For grouped data, arithmetic mean may be calculated by applying any of the following methods:

(i) Direct method,       (ii) Short-cut method ,    (iii) Step-deviation method

In the case of direct method, the formula x = åfm/n is used. Here m is mid-point of various classes, f is the frequency of each class and n is the total number of frequencies. The calculation of arithmetic mean by the direct method is shown below.

Example 2.3: The following table gives the marks of 58 students in Statistics. Calculate the average marks of this group.

 Marks No. of Students 0-10 4 10-20 8 20-30 11 30-40 15 40-50 12 50-60 6 60-70 2 Total 58

Solution:

Table 2.3: Calculation of Arithmetic Mean by Direct Method

 Marks Mid-point m No. of Students f fm 0-10 5 4 20 10-20 15 8 120 20-30 25 11 275 30-40 35 15 525 40-50 45 12 540 50-60 55 6 330 60-70 65 2 130 åfm = 1940

Where,

x = å fm/ n  1940 /58 = 33.45 marks or 33 marks approximately.

It may be noted that the mid-point of each class is taken as a good approximation of the true mean of the class. This is based on the assumption that the values are distributed fairly evenly throughout the interval. When large numbers of frequency occur, this assumption is usually accepted.

In the case of short-cut method, the concept of arbitrary mean is followed. The formula for calculation of the arithmetic mean by the short-cut method is given below:

x = A + å fd / n

Where A = arbitrary or assumed mean

f = frequency

d = deviation from the arbitrary or assumed mean

When the values are extremely large and/or in fractions, the use of the direct method would be very cumbersome. In such cases, the short-cut method is preferable. This is because the calculation work in the short-cut method is considerably reduced particularly for calculation of the product of values and their respective frequencies. However, when calculations are not made manually but by a machine calculator, it may not be necessary to resort to the short-cut method, as the use of the direct method may not pose any problem.

As can be seen from the formula used in the short-cut method, an arbitrary or assumed mean is used. The second term in the formula (åfd ¸ n) is the correction factor for the difference between the actual mean and the assumed mean. If the assumed mean turns out to be equal to the actual mean, (åfd ¸ n) will be zero. The use of the short-cut method is based on the principle that the total of deviations taken from an actual mean is equal to zero. As such, the deviations taken from any other figure will depend on how the assumed mean is related to the actual mean. While one may choose any value as assumed mean, it would be proper to avoid extreme values, that is, too small or too high to simplify calculations. A value apparently close to the arithmetic mean should be chosen.

For the figures given earlier pertaining to marks obtained by 58 students, we calculate the average marks by using the short-cut method.

Example 2.4:

Table 2.4: Calculation of Arithmetic Mean by Short-cut Method

 Marks Mid-point m f d fd 0-10 5 4 -30 -120 10-20 15 8 -20 -160 20-30 25 11 -10 -110 30-40 35 15 0 0 40-50 45 12 10 120 50-60 55 6 20 120 60-70 65 2 30 60 åfd = -90

It may be noted that we have taken arbitrary mean as 35 and deviations from midpoints. In other words, the arbitrary mean has been subtracted from each value of mid-point and the resultant figure is shown in column d.

x = A + å fd /n

35+(-90/58)

=35-1.55 = 33.45 or 33 marks approximately.

Now we take up the calculation of arithmetic mean for the same set of data using the step-deviation method. This is shown in Table 2.5.

Table 2.5: Calculation of Arithmetic Mean by Step-deviation Method

 Marks Mid-point f d d’= d/10 Fd’ 0-10 5 4 -30 -3 -12 10-20 15 8 -20 -2 -16 20-30 25 11 -10 -1 -11 30-40 35 15 0 0 0 40-50 45 12 10 1 12 50-60 55 6 20 2 12 60-70 65 2 30 3 6 åfd’ =-9

x = A + å fd ' / n * C

-35+ (-9* 10 /58) =  33.45 or 33 marks approximately.

It will be seen that the answer in each of the three cases is the same. The step- deviation method is the most convenient on account of simplified calculations. It may also be noted that if we select a different arbitrary mean and recalculate deviations from that figure, we would get the same answer.

Now that we have learnt how the arithmetic mean can be calculated by using different methods, we are in a position to handle any problem where calculation of the arithmetic mean is involved.

Example 2.6: The mean of the following frequency distribution was found to be 1.46.

 No. of Accidents No. of Days (frequency) 0 46 1 ? 2 ? 3 25 4 10

5                               5

Total         200 days

Calculate the missing frequencies.

Solution:

Here we are given the total number of frequencies and the arithmetic mean. We have to determine the two frequencies that are missing. Let us assume that the frequency against 1 accident is x and against 2 accidents is y. If we can establish two simultaneous equations, then we can easily find the values of X and Y.

Mean =

(0.46) + (1. x) + (2 . y) + (3 . 25) + (4 . l0) + (5 . 5)/ 200

1.46 = x + 2y + 140 /200

x + 2y + 140 = (200) (1.46)

x + 2y = 152

x + y=200- {46+25 + 1O+5}

x + y = 200 - 86

x + y = 114

Now subtracting equation (ii) from equation (i), we get

x + 2y   =         152

x + y     =         114

- -                  -

y          =         38

Substituting the value of y = 38 in equation (ii) above, x + 38 = 114 Therefore, x = 114 - 38 = 76

Hence, the missing frequencies are:

Against accident 1 : 76

Against accident 2 : 38