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# Arithmetic Mean

Top
 Sub Topics Arithmetic mean also known as arithmetic average is the value obtained by dividing the sum of various values of series by number of items in a series. It is the tool to measure central tendency of data. $\sum$x is written in place of 'n' items in given data. In case of continuous or grouped frequency distribution, value of 'x' is taken as mid point of corresponding class. Arithmetic mean is very simple and it can be calculated very easily. So, it is commonly used in study and analysis of business and economy. In case, we are dealing with certain data related to average income, average expenditure, average profit, average loss, wages, incomes and other fields like average score of child, then that the data used can be converted to their average means. Arithmetic mean is commonly referred to as "average" or simply as "mean".

## Arithmetic Mean Definition

Arithmetic mean of a given set of observations is their sum divided by the number of observations.

For example, arithmetic mean of 5, 6, 7, 12, 15 is $\frac{5 + 6 + 7 + 12 + 15}{5}$

= 9
In general, if $x_{1}, ......., x_{n}$ are the given n observations, then their arithmetic mean, usually denoted by $\bar{x}$ is given as follows:

$\bar{X}$ = $\frac{x_{1} + ........ + x_{n}}{n}$ = $\frac{\sum x}{n}$

where, $\sum$x is the sum of the observations.

In case of frequency distribution, we have

The arithmetic mean $\bar{X}$ is given by

$\bar{X}$ = $\frac{ (X_{1} + X_{1} + ....... + f_{1} times) + (X_{2} + X_{2} + .....+ f_{2} times) + ........... + (X_{n} + X_{n} + ........ + f_{n} times)}{f_{1} + f_{2} + .....+ f_{n}}$

$\Rightarrow$ $\frac{f_{1}X_{1} + ....... + f_{n}X_{n}}{f_{1} + .......+ f_{n}}$
= $\frac{\sum\ fx}{\sum\ f}$

= $\frac{\sum fx}{N}$

N = $\sum f$ is the total frequency.

In case of continuous or grouped frequency distribution, the value of X is taken as the mid value of the corresponding class.

The symbol $\sum$ is the letter capital sigma of the greek alphabet and is used in mathematics to denote the sum of values.

## Weighted Arithmetic Mean

If some items in a distribution are more important than others, then this point must be borne in mind, in order that average computed is representative of the distribution. In such cases, proper weightage is to be given to various items. The weights attached to each item is proportional to the important of the item in the distribution.

For example, if we want to have an idea of the change in cost of living of a certain group of people. Then, the sample mean of the prices of the commodities consumed by them will not do, since all the commodities are not equally important.

Examples: Pulses, gas, water, light are more important than tea, cosmetics, confectionery etc.,

Let $W_{1}$, ..........., $W_{n}$ be the weights attached to variable values $X_{1}$,......., $X_{n}$ respectively. Then, the weighted arithmetic mean usually denoted by $\bar{X_{W}}$ is given by

$\bar{X_{W}}$ = $\frac{W_{1}X_{1} + ..... + W_{n}X_{n}}{W_{1} + ..... + W_{n}}$

= $\frac{\sum\ WX}{\sum\ W}$

In case of frequency distribution, if $f_{1}$,.........,$f_{n}$ are the frequencies of the variable values $X_{1}$,......., $X_{n}$ respectively, then the weighted arithmetic mean is given as follows:

$\bar{X_{W}}$ = $\frac{W_{1}(f_{1}X_{1})+ ....... + W_{n}(f_{n}X_{n})}{W_{1} + ..... + W_{n}}$

= $\frac{\sum\ W(fX))}{\sum\ W}$

where, $W_{1}$,...................., $W_{n}$ are the respective weights of $X_{1}$, ................., $X_{n}$

## Arithmetic Mean Problems

Given below are some of the example problems on arithmetic mean.

### Solved Examples

Question 1: The following is the frequency distribution of the number of telephone calls received in 245 successive one minute intervals at an exchange:

 Number of calls 0 1 2 3 4 5 6 7 Frequency 14 21 25 43 51 40 39 12

Obtain the mean number of calls per minute.
Solution:
Let the variable X denote the number of calls received per minute at the exchange.

Computation of mean number of  calls is as follows:

 Number of calls (X) 0 1 2 3 4 5 6 7 Frequency (f) 14 21 25 43 51 40 39 12 N = 245 f X 0 21 50 129 204 200 234 84 $\sum$ f X = 922

From the above table, we see that $\sum$ f X = Mean number of calls = 922
Total N = 245

$\bar{X}$ = $\frac{\sum f X}{N}$ = $\frac{922}{245}$

= 3.763

Therefore, the mean number of calls is 3.763

Question 2: 25 students took a mathematics test. In that, 10 students had an average score of 45, while other students had an average score of 40. Find the average score of the whole class?
Solution:
Here, we multiply each average by the number of students that had that average and sum up to get the sum of weighted average.
That is,  45 x 10 + 40 x 15 = 1050

Given: total number of students = 25 = Total number of terms.

To find the weighted average, we use the formula

Weighted Average = $\frac{\text{Sum of Weighted Terms}}{\text{Total number of Terms}}$

= $\frac{1050}{25}$

= 42

Question 3: Comment on the performance of the students in 3 universities given below, using simple and weighted average.

 University Bombay Calcutta Madras Course of study % of Pass No of students % of Pass No of students % of Pass No of students M.A 71 3 82 2 81 2 M.Com 83 4 76 3 76 3.5 B.A 73 5 73 6 74 4.5 B.Com 74 2 76 7 58 2 B.Sc 65 3 65 3 70 7 M.Sc 66 3 60 7 73 2

Solution:
Computation of simple and weighted average is as follows

 University Bombay Calcutta Madras Course of Study % of Pass (X$_{1}$) No of students (W$_{1}$) $W_{1}X_{1}$ % of Pass (X$_{2}$) No of students (W$_{2}$) $W_{2}X_{2}$ % of Pass (X$_{3}$) No of students (W$_{3}$) $W_{3}X_{3}$ M.A 71 3 213 82 2 164 81 2 162 M.Com 83 4 332 76 3 228 76 3.5 266 B.A 73 5 365 73 6 438 74 4.5 333 B.Com 74 2 148 76 7 532 58 2 116 B.Sc 65 3 195 65 3 195 70 7 490 M.Sc 66 3 198 60 7 420 73 2 146 Total 432 20 1451 432 28 1977 432 21 1513

Simple Average:

Bombay

$\frac{432}{6}$ = 72

Calcutta

$\frac{432}{6}$ = 72

$\frac{432}{6}$ = 72

Weighted Average:

Bombay

$\frac{\sum\ W_{1}X_{1}}{\sum \ W_{1}}$
= $\frac{1451}{20}$ = 72.55

Calcutta

$\frac{\sum\ W_{2}X_{2}}{\sum \ W_{2}}$ = $\frac{1977}{28}$ = 70.61

$\frac{\sum\ W_{3}X_{3}}{\sum \ W_{3}}$ = $\frac{1513}{21}$ = 72.05