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Note on Mean deviation

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Mean deviation or the average deviation is the average of the deviation of the item from the measure of central tendency i.e. mean, median or mode. Mean deviation, however, gives best results when deviations are taken from the median. While computing the value of mean deviation, positive and negative signs are ignored, i.e., all value are treated as positive.

Theoretically, deviations can be taken from any of the three average mentioned above but in actual practice mean deviation is calculated either from mean or from the median. Between mean and median, the latter is supposed to be better than the former, because the sum of the deviations from the median is less than the sum of the deviations from the mean. Therefore, the value of the mean deviation from the median is always less than the value calculated from mean. So, as a choice between mean and median theoretically, the median should be preferred. But because of wide applications of mean in statistics, in practice, mean deviation is generally computed from mean. In aggregating deviations, the algebraic signs are not taken into account. There is no mathematical justification for ignoring the signs of the deviation. However, simple reasons may be that if we do not ignore the signs of deviation, then the mean deviation from mean becomes zero. Since the main objective of a measure of dispersion is to study the scatter of the given observations from a central value, the ignoring of the deviation does not matter much. However, leaving of plus and minus signs render mean deviation incapable of further algebraic treatment.

Calculation of Mean Deviation

Individual Series

Let X1, X2, X3, ...... Xn be the N-variate value and \(\overline {X}\) and Md be the ar thematic mean and median of the series. The mean deviation is defined by the following formula:

M.D. from mean = \(\frac{\sum|X - \overline{X}|}{N}\)

M.D. from median = \(\frac{\sum|X - Md|}{N}\)

 .

Discrete Series

Let X1, X2, X3, ....... Xn be the variate value and f1, f2, f3, ........... fn be their corresponding frequencies respectively. If \(\overline {X}\) and Md are arithmetic mean and median of the series, then mean deviation is defined by the following formula:

M.D. from mean = \(\frac {\sum{f}|X - \overline{X}|}{\sum{f}}\) = \(\frac {\sum{f}|X - \overline{X}|}{N}\)

M.D. from median = \(\frac {\sum{f}|X - Md|}{\sum{f}}\) = \(\frac {\sum{f}|X - Md|}{N}\)

where, N = \(\sum{f}\) = total frequency

Continuous Series

Let X1, X2, X3, ...... Xnbe the mid-value of corresponding classes and f1, f2, f3, ........ fn be their corresponding frequencies. If \(\overline {X}\) and Md are arithmetic mean and median, then mean deviation is defined by the following formula:

M.D. from mean = \(\frac{\sum{f}|m - \overline{X}|}{\sum{f}}\) = \(\frac{\sum{f}|m - \overline{X}|}{N}\)

M.D. from median = \(\frac{\sum{f}|m - Md|}{\sum{f}}\) = \(\frac{\sum{f}|m - Md|}{N}\)

where, N = \(\sum{f}\) = total frequency

Mean deviation is an absolute measure of dispersion. For compairing two or more series having different unit, we calculate coefficient of mean deviation. The relative measure of dispersion based on mean deviation is called coefficient of mean deviation.

Coefficient of M.D. from mean = \(\frac {M.D. from\;mean}{Mean}\)

Coefficient of M.D. from median = \(\frac {M.D. from\;median}{Median}\)

Merits and Demerits of Mean Deviation

Merits

  1. Mean deviation is based on all the observations and is thus definitely a better measure of dispersion than the range and quartile deviation.
  2. Mean deviation is rigidly defined and is easy to understand and calculate.
  3. Mean deviation is rigidly understood. It is the average of the deviations from a measure of central tendency.
  4. As compared with standard deviation, it is less affected by extreme observations.
  5. Since mean deviation is based on the deviation about an average, it provides a better measure for comparison about the formation of different distributions.

Demerits

  1. The strongest objection against mean deviation is that while computing its value we take the absolute value of the deviations about an average and ignore the signs of the deviations.
  2. The step of ignoring the signs of the deviations is mathematically unsound and illogical. It creates artificially and renders mean deviation useless for further mathematical treatment.

 

Individual Series:

M.D. from mean = \(\frac{\sum|X - \overline{X}|}{N}\)

M.D. from median = \(\frac{\sum|X - Md|}{N}\)

Discrete Series:

M.D. from mean = \(\frac {\sum{f}|X - \overline{X}|}{\sum{f}}\) = \(\frac {\sum{f}|X - \overline{X}|}{N}\)

M.D. from median = \(\frac {\sum{f}|X - Md|}{\sum{f}}\) = \(\frac {\sum{f}|X - Md|}{N}\)

Continuous Series:

M.D. from mean = \(\frac{\sum{f}|m - \overline{X}|}{\sum{f}}\) = \(\frac{\sum{f}|m - \overline{X}|}{N}\)

M.D. from median = \(\frac{\sum{f}|m - Md|}{\sum{f}}\) = \(\frac{\sum{f}|m - Md|}{N}\)

where, N = \(\sum{f}\) = total frequency

Coefficient of M.D.:

Coefficient of M.D. from mean = \(\frac {M.D. from\;mean}{Mean}\)

Coefficient of M.D. from median = \(\frac {M.D. from\;median}{Median}\)

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Very Short Questions

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  • Find the range and the coefficient of range from following data:

    Temp.o 0-10 10-20 20-30 30-40 40-50
    Days 5 12 25 18 2

    50, 1


    90, 1


    60,1


    65, 1


  • Compute mean deviation from mean:

    class 0-10 10-20 20-30 30-40
    f 2 4 5 1

    7.5,0.1


    11.125, 0.23


    9.1, 0.27


    M.D.(ar X) = 7.5, 0.39


  • x 10-19 20-29 30-39 40-49 50-59
    f 4 10 18 16 12

    8.4,0.23


    9.91, 0.26


    7.5, 0.37


    7.91, 0.19


  • comput

    class 0-10 10-20 20-30 30-40
    f 2 4 5 1

    e mean deviation from median. Also calculate coefficient of them.

     

    MD = 6.5,0.24


    8.1, 0.32


    7.1,026


    MD = 7.5, 0.37


  • class 10-20 10-30 10-40 10-50
    f 2 4 9 20

    7.62,0.23


    8.2, 0.11


    6.5, 0.145


    7.91, 0.19


  • Marks 0-10 0-20 0-30 0-40 0-50
    No. of students 4 12 24 44 62

    7.8, 0.45


    7.9, 0.56


    8.5, 0.23


    MD =9.7, 0.29


  • compute the mean deviation from the media and its coefficient from the data given below:

    x 1 2 3 4 5
    f 2 5 6 5 2

    0.6, 0.21


    0.8, 0.3


    0.9, 0.3


    0.5,0.8


  • Find the mean deviation from the mean  and its coeffiecent

    Marks  0-10 10-20 20-30 30-40 40-50
    No. of student 4 10 17 30 19

    8.5, 0.63


    7.2, 0.36


    10.09,0.32


    9.1, 00.22


  • Find the mean deviation and its coefficient from the following data:

    x 15 20 10 5 25 30
    f 4 6 5 8 4 4

    0.22


    0.81


    0.45


    0.31


  • Compute the mean deviation from media and its coefficient from the data given below:

    x 20 14 26 12 28
    f 18 12 12 4 4

    12, 5


    3, 1


    5, 0


    0, 0


  • Find the mean deviation from median and its coefficient from the given data:

    Marks obtained 20 30 50 40 60 70
    No. of Students 4 7 12 2 4 6

    21.75, 0.9


    12.29, 0.31


    13.72, 0.33


    4.0, 0.69


  • Find  mean, median  and its coefficient.

    x 10-19 20-29 30-39 40-49 50-59
    f 4 10 18 16 12

    1.02,2.2,0.33


    3.01,0.89,0.14


    M.D = 9.91, 0.27


    2.02,0.2,0.33


  • Compute mean deviation from mean. Also calculate the coefficient.

     

    x 15 20 10 5 25 30
    f 4 6 5 8 4 4

    0.8


    3


    0.1


    1


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riwaj

the mean of three different number is 5 and the mean of 4 different number is 12.find the value of 7 numbers


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