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: 0; the minimum value with sum of two dice would 2 (1 on each).
  2. getting a sum of 5?
    Answer: 4/36 = 1/8. [ 4 combinations: 1+4 ; 2+3 ; 3+1 ; 4+1 ]
  3. getting a sum of 12?
    Answer: 1/36; the minimum value with sum of two dice would 12 [ 1 combination: 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.

  1. Are living below the poverty line and speaking a foreign language at home disjoint?
    Answer: No, there 4.2% of of population falling in both categories.
  2. Draw a Venn diagram summarizing the variables and their associated probabilities.
    Answer: Venn Diagram Below:
#install.packages("VennDiagram")                       # Install VennDiagram package
library("VennDiagram")                                # Load VennDiagram package
## Loading required package: grid
## Loading required package: futile.logger
grid.newpage()
draw.pairwise.venn(area1 = 14.6,                        # Create pairwise venn diagram
                   area2 = 20.7,
                   cross.area = 4.2,
                   col = "red",
                   fill = c("green", "orange"),
                   category = c("below poverty line", "speak foreign language"))

## (polygon[GRID.polygon.11], polygon[GRID.polygon.12], polygon[GRID.polygon.13], polygon[GRID.polygon.14], text[GRID.text.15], text[GRID.text.16], text[GRID.text.17], text[GRID.text.18], text[GRID.text.19])
  1. What percent of Americans live below the poverty line and only speak English at home?
    Answer: from the Venn diagram we can see this is 14.6% - 4.2% = 10.4% .
  2. What percent of Americans live below the poverty line or speak a foreign language at home?
    Answer: As below:
    P(below below poverty line) + P(speaks foreign language) - P(both English speaking and below poverty line)
    (14.6% + 20.7% ) - 4.2%) = 31.1%.
  3. What percent of Americans live above the poverty line and only speak English at home?
    Answer: As below: P(not below poverty and not speaking foreign language) = 1 - P(below poverty and speaks foreign language)
    100% - 31.1% = 68.9%.
  4. Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?
    Answer: Yes its not independent: If below the poverty line and speaks a foreign language represent events from two different and independent processes, then the probability that both below the poverty line and speaks a foreign language occur can be calculated as the product of their separate probabilities:
    P(below the poverty line and speaks a foreign language) = P(below the poverty line)×P(speaks a foreign language)
    4.2% = 14.6% * 20.7%
    4.2% != 3%

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:
    = Probability (male blue eyes or female blue eyes)
    = Probability (male blue eyes ) + Probability ( female blue eyes) - (Probability (male blue eyes and female blue eyes) )

    114/204  + 108/204 - 78/204
    ## [1] 0.7058824

  2. What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?
    Answer:
    = Probability (male blue eyes and male blue with female blue eyes)
    = Probability ( male blue with female blue eyes) / Probability (male blue eyes)

    (78/204) / (114/204)
    ## [1] 0.6842105

  3. 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?
    Answer:
    Probability that a randomly chosen male respondent with brown eyes has a partner with blue eyes:
    = Probability (male brown eyes and male brown with female blue eyes)
    = Probability ( male brown with female blue eyes) / Probability (male brown eyes)

    (19/204) / (54/204)  
    ## [1] 0.3518519

    Probability of a randomly chosen male respondent with green eyes having a partner with blue eyes:
    = Probability (male green eyes and male green with female blue eyes)
    = Probability ( male green with female blue eyes) / Probability (male green eyes)

    (11/204) / (36/204)
    ## [1] 0.3055556

  4. Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.
    Answer: Yes its not independent: If male blue eyes and female blue eyes represent events from two different and independent processes, then the probability that both male blue eyes and female blue eyes occur can be calculated as the product of their separate probabilities:
    P(male blue eyes and female blue eyes) = P(male blue eyes)×P(female blue eyes)
    (78/204) = (114/204) x (108/204)
    0.38 = 0.55 x 0.53 0.38 != 0.29

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:
    = Probability (drawing a hardcover book and paperback fiction book second when drawing without replacement)
    = Probability (drawing a hardcover book ) * P(paperback fiction book second when drawing without replacement)

    (28/95) * (59/94)  #94 is with reduction of 1 for non-replacement  
    ## [1] 0.1849944

  2. Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
    Answer:
    = Probability (drawing a fiction book and hardcover book second, when drawing without replacement)
    = Probability (drawing a fiction book ) * P(hardcover book second, when drawing without replacement)

    (72/95) * (28/94)  #94 is with reduction of 1 for non-replacement  
    ## [1] 0.2257559

  3. 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:
    = Probability (drawing a fiction book and hardcover book second, when drawing with replacement)
    = Probability (drawing a fiction book ) * P(hardcover book second, when drawing with replacement)

    (72/95) * (28/95)  #95 on both as its with replacement  
    ## [1] 0.2233795

  4. The final answers to parts (b) and (c) are very similar. Explain why this is the case.
    Answer:
    The loss of 1 book is not significant against the larger set of 95 books. For this reason the change is not significant.

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: Average revenue per passenger is 15.7 with Standard Deviation of 19.95 [sqrt((0-15.7)^2*.54+(25-15.7)^2*.34+(60-15.7)^2*.12))]

    #vector for count of checked luggage
checked.bag.count <- c (0, 1, 2)
#vector for dollar value for checked luggage
checked.bag.value <- c (  0, 25, 60)
#vector for probability of checked baggage count
checked.bag.prob <- c (54, 34, 12)

#average revenue per customer is
avg.revenue.pc <- sum(checked.bag.value * checked.bag.prob) / sum(checked.bag.prob)
avg.revenue.pc
    ## [1] 15.7
    ##Standard Deviation
19.95
    ## [1] 19.95

  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.
    Answer: Revenue from 120 passenger will be $1884; And with standrad deviation the revenue will be $1884 +/- $19.95

    # 54% of 120 will checkin no luggage; each person pay $0. Getting the total to:
zero.luggage.revenue <- ((54/100)*120) * 0
# 34% of 120 will checkin one piece@25; each person pay $25. Getting the total to:
one.luggage.revenue <- ((34/100)*120) * 25
# 12% of 120 will checkin one piece@25 and second piece@35; each person pay $25+$35 =$60. Getting the total to:
two.luggage.revenue <- ((12/100)*120) * 60
#Total revenue will be
zero.luggage.revenue + one.luggage.revenue + two.luggage.revenue
    ## [1] 1884

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: We can use a Barplot here to see the distribution of the data; we can see two peaks in the plot. With majortiy / mean being at the middle in 15k to 65k range.

    barplot(c(2.2 , 4.7 , 15.8 , 18.3 , 21.2 , 13.9, 5.8, 8.4, 9.7), xlab="Income",ylab="Total",col="blue",main="Distribution of annual total personal income",border="red")

  2. What is the probability that a randomly chosen US resident makes less than $50,000 per year?
    Answer: Probability that a randomly chosen US resident makes less than $50,000 per year is 0.622 :

    (2.2 + 4.7 + 15.8 + 18.3 + 21.2 ) / 100
    ## [1] 0.622

  3. 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: Probability that a randomly chosen US resident makes less than $50,000 per year and is female is 0.25502 :

    ((2.2 + 4.7 + 15.8 + 18.3 + 21.2 ) / 100 ) * .41
    ## [1] 0.25502

  4. 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: Assumption in part c is not valid as 71% of 41% of overall population is not same as the probabilty calculated in part c. 25.5% [from part c] vs 29.11% [from below]

    (71/100) * ((41/100))
    ## [1] 0.2911