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Arithmetic Mean

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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

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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
Arithmetic Mean

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

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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

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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
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

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.