Data Analysis: Frequency Distribution and Outlier Analysis

This document presents a complete solution to a data analysis assignment. The solution includes the construction of a frequency distribution table and histogram using Excel, with appropriate class boundaries. It calculates the mean test score using Excel and determines the first, second (median), and third quartiles using the common-sense rule. Based on the results, the solution explains whether the distribution is bell-shaped or skewed. Finally, it identifies and calculates any outliers in the dataset using the 1.5 IQR rule, demonstrating the application of these statistical concepts to real-world data analysis. The assignment is based on an aptitude test with a dataset of test scores.

Running head: DATA ANALYSIS 1
Data Analysis
Student Name
Professor’s Name
University Name
Date

DATA ANALYSIS 2
Data Analysis
Solutions
a. The frequency distribution table created in excel is illustrated in the figure below:
The histogram created from the frequency distribution table above is shown below.
b. The mean test score can be calculated in excel by inserting the following formula in an
appropriate cell.
𝑚𝑒𝑎𝑛 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 (𝑏𝑖𝑛 𝑟𝑎𝑛𝑔𝑒)
The mean is 61.83 as shown below.
Frequency Distribution
Class Midpoint Upper limit Frequency Cumulative Freuency Relative Freuency
1≤x≤10 5.5 10 2 2 0.02
11≤x≤20 15.5 20 1 3 0.01
21≤x≤30 25.5 30 3 6 0.03
31≤x≤40 35.5 40 5 11 0.05
41≤x≤50 45.5 50 12 23 0.12
51≤x≤60 55.5 60 20 43 0.2
61≤x≤70 65.5 70 23 66 0.23
71≤x≤80 75.5 80 16 82 0.16
81≤x≤90 85.5 90 10 92 0.1
91≤x≤100 95.5 100 8 100 0.08
Sum 100 1
Summary Satistics
Min 1.50
Max 97.20
Average 61.83

DATA ANALYSIS 3
The common-sense method of determining the quartiles follows the following steps.
➢ Arrange the numbers in ascending order.
➢ Locate the center of the arranged data. In this case its between the 50 th and 51st
data points.
➢ From the center point determine the median of either half which will be the first
and the third quartile respectively (Selvanathan & Keller, 2017).
The first quartile is the median of the 1st data point to the 50th data point. Hence
𝑄1(𝑓𝑖𝑟𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒) = 50.8 + 51
2 = 50.9
Third quartile is the median of the 51st data point to the 100th data point. Hence
𝑄3(𝑡ℎ𝑖𝑟𝑑 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒) = 73.2 + 72.1
2 = 72.65
➢ The second quartile is the median of the overall arranged data which is the sum of
the 50th and 51st data points.
𝑄2(𝑚𝑒𝑑𝑖𝑎𝑛) = 62.4 + 62.8
2 = 62.6
c. No, the distribution is not bell shaped but negatively skewed since the left tail of the
histogram is longer and the mean is less than median (Q2).
d. The lower outliers below the first quartiles are values below the score given by:
𝐿𝑜𝑤𝑒𝑟 𝑂𝑢𝑡𝑙𝑖𝑒𝑟 = (𝑄1 − 1.5𝐼𝑄𝑅)
The interquartile range is given by:
𝐼𝑄𝑅 = 𝑄3 − 𝑄1
𝐼𝑄𝑅 = 72.65 − 50.9 = 21.75
Therefore,
𝐿𝑜𝑤𝑒𝑟 𝑂𝑢𝑡𝑙𝑖𝑒𝑟 = (𝑄1 − 1.5𝐼𝑄𝑅)

DATA ANALYSIS 4
𝐿𝑜𝑤𝑒𝑟 𝑂𝑢𝑡𝑙𝑖𝑒𝑟 = (50.9 − 1.5(21.75)) = 18.275
Scores below 18.275 are outliers and are highlighted in the table below.
The upper outliers are given by:
𝐿𝑜𝑤𝑒𝑟 𝑂𝑢𝑡𝑙𝑖𝑒𝑟 = (𝑄3 + 1.5𝐼𝑄𝑅)
𝐿𝑜𝑤𝑒𝑟 𝑂𝑢𝑡𝑙𝑖𝑒𝑟 = (72.65 + 1.5(21.75)) = 105.275
Scores above 105.275 are outliers. Since the maximum score is 97.20, there is no score
above the minimum upper limit for outliers which is 105.275 hence there is no upper
outliers in the distribution (Shao, 2010).
Lower outliers
1.5
6.7
11.2

DATA ANALYSIS 5
References
Selvanathan, E. A., & Keller, G. (2017). Business statistics abridged (7th ed). South Melbourne,
Victoria: Cengage Learning.
Shao, J. (2010). Mathematical statistics (2nd ed). New York: Springer.