2.6 Dice rolls.

If you roll a pair of fair dice, what is the probability of:

  1. getting a sum of 1?

    This is impossible so P(1) = 0

  2. getting a sum of 5?

    • Two combinations will come to 5, 1+4 and 2+3, so there are 4 possible rolls of the 36 so P(5) = 1/9

  3. getting a sum of 12?

    • Only one roll will hit 12 and it requires the same number on each die so P(12) = 1/36

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.

  1. Are living below the poverty line and speaking a foreign language at home disjoint?

    Yes

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

## (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])


Probability Matrix

##               Foreign Language English Total
## Below Poverty              4.2    10.4  14.6
## Above Poverty             16.5    68.9  85.4
## Total                     20.7    79.3 100.0

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

    10.4%

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

    31.1% (14.6% live below poverty line + 20.7% speak foreign language - 4.2% both)

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

    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?

    Yes

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 below summarizes the results. For simplicity, we only include heterosexual relationships in this exercise


Probability Matrix

##            Female Blue Female Brown Female Green Total
## Male Blue         38.2         11.3          6.4  55.9
## Male Brown         9.3         11.3          5.9  26.5
## Male Green         5.4          4.4          7.8  17.6
## Total             52.9         27.0         20.1 100.0

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

    Total Blue-eyed males (55.9%) + Total Blue-eyed females (52.9%)- Blue/Blue pairing (38.2%) = 70.6%

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

    Blue/Blue pairing (38.2) / Total Blue-eyed males (55.9) = 68.3%

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

    9.3%

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

5.4%

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

    Yes, they are independent. There is always some combination of male eye color with female eye color. More importantly, each person’s eye color has no impact on the eye color of another person

2.30 Books on a bookshelf.

The table below shows the distribution of books on a bookcase based on whether they are nonfiction or fiction and hardcover or paperback.

Frequency Matrix

##            Hardcover Paperback Total
## Fiction           13        59    72
## Nonfiction        15         8    23
## Total             28        67    95

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

    28/95 * 59/94 = 18.5%

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

    72/95 * 27/94 = 21.8%

  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.

    72/95 * 28/95 = 22.3%

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

    Because they are very similar situations, you’re just adjusting adding 1 to the numerator and denomenator of one of the factors.

2.38 Baggage fees.

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.

Probability Model

##                    No Bags 1 Bag  2 Bags
## x                     0.00 25.00   60.00
## P(X)                  0.54  0.34    0.12
## x*P(X=x)              0.00  8.50    7.20
## x-mu                -15.70  9.30   44.30
## (x-mu)^2            246.46 86.49 1962.49
## (x-mu)^2*P(X(X=x))  133.10 29.41  235.50
Standard Deviation = square root of $235.50 = $19.95

  1. About how much revenue should the airline expect for a flight of 120 passengers?

    $1,884

    With what standard deviation? Note any assumptions you make and if you think they are justified.

    Assuming that 54% / 34% / 12% equate to 54 of 100, 34 of 100, 12 of 100 you should just be able to multiply the SD for 100 passangers by 1.2 in order to get the SD for the larger data set.