Arithmetic Mean.
Arithmetic mean of a set of observations is their sum divided by the number of observations. e.g the arithmetic mean \[\bar{X}\] of n observations \[{{X}_{1}},\ldots ,{{X}_{n}}\] is given by \[\bar{X}=\frac{1}{n}({{X}_{1}},\ldots ,{{X}_{n}})=\frac{1}{n}\sum\limits_{i=1}^{n}{{{X}_{i}}}\] In case of frequency distribution. Xi where fi is the frequency of the variable Xi
Merits and Demerits of Arithmetic Mean
Merits
(i) It is rigidly defined.
(ii) It is easy to understand and easy to' calculate.
(iii) It is based upon all the observations.
(iv) It is amenable to algebraic treatment. The mean of the composite series in terms of the means and sizes of the component series is given by
\[\bar{X}=\frac{\sum\limits_{i=1}^{k}{{{n}_{i}}}{{{\bar{X}}}_{i}}}{(\sum\limits_{i}^{k}{{{n}_{i}}})}\]
(v) Of all the averages, arithmetic mean is affected least by fluctuations of sampling . This property is sometimes described by saying that arithmetic mean is, a stable average. Thus, we see that arithmetic mean satisfies all the properties laid down by Prof Yule for an ideal average.
Demerits
(i) it cannot be determined by inspection nor it can be located graphically.
(ii) Arithmetic mean cannot be used if we are dealing with qualitative characteristics which cannot be measured quantitatively ; such as, intelligence, honesty, beauty, etc. In such cases median (discussed later) is the only average to be used.
(iii) Arithmetic mean cannot be obtained if a single observation is missing or lost or is illegible unless we drop it out and compute the arithmetic mean of the remaining values.,
(iv). Arithmetic mean is affected very much by extreme values. In case of extreme items, arithmetic mean gives a distorted picture of the distribution and no longer remains representative of the distribution.
(v) Arithmetic mean may lead to wrong conclusions if the details of the data from which it is computed are not given. Let us consider the following marks obtained by two students A and B in three tests, viz., terminal test, half-yearly examination and annual examination respectively.
Marks in I Test II Test III Test Average marks
A 50% 60% 70% 60%
B 70% 60% 50% 60%
Thus average marks obtained by each of the two students at the end of the year If are 60%. we are given the average marks alone we conclude that the level of intelligence of both the students at the end of the year is same. This is a fallacious conclusion since we find from the data that student A has improved consistently while student B has deteriorated consistently.
(vi) Arithmetic mean cannot 'be calculated if the extreme class is open, e.g., below 10 or above 90.Moreover, even if a single observation is missing mean cannot be calculated.
(vii) In extremely asymmetrical (skewed) distribution, usually arithmetic mean is not a suitable measure of location.
Weighted Mean
In calculating arithmetic' mean we suppose that all the items in the distribution have equal importance. But in practice this may not be so. 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 weighted is to be given to various items the weights attached to each item being proportional to the importance 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 simple mean of the prices of the commodities consumed by them will not do, since all the commodities are not equally important, e.g., wheat, rice and pulses are more important than cigarettes, tea, confectionery, etc.
Let wi be the weight attached to the item Xi , i = I, 2, ... , n. Then we define:
Weighted arithmetic mean or weighted mean
\[\frac{\sum\limits_{i}{{{w}_{_{i}}}{{x}_{i}}}}{\sum\limits_{i}{{{w}_{i}}}}\] ………………………….(2.5)
It may be observed that the formula for weighted mean is the same as the formula for simple mean with fi (i = I, 2, ... , II), tile frequencies replaced by wi , (i = I, 2, ... , n), the weights.
Weighted mean gives the result equal to the simple mean if the weights assigned to each of the variate values arc equal. It results in higher value than the simple mean if smaller weights arc given to smaller items and larger weights to larger items. If the weights attached to larger items are smaller and those attached to smaller items arc larger, then the weighted mean results in smaller value than the simple mean.
Example
Find the simple and weighted arithmetic mean of the first n natural number, the weights being the corresponding numbers.
Solution-
The first natural numbers arc I, 2, 3……....n
X w wX
1 1 1^2
2 2 2^2
3 3 3^2
. . .
. . .
. . .
n n n
we know that
\[1+2+3+.....+n=\frac{n(n+1)}{2}\]
\[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+........+{{n}^{2}}=\frac{n(n+1)(2n+1)}{6}\]
\[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+........+{{n}^{2}}=\frac{n(n+1)(2n+1)}{6}\]
Simple A.M. is
\[\bar{X}=\frac{\sum{x}}{n}=\frac{1+2+3+........n}{n}=\frac{n+1}{2}\]
Weighted A.M. is
\[{{\bar{X}}_{w}}=\frac{\sum{w}X}{\sum{w}}=\frac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....{{n}^{2}}}{1+2+.....+n}=\frac{n(n+1)(2n+1)}{6}\frac{2}{n(n+1)}\]
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