Arithmetic Mean

Arithmetic mean is the most commonly used average or measure of the central tendency applicable only in case of quantitative data; it is also simply called the “mean”. Arithmetic mean is defined as:

“Arithmetic mean is a quotient of sum of the given values and number of the given values”.

Arithmetic mean can be computed for both ungrouped data (raw data: data without any statistical treatment) and grouped data (data arranged in tabular form containing different groups).

If $$X$$ is the involved variable, then the arithmetic mean of $$X$$ is abbreviated as $$A.M$$ of $$X$$ and denoted by $$\overline X $$. The arithmetic mean of $$X$$ can be computed with any of the following methods.

Method’s Name
Nature of Data
Ungrouped Data
Grouped Data
Direct Method
$$A.M{\text{ of }}X = \overline X = \frac{{\sum x}}{n}$$
\[A.M{\text{ of }}X = \overline X = \frac{{\sum fx}}{{\sum f}}\]
Indirect or
Short-Cut Method
$$A.M{\text{ of }}X = \overline X = A + \frac{{\sum D}}{n}$$
$$A.M{\text{ of }}X = \overline X = A + \frac{{\sum fD}}{{\sum f}}$$
Method of
Step-Deviation
$$A.M{\text{ of }}X = \overline X = A + \frac{{\sum u}}{n} \times c$$
$$A.M{\text{ of }}X = \overline X = A + \frac{{\sum fu}}{{\sum f}} \times h$$

Here
$$x:$$ indicates value of the variable $$X$$
$$n:$$ indicates number of values of $$X$$
$$f:$$ indicates frequency of different groups
$$A:$$ indicates assumed mean
$$D:$$ indicates deviation from $$A$$ i.e, $$D = (x – A)$$
$$u = \frac{{x – A}}{{c{\text{ or }}h}}$$
$$u:$$ indicates step-deviation and $$c:$$ indicates common divisor
$$h:$$ indicates size of class or class interval in case of grouped data
$$\sum :$$ indicates summation or addition

Example:
The one-way train fare of five selected BS students is recorded as follows $$(\$ ){\text{ :}}$$ $$10$$, $$5$$, $$15$$, $$8$$ and $$12$$. Calculate the arithmetic mean of the following data.

Solution:

Let train fare be indicated by $$x$$, then

$$x{\text{ }}(\$ )$$
$$10$$
$$5$$
$$15$$
$$8$$
$$12$$
$$\sum x = 50$$

 

The arithmetic mean of $$X = \overline X = \frac{{\sum x}}{n}$$, so we decide to use the above-mentioned formula. From the given data, we have $$\sum x = 50$$ and $$n = 5$$. Placing these two quantities in the above formula, we get the arithmetic mean for the given data.
\[\overline X = \frac{{50}}{5} = \$ 10\]

 

Example:

Provide the given distribution of the following frequency distribution of first year students of a particular college:

Age (Years)
$$13$$
$$14$$
$$15$$
$$16$$
$$17$$
Number of Students
$$2$$
$$5$$
$$13$$
$$7$$
$$3$$

 

Solution:

The given distribution is grouped data and the variable involved is ages of first year students, while the number of students represents frequencies.

Ages (Years)
$$x$$
Number of Students
$$f$$
$$fx$$
$$13$$
$$2$$
$$26$$
$$14$$
$$5$$
$$70$$
$$15$$
$$13$$
$$195$$
$$16$$
$$7$$
$$112$$
$$17$$
$$3$$
$$51$$
Total
$$\sum f = 30$$
$$\sum fx = 454$$

Now we will find the arithmetic mean as $$\overline X = \frac{{\sum fx}}{{\sum f}} = \frac{{454}}{{30}} = 15.13$$ years.

 

Example:

The following data shows the distance covered by $$100$$ people to perform their routine jobs.

Distance (Km)
$$0 – 10$$
$$10 – 20$$
$$20 – 30$$
$$30 – 40$$
Number of People
$$10$$
$$20$$
$$40$$
$$30$$

 

Solution:

The given distribution is grouped data and the variable involved is distance covered, while the number of people represents frequencies.

Distance (Km)
Number of People
$$f$$
Mid Points
$$x$$
$$fx$$
$$0 – 10$$
$$10$$
$$5$$
$$50$$
$$10 – 20$$
$$20$$
$$15$$
$$300$$
$$20 – 30$$
$$40$$
$$25$$
$$1000$$
$$30 – 40$$
$$30$$
$$35$$
$$1050$$
Total
$$\sum f = 100$$
$$\sum fx = 2400$$

Now we will find the arithmetic mean as $$\overline X = \frac{{\sum fx}}{{\sum f}} = \frac{{2400}}{{100}} = 24$$ Km.