Mean Deviation

Even though Range and Quartile Deviation give an idea about the spread of individual items of a series, they do not try to calculate their dispersion from its average. If the variations of items were calculated from the average, such a measure of dispersion would through light on the formation of the series and the spread of items round the central value. Mean deviation (M.D) is such a measure of dispersion.

Mean deviation of a series is the arithmetic average of the deviations of various items from a measure of central tendency. In aggregating the deviations, algebraic signs of the deviations are not taken into account. It is because, if the algebraic signs were taken into account, the sum of deviations from the mean should be zero and that from median is nearly zero. Theoretically the deviations can be taken from any of the three averages, namely, arithmetic mean, median or mode; but, mode is usually not considered as it is less stable. Between mean and median, the latter is supposed to be better because, the sum of the deviations from the median is less than the sum of the deviations from the méan.

While doing problems, if the type of the average is mentioned, we take that average: otherwise we consider mean or median as the case may be.

This measure of dispersion has found favour with economists and business men due to its simplicity in calculation. For forecasting of business cycles, this measure has been found more useful than others. it is also good for small sample studies where elaborate statistical analysis is not needed.

Where D represents deviations from mean or median, ignoring signs, and N the total number of items.

MD is an absolute measure of dispersion. The relative measure of MD is coefficient of MD, defined as:

$$ \mathbf {Coefficient\,of\,MD\,=\,{{{\frac{MD}{Mean}} }}} $$

Mean Deviation: Points to remember.
  1. It is based on all items
  2. A change in even one value will affect it
  3. Value will be least, if we are calculating it from median
  4. Value will be higher, if calculated from the mean
  5. Since it ignores signs of deviations, it is not suitable for open-end distribution

Mean Deviation from Arithmetic Mean

Individual Series

STEPS:-

  • Find Mean using the equation \( {{{\frac{ΣX}{N}} }} \)

  • Take deviations of individual values from mean, |d| (modulus) = (x - X̄), ignoring signs

  • MD = \( {{{\frac{Σ|D|}{N}} }} \) (N = number of items)

  • Relative measure of MD is coefficient of MD. Coefficient of MD = \( {{{\frac{MD}{Mean}} }} \)

  • Let us find the value of mean deviation and its coefficient from the following data.
  • Table 6.11
    Roll No. Marks
    1 12
    2 18
    3 23
    4 18
    5 25
    6 15
    7 9
    8 14
    9 6
    10 23
    11 19
    12 10

    $$N\,=\,12$$ $$ X̄\,=\, {{{\frac{ΣX}{N}} }} $$ $$ =\, {{{\frac{192}{12}} }} $$ $$ =\,16 $$ Now we need to find modulus d. For that we creates a table as shown below.

    Table 6.12
    Roll No. X (Marks) |D| = |X - X̄| = |X - 16|
    1 12 4
    2 18 2
    3 23 7
    4 18 2
    5 25 9
    6 15 1
    7 9 7
    8 14 2
    9 6 10
    10 23 7
    11 19 3
    12 10 6
    N = 12 Σ|D| = 60

    $$ MD \,from \,X̄ \,=\,{{{\frac{Σ|D|}{N}} }} $$ $$ =\,{{{\frac{60}{12}} }} $$ $$ =\,5 $$ $$ Coefficient \,of\, MD \,=\,{{{\frac{MD}{Mean}} }} $$ $$ =\,{{{\frac{5}{16}} }} $$ $$ =\,0.3125 $$

    Discrete Series

    STEPS:-

    • Find Mean using the equation \( {{{\frac{ΣfX}{Σf}} }} \)

    • Take deviations of individual values from mean, |d| (modulus) = (x - X̄), ignoring signs

    • Find f|d| and Σf|D|(modulus) = (x - X̄), ignoring signs

    • MD = \( {{{\frac{Σf|D|}{Σf}} }} \)

    • Coefficient of MD = \( {{{\frac{MD}{Mean}} }} \)

  • Let us find the value of mean deviation and its coefficient from the following data.
  • Table 6.13
    Value Frequency
    8 2
    13 5
    15 9
    21 14
    24 7
    28 7
    29 4
    30 2

    We need to find fx, |D| and f|D|. This is shown in the below given table.

    Table 6.14
    Value f fx |D| = |X - X̄| = |X - 21| f|D|
    8 2 16 13 26
    13 5 65 8 40
    15 9 135 6 54
    21 14 294 0 0
    24 7 168 3 21
    28 7 196 7 49
    29 4 116 8 32
    30 2 60 9 18
    N = 50 ΣfX = 1050 Σf|D| = 240

    $$ X̄\,=\, {{{\frac{ΣfX}{N}} }} $$ $$ =\, {{{\frac{1050}{50}} }} $$ $$ =\,21 $$ $$ MD \,from \,X̄ \,=\,{{{\frac{Σf|D|}{N}} }} $$ $$ =\,{{{\frac{240}{50}} }} $$ $$ =\,4.8 $$ $$ Coefficient \,of\, MD \,=\,{{{\frac{MD}{Mean}} }} $$ $$ =\,{{{\frac{4.8}{21}} }} $$ $$ =\,0.23 $$

    Continuous Series

    In order to calculate MD and its coefficient for continuous series, we use the same method described earlier. Here we the devition from midvalues of classes. That is, we take midpoint as X here.

    STEPS:-

    • Find Mean using the equation \( {{{\frac{Σfm}{Σf}} }} \)

    • Take deviations of mid points from mean, |d| (modulus) = (m - X̄), ignoring signs

    • Find f|d| and Σf|D|

    • MD = \( {{{\frac{Σf|D|}{Σf}} }} \)

    • Coefficient of MD = \( {{{\frac{MD}{Mean}} }} \)

  • Let us find MD from AM for the following series relates the marks of 20 students:

    Table 6.15
    Marks No. of Students
    0 - 10 2
    10 - 20 2
    20 - 30 5
    30 - 40 5
    40 - 50 3
    50 - 60 2
    60 - 70 1

    We need to find midvalue (m), fm, |d| and f|D|. This is shown in the below given table.

    Table 6.16
    Class f X = m fm |D| = |m - X̄| = |m - 32.5| f|D|
    0-10 2 5 10 27.5 55
    10-20 2 15 30 17.5 35
    20-30 5 25 125 7.5 37.5
    30-40 5 35 175 2.5 12.5
    40-50 3 45 135 12.5 37.5
    50-60 2 55 110 22.5 45
    60-70 1 65 65 32.5 32.5
    Σf = 20 Σfm = 650 Σf|D| = 255

    $$ X̄\,=\, {{{\frac{Σfm}{Σf}} }} $$ $$ =\, {{{\frac{650}{20}} }} $$ $$ =\,32.5 $$ $$ MD \,from \,X̄ \,=\,{{{\frac{Σf|D|}{Σf}} }} $$ $$ =\,{{{\frac{255}{20}} }} $$ $$ =\,12.75 $$

    Mean Deviation from Median

    Individual Series

    STEPS:-

    • Arrange the data in ascending order

    • Compute the median

      Median = Size of \( {{{\frac{N + 1}{2}} }}^{th} \) item

    • Take deviation of individual values from median. i.e., |d| = X - Median (ignoring signs)

    • MDMedian = \( {{{\frac{Σ|D|}{N}} }} \) ( N = Number of items)

      Coefficient of MD = \( {{{\frac{MD}{Median}} }} \)

  • Let us find the value of mean deviation and its coefficient from the following data. We can calculate mean deviation from median.
  • 4000, 4200, 4400, 4600, 4800

    $$ Median\, =\, {{{\frac{N + 1}{2}} }}^{th} item $$ $$ =\, {{{\frac{5 + 1}{2}} }}^{th} item $$ $$ =\, {{{\frac{6}{2}} }}^{th} item $$ $$ =\, 3^{rd} item $$ $$ =\, 4400 $$

    Table 6.17
    Deviation from Median 4400
    Income |D|
    4000 400
    4200 200
    4400 0
    4600 200
    4800 400
    N = 5 Σ|D| = 1200

    $$ MD_Median\, = \, {{{\frac{Σ|D|}{N}} }} $$ $$ = \, {{{\frac{1200}{5}} }} $$ $$ =\, 240 $$ $$ Coefficient \,of\, MD\, =\, {{{\frac{MD}{Median}} }} $$ $$ = \, {{{\frac{240}{4400}} }} $$ $$ =\, 0.054 $$

    Discrete Series

    STEPS:-

    • Arrange the data in ascending order

    • Find out cumulative frequency

    • Find median; Median = \( \Biggl[{{\frac{N + 1 }{2}}}\Biggl]^{th} item \)

    • Take deviation of individual values from median. i.e., |d| = X - Median (ignoring signs)

    • MDMedian = \( {{{\frac{Σf|D|}{Σf}} }} \)

      Coefficient of MD = \( {{{\frac{MD_{Median}}{Median}} }} \)

  • Let us find the value of mean deviation from the following data. We can calculate mean deviation from median.
  • Table 6.18
    x f
    2 1
    4 4
    6 6
    8 4
    10 1

    We need to find midvalue |d|, f|D| and cf. This is shown in the below given table.

    $$ Median\, =\, {{{\frac{N + 1}{2}} }}^{th} item $$ $$ =\, {{{\frac{16 + 1}{2}} }}^{th} item $$ $$ =\, {{{\frac{17}{2}} }}^{th} item $$ $$ =\, 8.5^{th} item $$ $$ =\, 6 $$ $$ ∴\, Median\,=\, 6 $$

    Table 6.19
    x f |D| f|D| cf
    2 1 4 4 1
    4 4 2 8 5
    6 6 0 0 11
    8 4 2 8 15
    10 1 4 4 16

    $$ MD_{Median}\, = \, {{{\frac{Σf|D|}{Σf}} }} $$ $$ = \, {{{\frac{24}{16}} }} $$ $$ =\, 1.5 $$

    Continuous Series

    STEPS:-

    • Find Median

    • Median class = Size of \( {{{\frac{N}{2}} }}^{th} item \)

    • Median = \( { L + \frac{\frac{N}{2} - {cf}}{f} × h} \)

    • Find out |d| = x - Median

    • Find out f|d|

      MDMedian = \( {{{\frac{Σf|D|}{Σf}} }} \)

      Coefficient of MD = \( {{{\frac{MD_{Median}}{Median}} }} \)

  • Let us find the value of mean deviation and coefficient of mean deviation from the following data. We can calculate mean deviation from median.
  • Table 6.20
    Age No. of Person
    0 - 10 6
    10 - 20 9
    20 - 30 20
    30 - 40 5
    40 - 50 10

    We need to find cf, midvalue (m), |d| and f|D|. This is shown in the below given table.

    Table 6.21
    Class f cf m |D| = |m - median| f|D|
    0-10 6 6 5 20 120
    10-20 9 15 15 10 90
    20-30 20 35 25 0 0
    30-40 5 40 35 10 50
    40-50 10 50 45 20 200
    Σf = 50 Σf|D| = 460

    $$ Median \,class \,= \,Size \,of \, {{{\frac{N}{2}} }}^{th} item $$ $$ = \,{{{\frac{50}{2}} }}^{th} item $$ $$ = \,25^{th} item $$ 25th item lies in the class 20 - 30

    $$ Median \,= \,{ L + \frac{\frac{N}{2} - {cf}}{f} × h} $$ $$ = \,{ 20 + \frac{{25} - {15}}{20} × 10} $$ $$ = \,{ 20 \,+ \,5 \,=\, 25} $$ $$ MD_{Median} \,=\, {{{\frac{Σf|D|}{Σf}} }} $$ $$ =\, {{{\frac{460}{50}} }} $$ $$ =\, 9.2 $$ $$ Coefficient \,of \,MD \,= {{{\frac{MD_{Median}}{Median}} }} $$ $$ =\, {{{\frac{9.2}{25}} }} $$ $$ =\, 0.368 $$

    MERITS OF MEAN DEVIATION

    • It is rigidily defined.
    • The calculation is very simple.
    • It is based on all values.
    • It is not affected by extreme items.
    • It truly represents the average deviations of the items.
    • It has practical utilities in the fields of Business and Commerce.

    DEMERITS OF MEAN DEVIATION

    • The algebraic signs are ignored while taking the deviation of items.
    • It is not capable of further algebraic tratment.
    • It is not often useful for statistical inference.
    • It will not give accurate result when deviations are taken from mode.
    • Very much affected by sampling fluctuations.