title: " Data 606: Chapter 3 - Probability"

author: “Sufian”

RPubs Link:

http://rpubs.com/ssufian/528292

output:

html_document:

df_print: paged

pdf_document:

extra_dependencies:

- geometry
- multicol
- multirow

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?

Ans:

You will always have sum >1, Not possible, so probability is zero

2. getting a sum of 5?

Ans:

0.11030 (see simulation below based on 100 thousand trials)

3. getting a sum of 12?

Ans:

0.02863 (see simulation below based on 100 thousand trials)

require(dice)

## Loading required package: dice

## Loading required package: gtools

two.dice <- function(){
  dice <- sample(1:6, size = 2, replace = TRUE)
  return(sum(dice))
}

sims <- replicate(100000, expr=two.dice())
# A table of simulation and its associated probablities
#table(sims)
table(sims)/length(sims)

## sims
##       2       3       4       5       6       7       8       9      10 
## 0.02775 0.05656 0.08280 0.11237 0.13848 0.16703 0.13706 0.11122 0.08264 
##      11      12 
## 0.05625 0.02784

---

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?

No, 4.2% fall into both categories

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

require(VennDiagram)

## Loading required package: VennDiagram

## Loading required package: grid

## Loading required package: futile.logger

## 
## Attaching package: 'futile.logger'

## The following object is masked from 'package:gtools':
## 
##     scat

# Pairwise Venn diagram

venn.plot <- draw.pairwise.venn(area1=.146,area2= .27,cross.area =0.042, c("Below proverty line",  "Foriegn Language"))
grid.draw(venn.plot)

grid.newpage()

```

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

ans:

P(Foreign Language)= 20.7%

P(English only) = 1-20.7% = 79.3%

P((under poverty and English only)) = P(poverty)XP(English only)=14.6%X79.3%=11.57%

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

P(poverty) = 14.6% and P(poverty and Foreign Language)= 4.2%

P(Foreign Language) = 20.7%

Ans:

P(Poverty or Foreign Language) = P(Foreign Language) + P(poverty) - P(poverty and Foreign Language)

                            = 14.6% + 20.7% - 4.2% = 31.1%

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

Ans:

P(above poverty and English ) = 1- P(poverty) = 14.6%

                          = 1- 14.6% 
                          
                          = (85.54%)X79.3%
                          
                          =67.7%
                          

6. Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?

Ans:

P(poverty) = 14.6%

P(Foreign Language) = 20.7%

P(Poverty and Foreign Language) = 4.2%

Test of independence : P(A and B) = P(A)*P(B)

In this case:

P(Poverty and Foreign Language) = 4.2% Not Equal to P(poverty) = 14.6% X P(Foreign Language) = 20.7%

0.042 Not Equal to 0.0302

Therefore these 2 events are dependent.

---

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?

ans:

P(self male with blue eyes) + P(partner female with blue eyes) - P(both male female and blue eyes) =

114/204 + 108/204 - 78/204 = 70.59%

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

ans:

P(female blue eyes|male blue eyes) = P(female blue eyes and male blue eyes) / p(male blue eyes)

                              = 68.84%
                              
                              

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

ans:

P(female blue eyes|male brown eyes) = P(female blue and male brown eyes) / p(male brown eyes)

                                = (19/204)/(54/204)
                                
                                = 35.18%
                                

P(female blue eyes|male green eyes) = P(female blue and male green eyes) / p(male green eyes)

                                = (11/204)/(36/204)
                                
                                = 11/36 = 30.55%

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

ans:

No, see illustraton below

P(A|B) = P(A), B has no effect on A

Using example from above:

P(female blue eyes|male green eyes) = P(female blue and male green eyes) / p(male green eyes)

                                = 30.55%
                                

P(female blue eyes|male green eyes) = P(female blue eyes)

                                = 108/204 = 52.9%, Male green eyes has no effect on female blue eyes
                                
                                

Because they are NOT equal, means they are dependent events

---

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.

ans:

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.

ans:

p(first: hard cover fiction)p(second: hard cover) + p(first: paper fiction)(p(second: hard cover)) =

13/9527/94 + 59/9528/94 = 22.4%

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.

ans:

With replacement

p(first: hard cover fiction)p(second: hard cover) + p(first: paper fiction)(p(second: hard cover)) =

13/9528/95 + 59/9528/95 = 22.3%

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

ans:

Because the percentage changes for replacement vs. without replacement is small

---

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.

ans:

Qty of bag P(X) Revs P(X)xRevs

zero bag 0.54 0 0

1 bag 0.34 25 8.5

2 bag 0.12 60 7.2

exp Rev= 15.7

x 0 25 35

prob 0.54 0.34 0.12

x*P 0 8.5 4.2

x-mean -15.7 9.3 19.3

(x-mean)^2 246.4 86.49 372.49

(x-mean)^2xP 133.10 29.40 44.69

vAR = 207.21

STD DEV = 14.39

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.

ans:

Passenger P(X) Split Qtybags Rev/Bag Rev 120 0.54 65 0 $0 $0 0.34 41 1 $25 $1,020 0.12 14 2 $60 $864

Expected rev = $1884

This is a really left-skewed distribution, most passengers do not carry bags and using std dev.

would be mis-leading

---

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.

A right skewed distribution

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

P(<50K) = 2.2+4.7+15.8+18.3+21.2 =62.2%

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.

P(<50K and female) = 62.2%x41% = 25.5%

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.

Use what is factually correct:

.41%x96,420,486 = 39,532,399

P(<50K and female) = .718

From this, P(<50K and female) = 39532399X.718 = 28,384,263

My estimate from part c is:

.255x39532399 = 24,597,224

No, I had under estimated in part c, because since its right skewed, the mean is disproportionatly higher than it should be, which makes the std dev much bigger than it should be.