Spring 2015 - University of Alabama

Grades

This is the student’s grade in the course with one dropped homework (HW2) and Test 1 being replaced with the Final Exam score.

## [1] "77.6%"


This is the code that produces this result.

student = c(HW1=30, HW2=0, HW3=0, HW4=17.5, HW5=23, HW6=23.5, HW7=20, HW8=16.5, HW9=27.5, HW10=27.5, Exam1=76, Exam2=82, Final=80)
HWs = student[c(1:10)]  #I am creating a separate vector for just the homework grades.
HWsWithDrop = HWs[-2]   #Since homeworks 2 & 3 are both zero, I chose to remove HW2.
HWAvg = ((mean(HWsWithDrop)/30)*.25)  #Weighted homework average.
Tests = (student[c(11,12)])    #Separate vector for tests.
FinalExam = student[13]
Tests[1] = FinalExam      #Since test 1 is the lowest I am replacing it the final exam score.
TestAvg = ((mean(Tests)/100)*.45)   #Weighted test average.
FinalExamAvg = (FinalExam/100*.3)   #Weighted final.
CourseAvg = sum(HWAvg, TestAvg, FinalExamAvg)   #Overall course average.


This percentage was found by computing the weighted average for the homeworks, tests, and final exam, which resulted in the following weighted values. The sum of these values equals the course average.

## Homework 
##  "0.172"
##    Test 
## "0.365"
## Final Exam 
##     "0.24"


The calculations that found these averages are as follows (assuming grade drops/replacements have occurred at this point):

HW Avg = [mean(HWsWithDrop)/Points Available] x Course Weight
HW Avg = (20.6/30) x .25
HW Avg = 0.172

Test Avg = [mean(Tests)/Points Available] x Course Weight
Test Avg = (81/100) x .45
Test Avg = 0.365

Final Exam Avg = (Final Exam Score/Points Available) x Course Weight
Final Exam Avg = (80/100) x .3
FInal Exam Avg = .24




Course Grade Summary


Summary

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   16.25   72.24   79.32   77.67   86.49   98.03

This is a summary of the class course grades. One thing to note is that the mean is smaller than the median, indicating that this data could be skewed left.

Std. Deviation of Grades

## [1] 13.61243

A std. deviation of 13.6 shows that 68% of the grades fall between +/- 13.6 points of the mean. Since the mean of the grades is ~77.7, we can assume that 68% of the grades are between 64.1 and 91.3.


As you can see in the histogram (above), this data is skewed left. This is the result of outliers being present toward and below the lower quartile of the values.


This boxplot shows the two outliers, one of them being small and not very influential, occuring just below a few other data points. The other, however, is very influential as it falls far below the mean and median.



This graph shows the frequency of the course letter grades. As seen on the graph, most of the students finished with a B in the class.


Percentage of class that will have to retake the course

## [1] "24.6%"


This percentage was found by taking the number of grades below 70 and dividing by the total number of grades in the data set.

Retake = sum(grades<70)/length(grades)
Retake = 16/65
Retake = 24.6 ~ 25%