Mean Deviation
This is the average sum of the deviations from the arithmetic mean (i.e. the sum of the differences between the scores and the mean) divided by the total frequency.
Example 1
Find the mean deviation in the following test scores,
| Score (x) | Frequency (f) |
| 95 | 1 |
| 85 | 1 |
| 80 | 1 |
| 75 | 4 |
| 70 | 1 |
| 65 | 3 |
| 55 | 1 |
| 40 | 3 |
Solution
The mean of the score X = 66.7
| 1
Scores (X) |
2
Frequency (f) |
3
X – X҄ = x = d |
4
Fx = fd |
| 95 | 1 | 95-66.7 = 28.3 | 1 X 28.3 = 28.3 |
| 85 | 1 | 85-66.7 = 18.3 | 1 X 18.3 = 18.3 |
| 80 | 1 | 80-66.7 = 13.3 | 1 X 13.3 = 13.3 |
| 75 | 4 | 75-66.7 = 8.3 | 4 X 8.3 = 33.2 |
| 70 | 1 | 70-66.7 = 3.3 | 1 X 3.3 = 3.3 |
| 65 | 3 | 65-66.7 = -1.7 | 3 X (-1.7) = -5.1 |
| 55 | 1 | 55-66.7 = -11.7 | 1(-11.7) = -11.7 |
| 40 | 3 | 40-66.7 = -26.7 | 3(-26.7) = -80.1 |
| ∑F = 15 | ∑FD = -0.5 |
So we subtract each score (X) from the mean (66.7) as shown in column above (3). Hence column (3) shows the deviation of each score form the mean. Column (4) shows the product of the deviations and their frequencies which we then sum as ∑fd. Hence the mean deviation is given by the formula ∑fd/∑f
where ∑fd = the sum of the products of deviations and the frequencies
∑f = total frequency
Mean deviation = ∑fd/∑f = -05/15 = -0.0333
We always take the positive sign regardless of the sign. The value of the mean deviation calculated above is so small that it is ignored usually in computing measures of dispersion which is not the true situation. So we talk about the Mean Absolute Deviation to give the true picture.
Mean Absolute Deviation
From the above mean deviation, the negative sign in column 3 were not ignored so we had an insignificant value for the mean deviation. This leads to the finding of the mean absolute deviation where the modulus (or absolute) of the values are considered in the computation.
Mean absolute deviation is defined as the arithmetic mean of the absolute (modulus) value of the difference of each score form the mean (that is adding the values of the deviations from the mean as if they were all positive) and dividing by their total frequency. That is the sum of the modulus of the deviations from the arithmetic mean divided by the total frequency. Hence the mean absolute deviation is given by the formula
MD = ∑f|x|/∑f or ∑f|d|/∑f
where |x| or |d| = modulus of the deviations
∑f|x| or ∑f|d| = sum of the product of the modulus of the deviation and their frequencies
∑f = total frequency
Standard Deviation
The standard deviation is the most important of all the measures of dispersion. Equally important is the variance which is the square of standard deviation. The standard deviation tells us how far or near a score is to the mean score. The standard deviation is denoted with s when we consider only a sample of a group to study or with σ (sigma) when we study the entire group. At times it is shortened as S.D. meaning standard deviation.
The formula for its computation is as follows:
S.D = √∑fd2/∑f or √∑fx2/∑f or √∑x2/N
and variance usually denoted by
s2 or σ2 = ∑fd2/∑f or ∑fx2/∑f
In other words S.D. = √v
where ∑f = sum of frequencies
N = sum of the scores, if each score occurs once only
d2 or x2 = square of the deviations from the area
i.e. d2 or x2 = (X – X҄)2, where X = score and X҄ = arithmetic mean.
Examples
Compute (i) the variance and (ii) the standard deviation of the following test scores:
| Score | 95 | 85 | 80 | 75 | 70 | 65 | 55 | 40 |
| Frequency | 1 | 1 | 1 | 4 | 1 | 3 | 1 | 3 |
Solution
The scores can be arranged in a frequency table as follows:
| (1)
Scores (X) |
(2)
Derivation from the mean d = X – X҄ = x |
(3)
Frequency (f) |
(4)
fd2 = f(X – X҄)2 |
| 95 | 95-66.7 = 28.3 | 1 | 1 x (28.3)2 = 806.89 |
| 85 | 85-66.7 = 18.3 | 1 | 1 x (18.3)2 = 334.84 |
| 80 | 80-66.7 = 13.3 | 1 | 1 x (13.3)2 = 176.89 |
| 75 | 75-66.7 = 8.3 | 4 | 4 x (8.3)2 = 275.56 |
| 70 | 70-66.7 = 3.3 | 1 | 1 x (3.3)2 = 8.69 |
| 65 | 65-66.7 = -1.7 | 3 | 3 x (-1.7)2 = 8.67 |
| 55 | 55-66.7 = -11.7 | 1 | 1(-11.7)2 = 136.89 |
| 40 | 40-66.7 = -26.7 | 3 | 3(-26.7)2 = 2138.67 |
In the above column (2) we use the value of the already computed mean X҄ = 66.6, hence so applying the formula for computing the standard deviation, √∑fd2/∑f = √∑(X – X҄)2/∑f
where √∑fd2 = 3883.35 and √∑f = 15
we have variance = 3883.25/15 = 258.89
SD = -√258.89 = 16.09
S the standard deviation for the rest scores is 16.09 while the variance, which is the square of the standard deviation = 258.89 i.e. (16.09)2 = 258.89 i.e. (S.D)2 = variance.
EXERCISES
Lets see how much you’ve learnt, attach the following answers to the comment below
| Score | 50-54 | 55-59 | 60-64 | 65-69 | 70-74 | 75-79 | 80-84 | 85-89 | 90-94 |
| Frequency | 3 | 5 | 8 | 10 | 7 | 6 | 3 | 2 | 1 |
Calculate for questions 1 – 5 using the table above
- Determine ∑f A. 45 B. 50 C. 55 D. 60
- ∑fd, if d = X – A A. -160 B. -167 C. -155 D. -201
- Mean X҄, as X҄ = A + ∑fd/∑f A. 70.05 B. 68.56 C. 23.55 D. 65.75
- What is the Variance? A. 97.03 B. 87.75 C. 92.03 D. 50.08
- What is the Standard Variation? A. 12.54 B. 9.59 C. 25.75 D. 32.23
