The Open - end classes are those in which the  lower limit of the first class and the upper limit of the last class are known. In such cases we cannot find out the arithmetic mean unless we make an assumption about the unknown limits. The assumption would generally depend upon the class interval following the first class and preceding the last class. For e.g. observe the data as shown below:

In the above case since the class interval is uniform, the exact assumption would be that the lower limit of the first class is zero and the upper limit of the last class is 60. The first class  would be 0-10 and the last class 50-60 observes the another case.

In the above case as the class interval is 20 in the second class 30 in the third class 40 in the fourth class it is increasing by 10. The exact assumption would be that the lower limit of the first class is zero and the upper limit of the last class 150. In another words first class is 0-10 and the last is  100-150.

If the class intervals are of varying width an effort should not be made to determine the lower limit of the lowest calls and the upper limit of the highest class. The use of median or mode would be excellent in such a case. Because of the difficulty of  ascertaining lower limit and upper limit in open-end distributions it is suggested that in such distributions the  arithmetic mean should not be used.

The Mathematical properties of an  arithmetic mean

The following are a few importunate mathematical properties of the arithmetic mean:

1. The sum of the deviations of the items from the arithmetic mean (taking signs into account) are always zero, Σ (X - X) = 0. This would be very clear from the example below:

Here X = ΣX / N = 150 / 5 = 30. When  sum of the deviations from the actual mean 30, is taken it comes out to be zero. It is because of this property that the mean is characterized as a point of balance the sum of the positive deviations from it is equal to the sum of the negative deviations from it.

2. The sum of the squared deviations of the items from the  arithmetic mean is minimum that is less than the sum of the squared deviations often items from any other value. The example below would verify the point.

The sum of the squared deviations is equal to 10 in the above case. If the deviations are taken from any other value, then the sum of the squared deviations would be greater than 10. For e.g. let us calculate the squares of the deviations of item form a value less than the arithmetic mean say

It is clear that Σ (X - X)² is greater. These properties that the sum of the squares of items is least form the means of immense use in regression anal sis which shall be described  later.

3. Since   X = ΣX / N or NX = Σ X

In other words if we replace every item in the series by the mean, then the sum of these substitutions will be equal to the sum of the individual items. For e.g. in the discussion of first property ΣX = 150  &  the arithmetic mean 30. If for every item we substitute 30 we get the same total 30+30+30+ 30 + 30 = 150.

This property is of very practical value for e.g. if we know that the average wage in a factory say $ 1,060 and the number of workers employed say 200. We can compute the total wages bill from the relation NX = ΣX. The total wage bill in this case would be 200X 1,060, $ 2, 12,000 which is equal to ΣX.

4. If we have the arithmetic means and number or items of two or more than two related groups we can compute them by the combined average of these groups by applying the following formula:

X₁₂ = N₁ X₁ + N₂ X₂ / N₁ + N₂

X₁₂ = combined mean of the two groups

X₁ = arithmetic mean of first group

X₂ = arithmetic mean of second group

N₁ = number of items in the first group

_N2 = number of items in the second group.

The following example  illustrate the application of the above formula

Illustration:

The mean heightt of 25 male workers in a factory is 61 inches and the mean height of 35 female workers in the same factory is 58 inches. Find the combined mean height of 60 workers in the factory.

Solution:

X₁₂ = N₁ X₁ + N₂ X₂ / N₁ + N₂

N₁ = 25 X 1 = 61 N₂ = 35 X2 = 58

X₁₂ = (25 X 61) + (35 X 58) 25 + 35 = 1525 + 2030 / 60 = 3555 / 60 = 59.25

Thus the combined mean height of 60 workers is 59.25 inches.

If we have to find out the combined mean to 3 sub-groups the above formula can be extended as follows:

X₁₂₃ = N₁ X₁ + N₂ X₂ + N₃ X₃ / N₁ + N₂ + N₃