Submission Instructions

  1. Complete all questions inside this R Markdown file.
  2. Knit the file to HTML (click Knit).
  3. Open the resulting HTML in your web browser.
  4. Save the HTML as a PDF:
    • File → Print → Save as PDF
  5. Submit both files on Blackboard:
    • Homework_4.Rmd
    • Homework_4.pdf

Instructions

1

Create a sequence of 20 numbers as follows: the first number is 5, and each number there after is found by multiplying the current number by 1.5 and then subtracting 1. Print the resulting sequence.

# Your code here
sequence <- numeric(20)

sequence[1]<- 5

for(x in 2:20){
  sequence[x]<- 1.5* sequence[x-1]
}

print(sequence)
##  [1]     5.00000     7.50000    11.25000    16.87500    25.31250    37.96875
##  [7]    56.95312    85.42969   128.14453   192.21680   288.32520   432.48779
## [13]   648.73169   973.09753  1459.64630  2189.46945  3284.20418  4926.30627
## [19]  7389.45940 11084.18910

2

Suppose you have the following vector of probabilities, each of which measures the probability a person has a particular disease:

0.25, 0.83, 0.76, 0.25, 0.33, 0.51, 0.67, 0.50, 0.54, 0.75, 0.71, 0.03, 0.38, 0.64, and 0.19.

Create a new variable which takes value:

  • 1 whenever the person is predicted to have the disease (probability ≥ 0.5).

  • 0 whenever the person is predicted not to have the disease (probability < 0.5).

Do this two different ways:

  • First, using a loop.

  • Second, using the ifelse() function.

Print the resulting vectors. They should contain the exact same values.

# Your code here


p <- c(0.25,0.83,0.76, 0.25, 0.33, 0.51, 0.67, 0.50, 0.54, 0.75, 0.71, 0.03, 0.38, 0.64,0.19)

prd <- numeric(length(p))

for (i in 1: length(p)) {
  if (p[i] >= 0.5 ){
    prd[i] <- 1 
  }else{
    prd[i] <- 0
  }
}
print(prd)
##  [1] 0 1 1 0 0 1 1 1 1 1 1 0 0 1 0
prdfelse <- ifelse(prd >= 0.5, 1, 0)

print(prdfelse)
##  [1] 0 1 1 0 0 1 1 1 1 1 1 0 0 1 0

3

In this problem you will work with the Cars2020 dataset, which you downloaded from
http://www.lock5stat.com/datapage3e.html and used on Homework 3. Please work with the CSV again here.

Create a new variable that classifies the fuel efficiencies of the cars in the dataset based on the following criteria:

  • Highway mpg under 20 is considered poor.
  • Highway mpg between 20 and 25 (inclusive) is considered acceptable.
  • Highway mpg above 25 and at most 35 is considered good.
  • Highway mpg above 35 is considered great.

Make sure the dataset contains the new variable, and then print only the following for every car in the dataset:

  • the make and model of the car and

  • the classified fuel efficiency.

#install.packages("Lock5Data")
install.packages("Lock5Data", repos = "https://cloud.r-project.org")
## Installing package into 'C:/Users/david/AppData/Local/R/win-library/4.5'
## (as 'lib' is unspecified)
## package 'Lock5Data' successfully unpacked and MD5 sums checked
## 
## The downloaded binary packages are in
##  C:\Users\david\AppData\Local\Temp\RtmpM9I4tE\downloaded_packages
library(Lock5Data)
## Warning: package 'Lock5Data' was built under R version 4.5.2
data("Cars2020")

Cars2020$fuel_efficiency <- ifelse(
  Cars2020$HwyMPG < 20, "poor",
  ifelse(Cars2020$HwyMPG <= 25, "acceptable",
  ifelse(Cars2020$HwyMPG <= 35, "good", "great"))
)

result <- data.frame(
  Make = Cars2020$Make,
  Model = Cars2020$Model,
  FuelEfficiency = Cars2020$fuel_efficiency
)

4

Use a loop to fill a 6 × 8 matrix (six rows and eight columns) as follows:

  • If the sum of the indices of an element’s position in the matrix is even, then the element is the product of the indices.
  • If the sum of the indices is odd, then the element is the sum of the indices plus three.

Print the resulting matrix.

Mat <- matrix(0, nrow = 6, ncol = 8)

for (i in 1:6) {
  for (j in 1:8){
    
    if ((i+ j) %% 2 == 0){
      Mat[i,j] <- i * j 
    }else {
      Mat[i,j]<- i+j +3 
    }
    }
  }
print(Mat)
##      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
## [1,]    1    6    3    8    5   10    7   12
## [2,]    6    4    8    8   10   12   12   16
## [3,]    3    8    9   10   15   12   21   14
## [4,]    8    8   10   16   12   24   14   32
## [5,]    5   10   15   12   25   14   35   16
## [6,]   10   12   12   24   14   36   16   48

5

Suppose I taught a course last semester that consisted of:

  • Homework (35%)

  • Midterm 1 (20%)

  • Midterm 2 (20%)

  • Final exam (25%)

Final letter grades are assigned as follows:

  • A = [90, 100]
  • B = [80, 90)
  • C = [70, 80)
  • D = [60, 70)
  • F = below 60

The grades for the students in the class can be found in the CourseGrades dataset on Blackboard.

Determine the letter grade each student earned in the course. Add that letter grade as a new variable to the data frame, and print the final data frame.

grades <- read.csv("CourseGrades.csv", header = TRUE)
#grades
grades$FinalScore <- 0.35 * grades$HW_Avg +
  0.20 * grades$Exam1 +
  0.20 * grades$Exam2 +
  0.25 * grades$FinalExam

grades$LetterGrade <- ifelse(grades$FinalScore >= 90, "A",
                             ifelse(grades$FinalScore >= 80, "B",
                                    ifelse(grades$FinalScore >= 70, "C",
                                           ifelse(grades$FinalScore >= 60, "D", "F"))))

print(grades)
##     Student HW_Avg Exam1 Exam2 FinalExam FinalScore LetterGrade
## 1      Bill  79.62    80    77        76    78.2670           C
## 2   Cameron  86.11    83    83        95    87.0885           B
## 3      Abby  94.53    94    92        86    91.7855           A
## 4      Will  78.22    77    76        96    81.9770           B
## 5     Janet  84.52    85    83        80    83.1820           B
## 6     Chris  85.79    85    83        81    83.8765           B
## 7     Laura  89.25    90    85        81    86.4875           B
## 8  Patricia  83.56    81    80        74    79.9460           C
## 9    George  96.91    97    92        76    90.7185           A
## 10   Joseph  84.17    85    80        96    86.4595           B
## 11  Diamond  87.51    87    84        78    84.3285           B
## 12     Jill  90.89    87    88        91    89.5615           B
## 13     T.J.  82.64    82    79        75    79.8740           C
## 14      Amy  78.76    77    74        68    74.7660           C

6

Suppose a friend of yours is currently a second-semester senior, and her GPAs for the previous semesters (starting first semester freshman year) are:

4.00, 3.50, 3.67, 3.33, 2.67, 3.42, and 3.56

Assuming she took the same number of credits each semester, write a loop that calculates and stores her cumulative GPA after each semester.

Then print the resulting vector of cumulative GPAs.

gpa <- c(4.00, 3.50, 3.67, 3.33, 2.67, 3.42, 3.56)

cgpa<- numeric(length(gpa))

for (i in 1:length(gpa)) {
  cgpa[i] <- mean(gpa[1:i])
  
}
print(cgpa)
## [1] 4.000000 3.750000 3.723333 3.625000 3.434000 3.431667 3.450000

7

Returning to Problem 6, suppose your friend was on an academic scholarship that requires the student to maintain a cumulative GPA of at least 3.6 in order to continue receiving financial support the following semester.

If at any point the student’s cumulative GPA drops below 3.6, they lose the scholarship permanently.

Write a while loop that counts the number of semesters your friend received financial support during her college career. Print that number.

Assume your friend received the scholarship starting first semester freshman year.

gpa <- c(4.00, 3.50, 3.67, 3.33, 2.67, 3.42, 3.56)

i <- 1 
count <- 0
currnt_sum <- 0

while (i <= length(gpa)) {
  currnt_sum <- currnt_sum + gpa[i]
  cgpa <- currnt_sum/ i
  
  if (cgpa < 3.6){
    break
  }
  
  count<- count + 1
  i <- i + 1 
  
}

print(count)
## [1] 4

8

The Fibonacci sequence is defined as follows:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

Each number after the first two is the sum of the two preceding numbers.

If you are not familiar with the Fibonacci sequence or how each number is generated, please search for it online. It is a fascinating sequence that appears frequently in mathematics and nature!

Write an R script that calculates and prints the first 40 numbers in the Fibonacci sequence.

You may start with the first two numbers (0 and 1). Use a loop or any method you prefer to generate the remaining numbers.

FS <- numeric(40)

FS[1]<-0
FS[2]<- 1

for (i in 3:40){
  FS [i] <- FS[i-1] + FS[i-2]
}

print(FS)
##  [1]        0        1        1        2        3        5        8       13
##  [9]       21       34       55       89      144      233      377      610
## [17]      987     1597     2584     4181     6765    10946    17711    28657
## [25]    46368    75025   121393   196418   317811   514229   832040  1346269
## [33]  2178309  3524578  5702887  9227465 14930352 24157817 39088169 63245986