c(0,25,35)
probability_baggage = c(.54,.34,.12)
df = data.frame(num_bags,cost_baggage,probability_baggage)
row.names(df) = c('Number of bags','cost of baggage','probability of event')
df['probability_event'] = df$cost_baggage*df$probability_baggage
event_prob = sum(df$probability_event)
variance_pe = var(df$probability_event)
sd_pe = sd(df$probability_event)
df
##                      num_bags cost_baggage probability_baggage
## Number of bags              0            0                0.54
## cost of baggage             1           25                0.34
## probability of event        2           35                0.12
##                      probability_event
## Number of bags                     0.0
## cost of baggage                    8.5
## probability of event               4.2

The average revenue per passenger would be 2.7 with a standard deviation of .25 and with a variances of 8.06.

(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.

We except the average revenue per flight of 120 passengers $1524 with a standard deviation of $510.01.

2.44

Income and gender. The relative frequency table 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.

library(ggplot2)
percent = c(2.2,4.7,15.8,18.3,21.2,13.9,5.8,8.4,9.7)
income_str <- ordered(c("$1 - $9,999 or less", 
            "$10,000 to $14,999", 
            "$15,000 to $24,999",
            "$25,000 to $34,999",
            "$35,000 to $49,999",
            "$50,000 to $64,000",
            "$65,000 to $74,999",
            "$75,000 to $99,999",
            "$100,000 or more"))
income = c(10,15,25,35,45,55,65,75,100)
data <- data.frame(percent,income)
colnames(data) <- c("Percentage","Income")
barplot(income , percent,xlab='Distribution')

It appears that this is unimodal with slight right skew.

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

P(Less than $50K ) = 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.*

P(make lesser than $50k and female) = \(0.41\times\) 62.2 = 25.5%.

(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.

Since only 6220% of all Americans earn under $50k, overall number of females in the samples is (41%), so We can say that there are more females earning under $50k than males. That is Why the statement in (c) above that gender and income are unlikey to be independent.