Week One Homework Assignment - Statistics and Probability
The head of the chemistry department at a local university has tasked a student teaching assistant to prepare a variety of statistics pertaining to the final course grades earned by students who have completed the university’s CHEM 103 course during the past ten years. These statistics will be used as a basis for predicting the potential performance of future students who enroll in the course. A total of 1,125 students have completed the course during the past ten years. Rather than trying to have the teaching assistant collect and analyze data for all 1,125 students, the department head has agreed to allow the teaching assistant to prepare the necessary statistics using a random sample consisting of 25 students who completed the course during the past ten years. The following table summarizes the final course grades earned by the 25 randomly selected students.
Student No. Course Grade
1 83.17
2 82.31
3 90.18
4 93.63
5 90.14
6 96.06
7 83.75
8 85.17
9 86.67
10 93.81
11 83.11
12 82.16
13 89.48
14 86.76
15 83.76
16 86.89
17 88.57
18 96.93
19 86.48
20 95.82
21 92.32
22 78.70
23 92.67
24 74.56
25 78.48
For the purposes of this assignment, assume that: (1) the sample consisting of 25 students is truly representative of the population of 1,125 students from which it was drawn; and (2) the 1,125 students who have completed the course during the past ten years constitute a truly representative sample of all future students who will eventually enroll in the course. Based upon these assumptions, use the sample data provided in the preceding table to answer the following questions.
1. What is the predicted range for the mean grade for an average future student enrolling in the CHEM 103 course?
o 27.65
o 22.37
o 19.87
o 34.68
Solution:
Range = Maximum – Minimum
Range = 96.93 – 74.56
Range = 22.37
2. What is the predicted mean grade for an average future student enrolling in the CHEM 103 course?
o 76.31
o 83.26
o 87.26
o 90.89
Solution:
Mean = ∑X/n
Mean = 2181.58/25
Mean = 87.26
3. What is the predicted median grade for an average future student enrolling in the CHEM 103 course?
o 86.76
o 91.94
o 79.16
o 90.91
Solution:
Median = (n +1)/2th observation
Median = (25 + 1)/2th observation
Median = 13th observation
Median = 86.76
4. What is the predicted standard error of the mean grade for an average future student enrolling in the CHEM 103 course?
o 1.58
o 2.29
o 2.11
o 1.18
Solution:
Standard deviation = √∑(X – mean)²/ (n – 1)
Standard deviation = √832.44/24
Standard deviation = 5.89
Standard error = s/√n
Standard error = 5.89/√25
Standard error = 1.18
5. Assuming the level of confidence for the interval estimate is not specified, what is the predicted int...
For the purposes of this assignment, assume that: (1) the sample consisting of 25 students is truly representative of the population of 1,125 students from which it was drawn; and (2) the 1,125 students who have completed the course during the past ten years constitute a truly representative sample of all future students who will eventually enroll in the course. Based upon these assumptions, use the sample data provided in the preceding table to answer the following questions.
1. What is the predicted range for the mean grade for an average future student enrolling in the CHEM 103 course?
o 27.65
o 22.37
o 19.87
o 34.68
Solution:
Range = Maximum – Minimum
Range = 96.93 – 74.56
Range = 22.37
2. What is the predicted mean grade for an average future student enrolling in the CHEM 103 course?
o 76.31
o 83.26
o 87.26
o 90.89
Solution:
Mean = ∑X/n
Mean = 2181.58/25
Mean = 87.26
3. What is the predicted median grade for an average future student enrolling in the CHEM 103 course?
o 86.76
o 91.94
o 79.16
o 90.91
Solution:
Median = (n +1)/2th observation
Median = (25 + 1)/2th observation
Median = 13th observation
Median = 86.76
4. What is the predicted standard error of the mean grade for an average future student enrolling in the CHEM 103 course?
o 1.58
o 2.29
o 2.11
o 1.18
Solution:
Standard deviation = √∑(X – mean)²/ (n – 1)
Standard deviati...
