# Arithmetic Mean

Arithmetic Mean
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 1 Introduction to Arithmetic Mean 2 Formula to find Arithmetic Mean 3 Tips on finding Arithmetic Mean 4 Properties of Arithmetic Mean 5 Merits and Demerits of Arithmetic Mean 6 Solved Examples on Arithmetic Mean 7 Challenging Question on Arithmetic Mean 8 Practice Questions on Arithmetic Mean 9 Maths Olympiad Sample Papers 10 Frequently Asked Questions (FAQs)

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## Introduction to Arithmetic Mean

We come across statements like "the average monthly income of a family is ₹15,000 or the average monthly rainfall of a place is 1000 mm" quite often.

Average is typically referred to as Arithmetic Mean but the fact is one of the type of average.

We will be focussing here only on Arithmetic Mean.

Let’s first understand the meaning of the term "Mean".

### Definition of Arithmetic Mean

Mean, often referred to as the arithmetic average or arithmetic mean, is calculated by adding all the numbers in a given set and then dividing by the total number of items within that set.

Let's discuss an example where we find use of arithmetic mean.

The arithmetic mean maintains its place in calculating a stock’s average closing price during a particular month.

Let's assume there are 24 trading days in a month.

How can we calculate the mean?

All you need to do is take all the prices, add them up, and divide by 24 to get the arithmetic mean.

Now look at the simulation below to see another example of why and how arithmetic mean is calculated.

## Formula to Find Arithmetic Mean

The general formula to find the arithmetic mean of a given data is:

 ​​$$\text{Mean, }\bar{x} =\dfrac{\text{ Sum of all observations}}{\text{Number of observations}}$$

It is denoted by $$\bar {x}$$, (read as x bar)

Data can be presented in different forms.

When we have raw data like the marks of a student in five subjects, we add the marks obtained in the five subjects and divide the sum by 5

Let us understand arithmetic mean in detail using the simulation shown below.

Here are six different kinds of land.

You want to decide for yourself how many trees should be in each piece of land.

How will this affect the mean number of trees?

Try finding mean using the following mean simulator.

Instruction: Move the green sliders in order to change the number of trees on each piece of land.

1. Observe how the mean changes as you move the green points.

Now consider a case where we have huge data like the heights of 40 students in a class or the number of people visiting an amusement park in a week.

Will it be convenient to find the arithmetic mean with the above method?

The answer is a big NO!

So, how can we find the mean?

We arrange the data in a form that is meaningful and easy to comprehend.

Let's understand how to compute the arithmetic average in such cases.

### I. Ungrouped Data

Here the arithmetic mean is calculated using the formula:

$$\text{Mean }\bar{x} =\dfrac{\text{ Sum of all observations}}{\text{Number of observations}}$$

Example

Compute the mean of the first 6 odd, natural numbers.

Solution:

The first 6 odd, natural numbers: 1, 3, 5, 7, 9, 11

\begin{align}  \bar{x} &=\dfrac{1+3+5+7+9+11 }{6 }\\&=\dfrac{36 }{6 }\\&=6\end{align}

Thus, the arithmetic mean is 6

### II. Grouped Data

There are three methods to find the mean of grouped data.

The choice of the method to be used depends on the numerical value of xi and fi

If xi and fi are sufficiently small, the direct method will work.

But, if they are numerically large, we use the assumed mean method or step-deviation method.

In this section, we will be studying all three methods along with examples.

### 1. Direct Method

Let x1, x2, x3 ……xn be the observations with the frequency f1, f2, f3 ……fn.

Then, mean is calculated using the formula:

 $$\bar{x} =\dfrac{x_1f_1+x_2f_2+......+x_nf_n}{\sum f_i}$$

Here, f1+ f2 + ....fn =  $$\sum f_i$$ indicates the sum of all frequencies.

Example

Find the mean of the following distribution:

 x 10 30 50 70 89 f 7 8 10 15 10

Solution:

 xi fi xi 10 7 $$10 \times 7 = 70$$ 30 8 $$30 \times 8 = 240$$ 50 10 $$50 \times 10 = 500$$ 70 15 $$70 \times 15 = 1050$$ 89 10 $$89 \times 10 = 890$$ Total $$\sum{f_i}=50$$ $$\sum{x_if_i}=2750$$

Add up all the (xifi ) values to obtain $$\sum x_if_i$$Add up all the fi values to get $$\sum f_i$$

Now, use the mean formula.

\begin{align}
\bar{x} &=\dfrac{\sum x_if_i}{\sum f_i} \\
&= \dfrac{2750}{50} \\
\\&= 55
\end{align}

 $$\therefore$$ $$\text {Mean} = 55$$

The above problem is an example of discrete grouped data.

Let's now consider an example where the data is present in the form of continuous class intervals.

Example:

Let's try finding the mean of the following distribution:

 Class-Interval 15-25 25-35 35-45 45-55 55-65 65-75 75-85 Frequency 6 11 7 4 4 2 1

Solution:

When the data is presented in the form of class intervals, the mid-point of each class (also called as class mark) is considered for calculating the mean.

The formula for mean remains the same as discussed above.

Note:

$$\text{Class Mark }= \dfrac{\text{Upper limit + Lower limit}}{2}$$

 Class- Interval Class Mark (xi) Frequency (fi) xi fi 15-25 20 6 120 25-35 30 11 330 35-45 40 7 280 45-55 50 4 200 55-65 60 4 240 65-75 70 2 140 75-85 80 1 80 Total 35 1390

We have, $$\sum f_i = 35$$ and $$\sum x_if_i = 35$$

\begin{align} \therefore\bar{x} &=\dfrac{\sum x_if_i}{\sum f_i} \\ &= \dfrac{1390}{35} \\ \\&=39.71 \end{align}

 $$\therefore$$ $$\text {Mean} = 39.71$$

### 2. Short-cut Method

This method is called as assumed mean method or change of origin method.

The following steps describe this method.

Step1: Calculate the class marks of each class (xi).

Step2: Let A denote the assumed mean of the data.

Step3: Find deviation (di) = xi – A

Step4: Use the formula:

 $$\bar{x}= A + (\dfrac{\sum f_id_i}{\sum f_i})$$

Let's understand this with the help of the following example.

Example:

Calculate the mean of the following using the short-cut method.

 Class-Intervals 45-50 50-55 55-60 60-65 65-70 70-75 75-80 Frequency 5 8 30 25 14 12 6

Solution:

Let us make the calculation table.

Let the assumed mean be A = 62.5

Note: A is chosen from the xi values.

Usually, the value which is around the middle is taken.

 Class- Interval Class mark/Mid-points (xi) fi di = (xi - A) fidi 45-50 47.5 5 47.5-62.5 =-15 -75 50-55 52.5 8 52.5-62.5 =-10 -80 55-60 57.5 30 57.5-62.5 =-5 -150 60-65 62.5 25 62.5-62.5 =0 0 65-70 67.5 14 67.5-62.5 =5 70 70-75 72.5 12 72.5-62.5 =10 120 75-80 77.5 6 77.5-62.5 =15 90 $$\sum f_i$$=100 $$\sum f_id_i$$= -25

Now we use the formula,

\begin{align} \bar{x} &=A + \dfrac{\sum f_id_i}{\sum f_i} \\ &= 62.5 + \frac{-25}{100} \\ &=62.5-0.25 \\&= 62.25\end{align}

 $$\therefore$$ Mean = 62.25

### 3. Step Deviation Method

This is also called change of origin or scale method.

The following steps describe this method:

Step1: Calculate the class marks of each class (xi).

Step2: Let A denote the assumed mean of the data.

Step3: Find $$u_i=\dfrac{x_i-A}{h}$$ where h is the class size.

Step4: Use the formula:

 $$\bar{x}= A + h\times (\dfrac{\sum f_iu_i}{\sum f_i})$$

Example

Consider the following example to understand this method.

Find the mean of the following using step-deviation method.

 Class Intervals 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Total Frequency 4 4 7 10 12 8 5 50

Solution:

To find the mean, we first have to find the class marks and decide A (assumed mean).

Let A = $$35$$

Here h (class width) = $$10$$

 C.I. xi fi ui= $$​​\dfrac{x_i - A}{h}$$ fiui 0-10 5 4 -3 4 x (-3)=-12 10-20 15 4 -2 4 x (-2)=-8 20-30 25 7 -1 7 x (-1)=-7 30-40 35 10 0 10 x 0= 0 40-50 45 12 1 12 x 1=12 50-60 55 8 2 8 x 2=16 60-70 65 5 3 5 x 3=15 Total $$\sum f_i=50$$ $$\sum f_iu_i=16$$

Using mean formula:

\begin{align}
\bar{x} &=A + h \times \dfrac{\sum f_iu_i}{\sum f_i} \\
&= 35+ \frac{16}{50} \times 10 \\
&=35 + 3.2 \\&= 38.2
\end{align}

 $$\text{Mean } = 38.2$$

Tips and Tricks
1. If the number of classes are less and the data has values with smaller magnitude, then direct method is preferred out of the three methods to find the arithmetic mean.
2. Step deviation works best when we have a grouped frequency distribution in which the width remains constant for every class interval and we have a considerably large number of class intervals.

## Properties of Arithmetic Mean

Let us have a look at some of the important properties of arithmetic mean.

Suppose we have $$n$$ observations denoted by x1, x2, x3, ….,xn and $$\bar{x}$$ is their arithmetic mean, then:

1. If all the observations in the given data set has a value say ‘m’, then their arithmetic mean is also ‘m’.

Consider the data having 5 observations: $$15,15,15,15,15$$

So, their total $$= 15+15+15+15+15= 15 \times 5 = 75$$

$$n = 5$$

Now, arithmetic mean $$= \dfrac{total}{n} =\dfrac{75}{5} = 15$$

2. The algebraic sum of deviations of a set of observations from their arithmetic mean is zero.

$$(\!x_1 \!-\! \bar{x}\!)\! +\! (\!x_2\! -\! \bar{x}\!) \!+\! (\!x_3\! -\!\bar{x}\!) \!+\!\! ...\!\! +\! (\!x_n\! - \!\bar{x}\!) \!\!= \!0$$

• For discrete data, $$\sum (x_i - \bar{x})$$ =  0

• For grouped frequency distribution, $$\sum f(x_i - \sum \bar{x}) = 0$$

3. If each value in the data increases or decreases by a fixed value, then the mean also increases/decreases by the same number.

Let the mean of x1, x2, x3 ……xn be $$\bar X$$, then the mean of x1+k, x2+k, x3 +k ……xn+k will be $$\bar X$$+k.

4. If each value in the data gets multiplied or divided by a fixed value, then the mean also gets multiplied or divided by the same number.

Let the mean of x1, x2, x3 ……xn be $$\bar X$$, then the mean of kx1, kx2, kx3  ……xn+k will be k$$\bar X$$.

Similarly, mean of $$\dfrac{x_1}{k}$$, $$\dfrac{x_2}{k}$$, $$\dfrac{x_3}{k}$$  ……$$\dfrac{x_n}{k}$$ will be$$\dfrac{\bar{X}}{k}$$.

Note: While dividing each value by k, it must be a non-zero number as division by 0 is not defined.

Check  out the simulation demonstrating the arithmetic mean property.

Input some values and see how adding a fix number to them would change the arithmetic mean.

## Merits and Demerits of Arithmetic Mean

The uses of arithmetic mean are not just limited to statistics and mathematics, but it is also used in experimental science, economics, sociology, and other diverse academic disciplines.

Listed below are some of the major advantages of arithmetic mean.

1. As the formula to find the arithmetic mean is rigid, the result doesn’t change.

Unlike median, it doesn’t get affected by the position of the value in the data set.

2. It takes into consideration each value of the data set.

3. Finding arithmetic mean is quite simple; even a common man having very less finance and math skills can calculate it.

4. It’s also a useful measure of central tendency, as it tends to provide useful results, even with large groupings of numbers.

5. It can be further subjected to many algebraic treatments unlike mode and median.

For example, the mean of two or more series can be obtained from the mean of the individual series.

6. The arithmetic mean is widely used in geometry as well.

For example, the coordinates of the “centroid” of a triangle (or any other figure bounded by line segments) are the arithmetic mean of the coordinates of the vertices.

After having discussed some of the major advantages of arithmetic mean, let's understand its limitations.

1.The strongest drawback of arithmetic mean is that it is affected by extreme values in the data set.

To understand this, consider the following example.

It’s Ryma’s birthday and she is planning to give return gifts to all who attend her party.

She wants to consider the mean age to decide what gift she could give everyone.

The ages (in years) of the invitees are as follows:

2, 3, 7, 7, 9, 10, 13, 13, 14, 14

Here, n = 10

Sum of the ages = $$2\!+\!3\!+\!7\!+\!7\!+\!9\!+\!10\!+\!13\!+\!13\!+\!14\!+\!14\! =\! 92$$

Thus, mean = 92/10 = 9.2

In this case, we can say that a gift that is desirable to a kid who is 9 years old may not be suitable for a child aged 2 or 14

2. In a distribution containing open-end classes, the value of mean cannot be computed without making assumptions regarding the size of the class.

 Class Interval Frequency Less than 15 20 15-25 12 25-35 3 35-45 12 More than 45 6

We know that to find the arithmetic mean of grouped data, we need the mid-point of every class.

As evident from the table, there are two cases (less than 15 and 45 or more) where it is not possible to find the mid-point and hence, arithmetic mean can’t be calculated for such cases.

3. Its practically impossible to locate the arithmetic mean by inspection or graphically.

4. It cannot be used for qualitative types of data such as honesty, favourite milkshake flavour, most popular product etc.

5. We can't find arithmetic mean, if a single observation is missing or lost.

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## Solved Examples

 Example 1

If the mean of 2m+3, m+2, 3m+4, 4m+5 is m+2, find m.

Solution:

The data contains 4 observations : $$2m+3, m+2, 3m+4, 4m+5$$

So, $$n= 4$$

Sum of 4 observations =

$$(\!2m\!+\!3\!)\!+\!(\!m\!+\!2\!)\!+\! (\!3m\!+\!4\!)\!+\! (\!4m\!+\!5\! )$$ = $$(\!10x\!+\!14\!)$$

\begin{align} \text {Mean  }&=\dfrac{10m+14}{4}\\\therefore m+2&= \dfrac{10m+14}{4}\\4\times (m+2)&=10m+14\\4m+8&=10m+14\\  -6m&=6\\\Rightarrow  m&=-1\end{align}

 $$\therefore$$ $$m = -1$$
 Example 2

Find the missing frequencies in the following frequency distribution if its mean is 1.46

 Number of accidents (xi) 0 1 2 3 4 5 Total Frequency (fi) 46 a b 25 10 5 200

Solution:

We will be using direct method to find the missing frequency.

 xi fi xifi 0 46 0 1 a a 2 b 2b 3 25 75 4 10 40 5 5 25 Total $$\sum f_i= 86+a+b$$ $$\sum x_if_i$$ = $$140+a+2b$$

\begin{align} Given: \sum f_i&= 200\\\therefore 86+a+b&= 200\\a+b&=114.....Eqn(1) \\ \text{Also, Mean } &=1.46\\1.46 &=\dfrac{\sum x_if_i}{\sum f_i}\\1.46&=\dfrac{140+a+2b}{200}\\292&=140+a+2b\\\Rightarrow a+2b&=152.....Eqn(2)\end{align}

On solving Eqn (1) & (2) we get:

 $$\therefore$$ $$a=76,b=38$$
 Example 3

The mean monthly salary of 10 workers of a group is ₹1445

One more worker whose monthly salary is ₹1500 has joined the group.

Find the mean monthly salary of 11 workers of the group.

Solution:

Here, $$n=10, \bar{x}=1445$$

Using formula,

\begin{align} \bar{x}&=\dfrac{\sum x_i}{n} \\\therefore \sum x_i&= \bar{x} \times n\\\sum x_i&= 1445 \times 10\\&=14450\end{align}

\begin{align}\text {(total salary of 10 workers = ₹14450)}\end{align}

\begin{align} \text {total salary of 11 workers} \end{align}  = \begin{align} 14450+1500 = ₹15950\end{align}

\begin{align} \text {Average salary of 11 workers} &=\dfrac{15950}{11} \\&= 1450\end{align}

 $$\therefore$$ Average monthly salary of 11 workers =  $$₹1450$$
 Example 4

Find the mean of the following distribution using step deviation method:

 C.I 0-400 400-800 800-1200 1200-1600 1600-2000 2000-2400 f 142 265 560 271 89 16

Solution:

Let $$\text{A= 1400, h = 400}$$

 C.I Class marks (xi) fi $$u_i= \dfrac{x_i-A}{h}$$ fiui 0-400 200 142 -3 -426 400-800 600 265 -2 -530 800-1200 1000 560 -1 -560 1200-1600 1400 271 0 0 1600-2000 1800 89 1 89 2000-2400 2200 16 2 32 $$\sum f_i$$ = $$1343$$ $$\sum f_iu_i$$=$$-1395$$

\begin{align} \text {Mean, }\bar{x} &=A+\dfrac{\sum f_iu_i}{\sum f_i}\times h\\&=1400 + \dfrac{-1395}{1343}\times 400 \\&=1400-415.49=984.51\end{align}

 $$\therefore$$ Mean = $$984.51$$

Challenging Question
1. Nine friends went to a restaurant for dinner.

Eight of them spent ₹120 each on their meals and the ninth spent ₹80 more than the average expense of all nine.

What was the total money spent by them?

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## Practice Questions

Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.

IMO (International Maths Olympiad) is a competitive exam in Mathematics conducted annually for school students. It encourages children to develop their math solving skills from a competition perspective.

## 1. What is the formula of arithmetic mean?

Arithmetic mean is given by the formula:

$$\text{Mean }\bar{x} =\dfrac{\text{ Sum of all observations}}{\text{Number of observations}}$$

If we have grouped data, the formula used is:

$$\bar{x} =\dfrac{x_1f_1+x_2f_2+......+x_nf_n}{\sum f_i}$$

## 2. What are the types of arithmetic mean?

There are two types of arithmetic mean.

1. Simple Arithmetic Mean: In simple arithmetic mean, all items of a series are given equal importance.

2. Weighted Arithmetic Mean: Under weighted arithmetic mean, different items of the series are assigned weights according to their relative importance.

So, instead of each of the data points contributing equally to the final average, some data points contribute more than others.

When all points have equal weight, the two types of mean are the same.

## 3. When is arithmetic mean used?

Use of arithmetic mean is not just limited to maths but in everyday life as well.

Average income of a family,

Taking into consideration the merits and demerits of arithmetic mean, it is important to understand the kind of data it should be used for.

Arithmetic mean gives a good approximation about data when the values do not have extremes.

It  is not a suitable average for distributions with extreme items or open-end classes.

For example: Consider two data sets
A= {34, 45, 43, 23, 40} and B= {34, 56, 90, 120, 2}

Out of the two sets, A is more suitable to be considered as average as the values do not show much variation.

In the case of set B, we have extreme values 2 and 120

Hence, finding arithmetic mean won’t serve any purpose.

## 4. Are arithmetic mean and average the same?

The words Mean and Average are used interchangeably, but they are not same.

Average basically means central value.

In Mathematics, it is the representation of the typical value of the data.

Averages are of different types.

Arithmetic mean is one of the averages.

Median and mode are the other two widely used averages.

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