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Unit 4: Central Tendency: Mean, Median and Mode and their Properties

The sum of the squared deviations is equal to 10 in the above case. If the deviations are taken Notes
from any other value the sum of the squared deviations would be greater than 10. This is
known as least squares property of the arithmetic mean.
3. The standard error of arithmetic mean is less than that of any other measure of central tendency.

∑X
4. Since X = N

NX = ∑X .
In other words, if we replace each item in the series by the mean, then the sum of these
substitutions will be equal to the sum of the individual items. For example, in the discussion of
first property ∑X = 150 and the arithmetic mean = 30. If for each item we substitute 30, we get
the same total i.e., 150 = [30 + 30 + 30 + 30 + 30].
* Algebraically the property ( ∑ − X = 0 is derived from the fact that NX = ∑X
) X
This property is of great practical value. For example, if we know the average wage in a factory,
say, Rs. 200, and the number of workers employed, say, 50, we can compute total wage bill
from the relation NX = ∑X . The total wage bill in this case would be 200 × 50 = 10,000 which
is equal to ∑X .
5. If we have the arithmetic mean and number of items of two or more than two related groups,
we can compute combined average of these groups by applying the following formula:

NX 1 + N X 2
2
1
X 12 = N 1 + N 2
X 12 = combined mean of the two groups
X 1 = arithmetic mean of first group

X 2 = arithmetic mean of second group

N 1 = number of items in the first group
N 2 = number of items in the second group.

6. If the given observations on X be changed to observation on Y, where Y = a + bX, then Y =
+ X .
ab
The following example shall illustrate the application of the above formula:
Example 1 : A factory employs 100 workers of whom 60 work in the first shift and 40 work in the
second shift. The average wage of all the 100 workers is Rs. 38. If the average wage of
60 workers of the first shift is Rs. 40, find the average wage of the remaining 40 workers
of the second shift.
Solution : Total no. of employees = 100
No. of employees in the first shift, i.e., N = 60
1
No. of employees in the second shift, i.e., N = 40
2
X 12 = 38, X = 40
1
NX 1 + N X 2
2
1
X 12 = N 1 + N 2

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