Data606Week03
Monu Chacko
February 16, 2019
2.6 Dice rolls. If you roll a pair of fair dice, what is the probability of
- getting a sum of 1?
Answer
The probability of getting a sum of 1 is 0
- getting a sum of 5?
Answer
Total Possibilities of getting 5 is 4
Data:
1 + 4
2 + 3
3 + 2
4 + 1
Total Possibilities is 6 x 6 = 36
So the probability of getting 5 is
4/36## [1] 0.1111111#or
1/9## [1] 0.1111111- getting a sum of 12?
Answer
Total Possibilities of getting 12 is 1
Data: 6 + 6
Total Possibilities is 6 x 6 = 36
So the probability of getting 12 is
1/36## [1] 0.027777782.8 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?
Answer
No they are not disjoint. You can be under poverty line and speak foreign language
- Draw a Venn diagram summarizing the variables and their associated probabilities.
library(VennDiagram)
belowPovertyLine <- 14.6
speakForeignLanguage <- 20.7
bothCategories <- 4.2
venn.plot <- draw.pairwise.venn(belowPovertyLine, speakForeignLanguage, cross.area=bothCategories, c("Below Poverty", "Speak Only Foreign Language"), fill=c("green", "yellow"), cat.dist=-0.08, ind=FALSE)
grid.draw(venn.plot)
Answer
- What percent of Americans live below the poverty line and only speak English at home?
Answer
povertyOnly <- belowPovertyLine - bothCategories
povertyOnly## [1] 10.4- What percent of Americans live below the poverty line or speak a foreign language at home?
Answer
speakForeignLanguage + belowPovertyLine - bothCategories## [1] 31.1- What percent of Americans live above the poverty line and only speak English at home?
Answer
povertyOnly <- belowPovertyLine - bothCategories
100 - speakForeignLanguage - povertyOnly ## [1] 68.9- Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?
Answer
(belowPovertyLine/100) * (speakForeignLanguage / 100)## [1] 0.030222Answer
P(A or B) = P(A)+P(B)-P(A and B)
((114+108)/204) - (78/204)## [1] 0.7058824- What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?
Answer
P(A|B) = p(A and B)/P(B)
78/114## [1] 0.6842105- 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
19/54## [1] 0.3518519#probability of a randomly chosen male respondent with green eyes having a partner with blue eyes
11/36## [1] 0.3055556- Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.
Answer
Eye colors of male respondents and their partners does not appear independent. Data for the same color seem to be larger in all categories.
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.
Format Hardcover Paperback Total Type Fiction 13 59 72 Nonfiction 15 8 23 Total 28 67 95
- Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement.
Answer
(28/95) * (59/94)## [1] 0.1849944- Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
Answer
(72/95) * (28/94)## [1] 0.2257559- 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
((72/95)*(28/95))## [1] 0.2233795- The final answers to parts (b) and (c) are very similar. Explain why this is the case.
Answer
This is because of the total data is large i.e 95. Taking one book is less significant. Take the example below with a smaller size. The difference is larger.
((2/5)*(3/4))## [1] 0.3((2/5)*(3/5))## [1] 0.242.38 Baggage fees. An airline charges the following baggage fees: $25 for the first bag and $35 for the second. Suppose 54% of passengers have no checked luggage, 34% have one piece of checked luggage and 12% have two pieces. We suppose a negligible portion of people check more than two bags.
- Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation.
Answer
#Probability of checked luggage - 0 bags 54%, 1 bag - 34%, 2 bags - 12%
probabilityCheckedLuggage <- c(0.54, 0.34, 0.12)
#Number of bags
bagQuantity <- c(0, 1, 2)
#Baggage Fee
baggageFee <- c(0, 25, 25 + 35)
dfCheckinBags <- data.frame(probabilityCheckedLuggage, bagQuantity, baggageFee)
dfCheckinBags$probabilityFee <- dfCheckinBags$probabilityCheckedLuggage * dfCheckinBags$baggageFee
dfCheckinBags## probabilityCheckedLuggage bagQuantity baggageFee probabilityFee
## 1 0.54 0 0 0.0
## 2 0.34 1 25 8.5
## 3 0.12 2 60 7.2#Revenue per passenger
revenuPassenger <- sum(dfCheckinBags$probabilityFee)
revenuPassenger## [1] 15.7#Variance
dfCheckinBags$MeanDf <- dfCheckinBags$probabilityFee - revenuPassenger
dfCheckinBags$Sqr <- dfCheckinBags$MeanDf ^ 2
dfCheckinBags$sp <- dfCheckinBags$Sqr * dfCheckinBags$probabilityCheckedLuggage
dfCheckinBags## probabilityCheckedLuggage bagQuantity baggageFee probabilityFee MeanDf
## 1 0.54 0 0 0.0 -15.7
## 2 0.34 1 25 8.5 -7.2
## 3 0.12 2 60 7.2 -8.5
## Sqr sp
## 1 246.49 133.1046
## 2 51.84 17.6256
## 3 72.25 8.6700#SD
var <- sum(dfCheckinBags$sp)
stdDev <- sqrt(var)
stdDev## [1] 12.62538- 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
#120 passengers
noOfPassengers <- 120
avgerageInc <- revenuPassenger * noOfPassengers
avgerageInc## [1] 1884#SD
var120 <- (noOfPassengers ^ 2) * var
sd120 <- sqrt(var120)
sd120## [1] 1515.0462.44 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.
Answer
- Describe the distribution of total personal income.
Answer
income <- c("$1 - $9,999 or loss","$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")
total <- c(2.2,4.7,15.8,18.3,21.2,13.9,5.8,8.4,9.7)
#incomelowerrange <- c(1,10000, 15000, 25000, 35000, 50000, 65000, 75000, 100000)
#incomeupperrange <- c(9999,14999, 24999, 34999, 49999, 64000, 74999, 99999, 199999)
#incomeuppermiddle <- (incomelowerrange + incomeupperrange) /2
#incometotal <- incomeuppermiddle * total
df_incomegender <- data.frame(income, total)
df_incomegender## income total
## 1 $1 - $9,999 or loss 2.2
## 2 $10,000 to $14,999 4.7
## 3 $15,000 to $24,999 15.8
## 4 $25,000 to $34,999 18.3
## 5 $35,000 to $49,999 21.2
## 6 $50,000 to $64,000 13.9
## 7 $65,000 to $74,999 5.8
## 8 $75,000 to $99,999 8.4
## 9 $100,000 or more 9.7hist(df_incomegender$total)
barplot(df_incomegender$total, names.arg=income)
- What is the probability that a randomly chosen US resident makes less than $50,000 per year?
Answer
#First 5 rows / total
probab50000 <- sum(df_incomegender[1:5,]$total) / sum(df_incomegender$total)
probab50000## [1] 0.622- 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
probab50000female <- 0.41 * probab50000
probab50000female## [1] 0.25502- 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
female50000 <- 0.718 * .41
female50000## [1] 0.29438There is a small variation from (c). The assumption is not valid.