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. |
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.
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 $X_{1}$, ................., $X_{n}$ Given below are some of the example problems on arithmetic mean.
Solved Examples
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:
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
Solution:
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
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:
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
Madras
$\frac{432}{6}$ = 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
Madras
$\frac{\sum\ W_{3}X_{3}}{\sum \ W_{3}}$ = $\frac{1513}{21}$ = 72.05
Comment: On the basis of simple average, which comes out to be same for each university, we cannot distinguish between the pass percentage of the students in the 3 universities. However, the weighted average shows that the results are best in the Bombay university, followed by Madras university, while Calcutta university shows the lowest performance.