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
ShortCut 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
StepDeviation 
$$A.M{\text{ of }}X = \overline X = A + \frac{{\sum u}}{n} \times c$$

$$A.M{\text{ of }}X = \overline X = A + \frac{{\sum fu}}{{\sum f}} \times h$$

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 stepdeviation 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 oneway 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 abovementioned 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.
Peter
June 15 @ 5:10 pm
How can l download this write up?