Example:
Calculate the Mean Deviation from the data given in example, by using the short-cut method and assuming the mean value as 25 marks:
Solution:
Marks (X) | Mid Values (w.v) (ii) | No.of students (f) (iii) | Dev. from assumed mean 25 (dx) (ignoring + and –sings) (iv) | Step Dev dx/I (v) | Total dev.from assumed Av. fdx. (vi) |
0-10 | 5 | 6 | 20 | 2 | 12 |
10-20 | 15 | 5 | 10 | 1 | 5 |
20—30 | 25 | 8 | 0 | 0 | 0 |
30-40 | 35 | 15 | 10 | 1 | 15 |
40-50 | 45 | 7 | 20 | 2 | 14 |
50-60 | 55 | 6 | 30 | 3 | 18 |
60-70 | 65 | 3 | 40 | 4 | 12 |
∑f = 50 | 76X10 = 760 |
Total Deviations from assumed mean (+ & — signs ignored) = 760
Note: Where actual and assumed averages are in different class intervals a special adjustment is necessary. In such cases the frequency of the class in which the actual mean lies is treated separately. It is multiplied by the difference of the deviations of the mid-value from the actual and assumed averages. The product so obtained is subtracted from the total deviation from the assumed mean.
(i) Now the number of items smaller than the actual mean (33.4 as calculated in example 15) = 19 (frequency of the mean class being ignored).
(ii) Number of items bigger than the actual mean = 16 (frequency of the mean class being ignored).
(iii) Frequency of the Mean class = 15
(iv) Difference between Actual and Assumed means = (33.4-25) = 8.4
(v) Difference of Deviations of mid-value of mean class from the actual and assumed average
= (35-33.4) — (35—25)
= 1.6—10 = 8.4 (+ and—signs ignored)
(vi) Total Deviation from actual Mean
= 760+ (19 x 8.4) — (16 X 8.4) — (15 x 8.4) = 659.2
(vii) Mean Deviation = 569.2/50 = 13.18 marks
Thus the answer derived in this manner is the same as we got in example No. 14.
It should be noted that the difference between the values of actual and assumed mean (8.4 in this case) and the 'difference of deviations, of mid-values of mean class from the actual and assumed average' [(32—33.4) (35—25) = 8.41 need not necessarily be the same figure. It is a coincidence that the two values are the same in the problem.
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Short-cut Method
While calculating the mean deviation from the mean it may be more convenient to use a short-cut method by assuming an arithmetic average. The process of calculating mean deviation by the short-cut method involves the following steps:—
(1) Deviations of items are taken from an assumed mean and multiplied by their respective frequencies and the products so obtained are totaled.
(2) Number of items less than the actual arithmetic average are multiplied by the difference between the actual and the assumed mean.
(3) Similarly, the number of items more than the arithmetic average are multiplied by the difference between actual and the assumed mean.
(4) The latter (No. 3) is deducted from the former (No. 2) and the balance is added to the sum of the products of deviations from the assumed mean and their frequencies (No. 1).
(5) The resulting figure is divided by the number of items and it is the value of the mean deviation.
Example is solved below by this short-cut method.
Marks (X) | Step Deviation From as. Av (25) (f) | No. of Student’s (f) (fd) | Number of (f) | Dev. From as. Av. 25 (fd) (+ & - signs (with + & -signs) | |
5 | —2 | 5 | 10 | 8 | —10 |
15 | —1 | 8 | 15 | 0 | —8 |
15 | 0 | 15 | 0 | —8 | 30 |
35 | +1 | 16 | 16 | +16 | |
45 | +2 | 6 | +12 | 12 | + 12 |
4 | 17 | ∑f= 50 | ∑fd’ = 46 | ∑fd = +10 |
Total deviations from assumed average of 25 = 46 x 10=460
Adjustments
Number of items less than the actual arithmetic average (27) =28
Number of items more than the actual arithmetic average (27) =22
Difference between actual and assumed average =2
Total deviations from actual average or (∑fdX)
= 460 + (28 x2) — (22 x2)
= 460 + 56 — 44=472
Mean Deviation = (∑|fdX)/N = 472/50
= 9.44 marks.
(3) Calculation of mean deviation in continuous series. The calculation of mean deviation in continuous series is done by the same procedure by which it is done in discrete series. In the short-cut method also the same procedure is -followed provided the assumed mean or median is in the same class-interval in which the actual mean or median is. If the assumed average is in a different class interval, further adjustments are necessary.
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