Data 606 Homework 2

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Problem 2.6

Dice rolls.

P(sum of 2 dice is 1) = 0
The probability of getting a sum of 1 when you roll a pair of dice is not a possibility.

P(sum of 2 dice is 5)

1st die	2nd die	Probability
2	3	(1/6)*(1/6)=(1/36)
3	2	(1/6)*(1/6)=(1/36)
4	1	(1/6)*(1/6)=(1/36)
1	4	(1/6)*(1/6)=(1/36)

Rolling of first and second die are independent of each other. Each possible outcome are mutually exclusive.

P(2 and 3) or P(3 and 2) or P(4 and 1) or P(1 and 4) = (1/36) + (1/36) + (1/36) + (1/36) = (4/36) = 1/9 ~ 11.11%

P_sumOf2Dice_5 = (1/36) + (1/36) + (1/36) + (1/36)
P_sumOf2Dice_5

## [1] 0.1111111

P(getting a sum of 12)

1st die	2nd die	Probability
6	6	(1/6)*(1/6)=(1/36)

So, there’s only one way to get a sum of 12 when you roll 2 dice.

P(getting a sum of 12) = 1/36

Problem 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. 4.2% of the American survyed in 2010 both are living below the poverty line and speak a language other than English.

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

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

10.4% of Americans live below poverty line and only speak English at home.

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

P(live below poverty line) + P(Speak foreign language at home) - P(below poverty line and speak foreign language at home) =
14.6 + 20.7 - 4.2 = 31.1.

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

1 - [P(live below poverty line) + P(Speak foreign language at home) - P(below poverty line and speak foreign language at home)]
100 - 31.1 = 68.9

(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 are not mutually exclusive events. Knowing that an individual speaks a foreign language at home would give us information regarding the probability that this individual is living below the poverty line.

Problem 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.

Total Respondents = 204

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

Males with blue eyes = 114

Males with female partners with blue eyes = 108

Males with blue eyes and with female partners with blue eyes = 78

P(male has blue eyes) + P(female partner has blue eys) - P(couple both have blue eyes)
(114/204) + (108/204) - (78/204) = (144/204) = (12/17) ~ 70.59%

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

with blue eyes?

Males with blue eyes = 114

Males with blue eyes and with female partners with blue eyes = 78

P(male has blue eyes and partner has blue eyes | male has blue eyes)
78/114 = 13/19 ~ 68.42%

(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?

Males with brown eyes = 54

Males with brown eyes and with female partners with blue eyes = 19

P(male with brown eyes and female partner with blue eyes | male has brown eyes)
19/54 ~ 35.19%

Males with green eyes = 36

Males with green eyes and with female partners with blue eyes = 11

P(male with green eyes and female partner with blue eyes | male has green eyes)
11/36 ~ 30.56%

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

It appears that the colors of male respondents and their partners eyes are not independent of each other because the conditional probability of men with blue, brown, and green eyes having a partner with blue appear to be different enough. In this study, the probability of a blue eyed man having a female partner with blue eyes as well is 68%, which is a lot higher compared to about 35% for brown eyed men and to about 30% for green eyed men. This difference could either be because of an underlying association or just by chance. A null hypothesis testing would need to be performed.

Problem 2.30

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

Total Books = 95

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

Hardcover = 28

Paperback Fiction = 59

(28/95) * (59/94) = (1652/8930) = (826/4465) ~ 18.50%

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

Fiction = 72

Hardcover = 28

(72/95) * (28/94) ~ 22.58%

(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.

(72/95) * (28/95) ~ 22.34%

Problem 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.

(a) Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation.

The expected revenue per passenger is $15.70.

The standard deviation is $19.95.

(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.

The expected revenue for 120 passenger is 120 x 15.70 = $1,884.

The standard deviation is $19.95.

1 standard deviation from the expected revenue for 120 passengers would be (15.70 + 19.95) x 120 = $4,278.

The opposite direction of 1 standard deviation from the expected mean, it appears that the airline would lose (15.70 - 19.95 = -4.25) $4.25 per passenger. I’m not sure what this means. I rechecked my calculations three times, and the standard deviation is larger than the expected mean. So this would be -4.25 x 120 = -510.

Problem 2.44

Income and gender.
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.

(a) Describe the distribution of total personal income.

The distribution of personal income appears to be symmetric with a peak around the 35,000 to 49,000 range.

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

The probability that a randomly chosen US resident makes less than $50,000 a year is 62.20%.

(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.

The assumption I’m making is the proportion of male and female per category of income is consistent at 59% males and 41% females.
P(make less than 50K and female) = P(make less than 50k) x P(female) = .6220 x .41 = .2550 or 25.50%

(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.

The sample has 41% women. This means that .41 x .718 = .29 of the sample population who are women have income less than 50,000 a year.
So, So P(make less than 50k and female) = .6220 x .29 = 18.04%. So given this extra information regarding the female’s salary the chance of a randomly selecting someone who makes less than 50k and is also a female is actually less likely - from 25.50% to 18%.