Before the discussion of arithmetic mean, we shall introduce certain notations. It will be assumed that there are n observations whose values are denoted by X_{1},X_{2}, ..... X_{n} respectively. The sum of these observations X_{1} + X_{2} + ..... + X_{n} will be denoted in abbreviated form as
where S (called sigma) denotes summation sign.
The subscript of X, i.e., 'i' is a positive integer, which indicates the serial number of the observation. Since there are n observations, variation in i will be from 1 to n. This is indicated by writing it below and above S, as written earlier. When there is no ambiguity in range of summation, this indication can be skipped and we may simply write X_{1} + X_{2} + ..... + _{Xn} = SX_{i}.
Arithmetic Mean is defined as the sum of observations divided by the number of observations. It can be computed in two ways:
In case of simple arithmetic mean, equal importance is given to all the observations while in weighted arithmetic mean, the importance given to various observations is not same.
Calculation of Simple Arithmetic Mean
(a) When Individual Observations are given.
Let there be n observations X_{1}, X_{2} ..... X_{n}. Their arithmetic mean can be calculated either by direct method or by short cut method. The arithmetic mean of these observations will be denoted by X
Direct Method: Under this method, X is obtained by dividing sum of observations by number of observations, i.e.,
Shortcut Method: This method is used when the magnitude of individual observations is large. The use of shortcut method is helpful in the simplification of calculation work. Let A be any assumed mean. We subtract A from every observation. The difference between an observation and A, i.e., Xi  A is called the deviation of i th observation from A and is denoted by di. Thus, we can write ; d_{1 = X1  A, d2 = X2  A, ..... dn = Xn  A. On adding these deviations and dividing by n we get}
This result can be used for the calculation of X .
Remarks: Theoretically we can select any value as assumed mean. However, for the purpose of simplification of calculation work, the selected value should be as nearer to the value of X as possible.
Example : The following figures relate to monthly output of cloth of a factory in a given year
(c) When data are in the form of a grouped frequency distribution
In a grouped frequency distribution, there are classes along with their respective frequencies. Let li be the lower limit and ui be the upper limit of ith class. Further, let the number of classes be n, so that i = 1, 2,.....n. Also let fi be the frequency of ith class. This distribution can written in tabular form, as shown.
Note: Here u1 may or may not be equal to l2, i.e., the upper limit of a class may or may not be equal to the lower limit of its following class.
It may be recalled here that, in a grouped frequency distribution, we only know the number of observations in a particular class interval and not their individual magnitudes. Therefore, to calculate mean, we have to make a fundamental assumption that the observations in a class are uniformly distributed.
Under this assumption, the midvalue of a class will be equal to the mean of observations in that class and hence can be taken as their representative. Therefore, if Xi is the midvalue of i th class with frequency fi , the above assumption implies that there are fi observations each with magnitude Xi (i = 1 to n). Thus, the arithmetic mean of a grouped frequency distribution can also be calculated by the use of the formula, given in § below.
Remarks: The accuracy of arithmetic mean calculated for a grouped frequency distribution depends upon the validity of the fundamental assumption. This assumption is rarely met in practice. Therefore, we can only get an approximate value of the arithmetic mean of a grouped frequency distribution.
Example : Calculate arithmetic mean of the following distribution :
Solution: Here only shortcut method will be used to calculate arithmetic mean but it can also be calculated by the use of directmethod
Example : The following table gives the distribution of weekly wages of workers in a factory. Calculate the arithmetic mean of the distribution.
Solution: It may be noted here that the given class intervals are inclusive. However, for the computation of mean, they need not be converted into exclusive class intervals.
Step deviation method or coding method
In a grouped frequency distribution, if all the classes are of equal width, say 'h', the successive midvalues of various classes will differ from each other by this width. This fact can be utilised for reducing the work of computations.
Weighted Arithmetic Mean
In the computation of simple arithmetic mean, equal importance is given to all the items. But this may not be so in all situations. If all the items are not of equal importance, then simple arithmetic mean will not be a good representative of the given data. Hence, weighing of different items becomes necessary. The weights are assigned to different items depending upon their importance, i.e., more important items are assigned more weight.
For example, to calculate mean wage of the workers of a factory, it would be wrong to compute simple arithmetic mean if there are a few workers (say managers) with very high wages while majority of the workers are at low level of wages. The simple arithmetic mean, in such a situation, will give a higher value that cannot be regarded as representative wage for the group. In order that the mean wage gives a realistic picture of the distribution, the wages of managers should be given less importance in its computation.
The mean calculated in this manner is called weighted arithmetic mean. The computation of weighted arithmetic is useful in many situations where different items are of unequal importance, e.g., the construction index numbers, computation of standardized death and birth rates, etc.
Merits and Demerits of Arithmetic Mean
Merits:
Demerits:
Exercise with Hints
Hint : Take the midvalue of a class as the mean of its limits and find arithmetic mean by the stepdeviation method.
Hint : This distribution is with open end classes. To calculate mean, it is to be assumed that the width of first class is same as the width of second class. On this assumption the lower limit of the first class will be 0. Similarly, it is assumed that the width of last class is equal to the width of last but one class. Therefore, the upper limit of the last class can be taken as 6,000.
Hint: First convert the distribution into class intervals and then calculate X .
Find average profit per shop.
Hint: This is a less than type cumulative frequency distribution.
The actual salaries of staff members are as given below :
1120, 1200, 1500, 4500, 4250, 3900, 3700, 3950, 3750, 2900, 2500, 1650, 1350, 4800, 3300, 3500, 1100, 1800, 2450, 2700, 3550, 2400, 2900, 2600, 2750, 2900, 2100, 2600, 2350, 2450, 2500, 2700, 3200, 3800, 3100.
Determine (i) Total amount of bonus paid and (ii) Average bonus paid per employee.
Hint: Find the frequencies of the classes from the given information.
Hint: Rearrange this in the form of frequency distribution by taking class intervals as 90  100, 100  110, etc.
Hint: Take class intervals as 0  6, 6  10, 10  14, 14  20, etc.
Hint: Use the formula X = A + å fu/ N× h to find the class width h.
Hint: The no. of shares of each type = no. of companies ´ average no. of shares.
Hint: Take n1 as the no. of boys and 150  n1 as the no. of girls.
Hint: See example above.
Hint: See example above.
Hint: See previous example.
Hint: See previous example.
Hint: See above example.
If 100 students took the examination and their mean marks were 51, calculate the mean marks of students who failed.
Hint: See above example .
Hint: Let x be the percentage of marks in third test. Then the weighted average of 75, 60 and x should be 60, where weights are 20, 50 and 30 respectively.
Hint: Correct average is weighted arithmetic average.
Calculate the average rate of conveyance allowance given to each salesman per kilometre by the company.
Hint: Obtain total number of kilometre travelled for each rate of conveyance allowance by multiplying midvalues of column 1 with column 2. Treat this as frequency 'f' and third column as 'X' and find X .
Find: (i) Average income per member for the entire group of families.
(ii) Average expenditure per family.
(iii) The difference between actual and average expenditure for each family.
Hint: (i) Average income per member = Total income of the group of families/Total no of members in the group
(ii) Average expenditure per family = Total expenditure of the group/No of families
Estimate:
(i) Mean income of an employee per month.
(ii) Monthly contribution to welfare fund if every employee belonging to the top 80% of the earners is supposed to contribute 2% of his income to this fund.
Hint: The distribution of top 80% of the wage earners can be written as
By taking midvalues of class intervals find Sfx, i.e., total salary and take 2% of this.
Which of these two diseases has more incidence in April 1991? Justify your conclusion.
Hint: The more incidence of disease is given by higher average number of patients.
Hint: Since the weight of the largest wage is less in 1995, the increase in average wage will be less than 15%.
Hint: See above example .


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