HOW TO CALCULATE ARITHMETIC MEANS

In the previous post Introduction to The Measures in Statistics we have been introduced to the concept of measures of location. One of such measures is the mean, or more specifically the arithmetic mean. In the present post we elaborate more on the formulae used to find the mean. It is worth pointing out that the selection of the proper formula depends on how the data in hand are presented. Throughout this post, to shorten the phrase arithmetic mean we continue to use the terms mean or average.

Case 1: The data are not grouped

In this case, we are given n data, namely X₁, X₂, X₃, …, X_n. To calculate the mean, apply the following formula.

$\bar{X} = \frac{X_1 + X_2 + X_3 + \cdots X_n}{n}$

Example 1

A sample of 5 customers of a post office revealed the following waiting time (in minutes): 3, 5, 10, 6, 6. Find the mean waiting time.

Answer

In this case n = 5. Let x₁`= 3 minutes, x₂ = 5 minutes, x₃ = 10 minutes, x₄ = x₅ = 6 minutes. The mean waiting time is:

Case 2: n data consist of k distinct values with k < n

Suppose that we have n data with k distinct values x₁, x₂, …, x_k, where k < n. For every i ∈ {1, 2, 3, …, k} let the frequency of occurrence of x_i be f_i. The average can be determined by applying the formula below.

$\bar{X} = \frac{\sum_{i=1}^k f_i \cdot x_i}{n}$ where $n = \sum_{i=1}^k f_i$

This kind of average is a special case of the weighted mean, where each f_i acts as the weight of x_i.

Example 2

A sample of 20 students revealed that 8, 7, and 5 of them got the IQ scores of 90, 100, and 120, respectively. Find the average IQ score.

Answer

The data can be presented in the following table.

The average IQ score is $\bar{X} = \frac{8 \cdot 90 + 7 \cdot 100 + 5 \cdot 120}{20}=101$ .

The calculation can also be tabulated as follows.

Case 3: The data are presented in a frequency distribution table

In a frequency distribution table, data are classified into several classes. To each class there is assigned a symbol called class interval. In addition, every class interval has end numbers, which are called class limits. The smaller number is called the lower class limit and the larger is called the upper class limit. More specifically, each class interval is of the form “LCL – UCL” where LCL and UCL denote the lower class limit and the upper class limit, respectively. The table also shows the class frequency of each class, which is the number of observations in each class. A typical example of the frequency distribution table is shown in Example 3. To calculate the mean of the data presented as such, we can apply the following formula.

$\bar{X} = \frac{\sum_{i=1}^k f_i \cdot M_i}{n}$ …………………………………………………………………………………………………………………………………………………………………………… (*)

where $n = \sum_{i=1}^k f_i$ and for every i ∈ {1, 2, …, k} $M_i = \frac{UCL_i + LCL_i}{2}$

In the formula, f_i is the frequency of class i, whereas UCL_i and LCL_i are the the upper and lower class limits of class i, respectively. The notation M_i means the class mark of class i.

Example 3

The following table shows the distribution of the distance from home to work travelled by 50 employees every day. Find the average distance.

Answer

In the frequency distribution table there are 5 classes. The class interval of the first class is 1-3 and its class mark is $M_1 = \frac{1+3}{2} \: km = 2 \: km$ . The class interval of the second class is 4-6 and its class mark is $M_2 = \frac{4+6}{2} \: km = 5 \: km$ . Continuing this way, we have the following.

To apply (*) we can extend the table to the right, adding the column for f_i⋅M_i. See the table below.

(For brevity, we have deleted the columns for UCL_i and LCL_i as they are no longer needed.)

Now, get the sums of all f_i and f_i⋅M_i. This gives:

As the table shows, $n = \sum_{i=1}^5 f_i = 50$ and $\sum_{i=1}^5 f_i \cdot M_i = 313$ . By (*), we have $\bar{X} = \frac{313}{50} \: km = 6.26 \: km$ .