Midterm1
Diego De Armas
2023-02-09
Q1
install.packages("tidyverse", repos = "https://cloud.r-project.org" )##
## The downloaded binary packages are in
## /var/folders/5_/389qrkvs1sd7nkp792bslx5r0000gn/T//RtmpdBnz7E/downloaded_packageslibrary(tidyverse)## ── Attaching packages ─────────────────────────────────────── tidyverse 1.3.2 ──
## ✔ ggplot2 3.4.0 ✔ purrr 1.0.1
## ✔ tibble 3.1.8 ✔ dplyr 1.1.0
## ✔ tidyr 1.3.0 ✔ stringr 1.5.0
## ✔ readr 2.1.3 ✔ forcats 1.0.0
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()mydata <- read_csv("challenger-1.csv")## Rows: 23 Columns: 4
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr (1): incident
## dbl (3): launch, temp, o_ring_probs
##
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.mydata## # A tibble: 23 × 4
## launch temp incident o_ring_probs
## <dbl> <dbl> <chr> <dbl>
## 1 1 53.6 Yes 3
## 2 2 57.2 Yes 1
## 3 3 57.2 Yes 1
## 4 4 62.6 Yes 1
## 5 5 66.2 No 0
## 6 6 66.2 No 0
## 7 7 66.2 No 0
## 8 8 66.2 No 0
## 9 9 66.2 No 0
## 10 10 68 No 0
## # … with 13 more rowshead(mydata)## # A tibble: 6 × 4
## launch temp incident o_ring_probs
## <dbl> <dbl> <chr> <dbl>
## 1 1 53.6 Yes 3
## 2 2 57.2 Yes 1
## 3 3 57.2 Yes 1
## 4 4 62.6 Yes 1
## 5 5 66.2 No 0
## 6 6 66.2 No 0tail(mydata)## # A tibble: 6 × 4
## launch temp incident o_ring_probs
## <dbl> <dbl> <chr> <dbl>
## 1 18 75.2 Yes 2
## 2 19 75.2 No 0
## 3 20 75.2 No 0
## 4 21 78.8 No 0
## 5 22 78.8 No 0
## 6 23 80.6 No 0install.packages("pscyh", repos = "https://cloud.r-project.org")## Warning: package 'pscyh' is not available for this version of R
##
## A version of this package for your version of R might be available elsewhere,
## see the ideas at
## https://cran.r-project.org/doc/manuals/r-patched/R-admin.html#Installing-packageslibrary(psych)##
## Attaching package: 'psych'## The following objects are masked from 'package:ggplot2':
##
## %+%, alpha1-A)
str(mydata)## spc_tbl_ [23 × 4] (S3: spec_tbl_df/tbl_df/tbl/data.frame)
## $ launch : num [1:23] 1 2 3 4 5 6 7 8 9 10 ...
## $ temp : num [1:23] 53.6 57.2 57.2 62.6 66.2 66.2 66.2 66.2 66.2 68 ...
## $ incident : chr [1:23] "Yes" "Yes" "Yes" "Yes" ...
## $ o_ring_probs: num [1:23] 3 1 1 1 0 0 0 0 0 0 ...
## - attr(*, "spec")=
## .. cols(
## .. launch = col_double(),
## .. temp = col_double(),
## .. incident = col_character(),
## .. o_ring_probs = col_double()
## .. )
## - attr(*, "problems")=<externalptr>summary(mydata)## launch temp incident o_ring_probs
## Min. : 1.0 Min. :53.60 Length:23 Min. :0.0000
## 1st Qu.: 6.5 1st Qu.:66.20 Class :character 1st Qu.:0.0000
## Median :12.0 Median :69.80 Mode :character Median :0.0000
## Mean :12.0 Mean :69.02 Mean :0.4348
## 3rd Qu.:17.5 3rd Qu.:74.30 3rd Qu.:1.0000
## Max. :23.0 Max. :80.60 Max. :3.0000describe(mydata$launch)## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 23 12 6.78 12 12 8.9 1 23 22 0 -1.36 1.41describe(mydata$temp)## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 23 69.02 6.97 69.8 69.33 5.34 53.6 80.6 27 -0.4 -0.44 1.45describe(mydata$o_ring_probs)## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 23 0.43 0.79 0 0.26 0 0 3 3 1.81 2.69 0.161- b)
The variable temperature is interval level due to the fact tha you can categorize, rank and, infer equals intervals.
the variable launch is ordinal level due to the fact that can categorize and rank the data.
the variable o ring probs is ratio level because you can categorize, rank, infer equals intervals and, there is a true zero.
the variable incident is nominal level due to the fact that you can only categorize the data by labellinng.
1- C)
?hist
hist(mydata$o_ring_probs,
main = "Histogram",
xlab = "O Ring Probs",
col = "purple"
)
## 1- d)
boxplot( formula(mydata$temp~mydata$incident),
notch=F,
horizontal=TRUE,
main = "Box Plot",
xlab = "Tempeture",
ylab = "Incident",
col = c("pink", "purple")
)
If the tempeture for the day of the launch is 36 degrees fahrenheir this might cause a concern because as we can see in the box plot all the launches with a temperature lower of 62 fahrenheit resulted on an incident.
1- E)
flaunch <- order(mydata$temp)
flaunch ## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23Looking at the table we can see that the first successfull launch ocuured on lauch number 5 at a temperature of 66.2
1- F)
length((mydata$temp>=65)==TRUE)## [1] 23Q2- A)
#Identify the variables
# Telling the truth
a <- 0.8
# not telling the truth
a2 <- 0.2
# Detecting actual truth teller
b <- 0.90
# Detecting actual liar
b2 <- 0.59
#calculate the rest of the variables
notbgivena <- 1 - b
notbgivena2 <- 1 -b2
#Joint Probabilities
ab <- a * b
abp <- a * notbgivena
ab2 <- a2 * b2
ab2p <- a2 * notbgivena2bayes theorem
bt <- (a2*b2)/(a*b+a2*b2)
bt## [1] 0.1408115print(bt * 100)## [1] 14.08115The probability of an individual is actually a liar given the polygraph detection is 14%
Q2- B)
result <- a2 * a
result ## [1] 0.16the probability of ramdon individual to be a liar is 16%
result2 <- 0.59 * 0.41
result2## [1] 0.2419The probability of the polygraph to accuarate identified the ramdon individual as a liar is 24%
Q3- A)
# model as Poisson
# Identify the variables
# what is n ?
na <- 9
# probability of success
pi <- 0.5
lambda3a <- na * pi
sd3a <- sqrt(lambda3a)
sd3a## [1] 2.12132dpois(na, lambda3a)## [1] 0.02316458Q3- B)
#Model As Binomial
n3b <- 8
pi3b <- 0.5
result3b <- dbinom(0,8,p=0.5)
result3b## [1] 0.00390625sd3b <- n3b*pi3b*(1-pi3b)
sd3b## [1] 2ev <- n3b * pi3b
ev## [1] 4Q4- A)
# 0.75 is equal to wrong result
#0.25 is equal to right result
result4a <- 0.75 * 0.75 * 0.25
result4a## [1] 0.140625Q4 - B)
#Binomial
# math equation
#P(3 <= x <= 4| n = 5, pi = 0.25)
n4 <- 5
pi4 <- 0.25
x4 <- 3:4
result4b <- sum(dbinom(x4, n4, pi4))
result4b## [1] 0.1025391Q4- C)
#identify the variables
x4c <- 2.5
result4c <- 1-pbinom(x4c, n4, pi4, lower.tail = TRUE)
result4c## [1] 0.1035156#double checking
result4c2 <- pbinom(x4c, n4, pi4, lower.tail=FALSE)
result4c2## [1] 0.1035156Q5- A1)
#identify the variables
mean5a <- 72.6
sd5a <- 4.78
#math equation
#P(x < 80 | mean = 72.6, sd = 4.78)
result5a1 <- pnorm( q = 80, mean = mean5a, sd = sd5a)
result5a1 ## [1] 0.939203Q5- A2)
#math equation
# P (68 < x < 78 | mean = 72.6, sd = 4.78)
result5a2 <- pnorm( q = 78, mean = 72.6, sd = 4.78) - pnorm( q= 68, mean = 72.6 , sd = 4.78)
result5a2## [1] 0.7027615Yes as we can see the median velocity of a car in the interstate 5 is 72.6 miles. As we can see in the result for question 5-a2, we had that a total of 70% of cars travel between 68 miles and 78 miles. Both velocities are close to the mean for which we can say that most of the cars are driving at close number to 72.6 miles. This can be more easily appreciating if we created a bell curve graph.
Q5- A3)
#math equation
# P(x > 70| mean = 72.6, sd = 4.78)
#adjust for discrete variable
result5a3 <- 1- pnorm( q = 70, mean = mean5a, sd = sd5a, )
result5a3## [1] 0.7067562Q5- b1)
#Identify the variables
mean5ba <- 4313
sd5ba <- 583
result5ba <- qnorm(p = 0.05, mean = 4313, sd = 583)
result5ba## [1] 3354.05Q5- b2)
#identify the variables
mean5b2 <- 5261
sd5b2 <- 807
result5b2 <- qnorm(p = 0.9, mean = 5261, sd = 807)
result5b2## [1] 6295.212